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# 8.2: Area of a Surface of Revolution - Mathematics

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. We wish to find the surface area of the surface of revolution created by revolving the graph of (y=f(x)) around the (x)-axis as shown in the following figure.

As we have done many times before, we are going to partition the interval ([a,b]) and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For (i=0,1,2,…,n), let (P={x_i}) be a regular partition of ([a,b]). Then, for (i=1,2,…,n,) construct a line segment from the point ((x_{i−1},f(x_{i−1}))) to the point ((x_i,f(x_i))). Now, revolve these line segments around the (x)-axis to generate an approximation of the surface of revolution as shown in the following figure.

Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.

To find the surface area of the band, we need to find the lateral surface area, (S), of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let (r_1) and (r_2) be the radii of the wide end and the narrow end of the frustum, respectively, and let (l) be the slant height of the frustum as shown in the following figure.

We know the lateral surface area of a cone is given by

[ ext{Lateral Surface Area } =πrs,]

where (r) is the radius of the base of the cone and (s) is the slant height (Figure (PageIndex{7})).

Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (Figure (PageIndex{8})).

The cross-sections of the small cone and the large cone are similar triangles, so we see that

[ dfrac{r_2}{r_1}=dfrac{s−l}{s}]

Solving for (s), we get =s−ls

[egin{align*} dfrac{r_2}{r_1} &=dfrac{s−l}{s} r_2s &=r_1(s−l) r_2s &=r_1s−r_1l r_1l &=r_1s−r_2s r_1l &=(r_1−r_2)s dfrac{r_1l}{r_1−r_2} =s end{align*}]

Then the lateral surface area (SA) of the frustum is

[egin{align*} S &= ext{(Lateral SA of large cone)}− ext{(Lateral SA of small cone)} [4pt] &=πr_1s−πr_2(s−l) [4pt] &=πr_1(dfrac{r_1l}{r_1−r_2})−πr_2(dfrac{r_1l}{r_1−r_2−l}) [4pt] &=dfrac{πr^2_1l}{r^1−r^2}−dfrac{πr_1r_2l}{r_1−r_2}+πr_2l [4pt] &=dfrac{πr^2_1l}{r_1−r_2}−dfrac{πr_1r2_l}{r_1−r_2}+dfrac{πr_2l(r_1−r_2)}{r_1−r_2} [4pt] &=dfrac{πr^2_1}{lr_1−r_2}−dfrac{πr_1r_2l}{r_1−r_2} + dfrac{πr_1r_2l}{r_1−r_2}−dfrac{πr^2_2l}{r_1−r_3} [4pt] &=dfrac{π(r^2_1−r^2_2)l}{r_1−r_2}=dfrac{π(r_1−r+2)(r1+r2)l}{r_1−r_2} [4pt] &= π(r_1+r_2)l. label{eq20} end{align*}]

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the (x-axis). A representative band is shown in the following figure.

Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have

[egin{align*} S &=π(r_1+r_2)l &=π(f(x_{i−1})+f(x_i))sqrt{Δx^2+(Δyi)^2} &=π(f(x_{i−1})+f(x_i))Δxsqrt{1+(dfrac{Δy_i}{Δx})^2} end{align*}]

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select (x^∗_i∈[x_{i−1},x_i]) such that (f′(x^∗_i)=(Δy_i)/Δx.) This gives us

[S=π(f(x_{i−1})+f(x_i))Δxsqrt{1+(f′(x^∗_i))^2} onumber]

Furthermore, since(f(x)) is continuous, by the Intermediate Value Theorem, there is a point (x^{**}_i∈[x_{i−1},x[i]) such that (f(x^{**}_i)=(1/2)[f(xi−1)+f(xi)],

so we get

[S=2πf(x^{**}_i)Δxsqrt{1+(f′(x^∗_i))^2}. onumber]

Then the approximate surface area of the whole surface of revolution is given by

[ ext{Surface Area} ≈sum_{i=1}^n2πf(x^{**}_i)Δxsqrt{1+(f′(x^∗_i))^2}. onumber]

This almost looks like a Riemann sum, except we have functions evaluated at two different points, (x^∗_i) and (x^{**}_{i}), over the interval ([x_{i−1},x_i]). Although we do not examine the details here, it turns out that because (f(x)) is smooth, if we let n(→∞), the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both (x^∗_i) and x^{**}_i) are in the interval ([x_{i−1},x_i]), so it makes sense that as (n→∞), both (x^∗_i) and (x^{**}_i) approach (x) Those of you who are interested in the details should consult an advanced calculus text.

Taking the limit as (n→∞,) we get

[ egin{align*} ext{Surface Area} &=lim_{n→∞}sum_{i=1}n^2πf(x^{**}_i)Δxsqrt{1+(f′(x^∗_i))^2} [4pt] &=∫^b_a(2πf(x)sqrt{1+(f′(x))^2}) end{align*}]

As with arc length, we can conduct a similar development for functions of (y) to get a formula for the surface area of surfaces of revolution about the (y-axis). These findings are summarized in the following theorem.

Surface Area of a Surface of Revolution

Let (f(x)) be a nonnegative smooth function over the interval ([a,b]). Then, the surface area of the surface of revolution formed by revolving the graph of (f(x)) around the x-axis is given by

[ ext{Surface Area}=∫^b_a(2πf(x)sqrt{1+(f′(x))^2})dx]

Similarly, let (g(y)) be a nonnegative smooth function over the interval ([c,d]). Then, the surface area of the surface of revolution formed by revolving the graph of (g(y)) around the (y-axis) is given by

[ ext{Surface Area}=∫^d_c(2πg(y)sqrt{1+(g′(y))^2}dy]

Example (PageIndex{4}): Calculating the Surface Area of a Surface of Revolution 1.

Let (f(x)=sqrt{x}) over the interval ([1,4]). Find the surface area of the surface generated by revolving the graph of (f(x)) around the (x)-axis. Round the answer to three decimal places.

Solution

The graph of (f(x)) and the surface of rotation are shown in Figure (PageIndex{10}).

We have (f(x)=sqrt{x}). Then, (f′(x)=1/(2sqrt{x})) and ((f′(x))^2=1/(4x).) Then,

[egin{align*} ext{Surface Area} &=∫^b_a(2πf(x)sqrt{1+(f′(x))^2}dx [4pt] &=∫^4_1(sqrt{2πsqrt{x}1+dfrac{1}{4x}})dx [4pt] &=∫^4_1(2πsqrt{x+14}dx. end{align*}]

Let (u=x+1/4.) Then, (du=dx). When (x=1, u=5/4), and when (x=4, u=17/4.) This gives us

[egin{align*} ∫^1_0(2πsqrt{x+dfrac{1}{4}})dx &= ∫^{17/4}_{5/4}2πsqrt{u}du [4pt] &= 2πleft[dfrac{2}{3}u^{3/2} ight]∣^{17/4}_{5/4} [4pt] &=dfrac{π}{6}[17sqrt{17}−5sqrt{5}]≈30.846 end{align*}]

Exercise (PageIndex{4})

Let ( f(x)=sqrt{1−x}) over the interval ( [0,1/2]). Find the surface area of the surface generated by revolving the graph of ( f(x)) around the (x)-axis. Round the answer to three decimal places.

Hint

Use the process from the previous example.

[ dfrac{π}{6}(5sqrt{5}−3sqrt{3})≈3.133]

Example ( PageIndex{5}): Calculating the Surface Area of a Surface of Revolution 2

Let ( f(x)=y=dfrac[3]{3x}). Consider the portion of the curve where ( 0≤y≤2). Find the surface area of the surface generated by revolving the graph of ( f(x)) around the ( y)-axis.

Solution

Notice that we are revolving the curve around the ( y)-axis, and the interval is in terms of ( y), so we want to rewrite the function as a function of ( y). We get ( x=g(y)=(1/3)y^3). The graph of ( g(y)) and the surface of rotation are shown in the following figure.

We have ( g(y)=(1/3)y^3), so ( g′(y)=y^2) and ( (g′(y))^2=y^4). Then

[egin{align*} ext{Surface Area} &=∫^d_c(2πg(y)sqrt{1+(g′(y))^2})dy [4pt] &=∫^2_0(2π(dfrac{1}{3}y^3)sqrt{1+y^4})dy [4pt] &=dfrac{2π}{3}∫^2_0(y^3sqrt{1+y^4})dy. end{align*}]

Let ( u=y^4+1.) Then ( du=4y^3dy). When ( y=0, u=1), and when ( y=2, u=17.) Then

[egin{align*} dfrac{2π}{3}∫^2_0(y^3sqrt{1+y^4})dy &=dfrac{2π}{3}∫^{17}_1dfrac{1}{4}sqrt{u}du [4pt] &=dfrac{π}{6}[dfrac{2}{3}u^{3/2}]∣^{17}_1=dfrac{π}{9}[(17)^{3/2}−1]≈24.118. end{align*}]

Exercise (PageIndex{5})

Let ( g(y)=sqrt{9−y^2}) over the interval ( y∈[0,2]). Find the surface area of the surface generated by revolving the graph of ( g(y)) around the ( y)-axis.

Hint

Use the process from the previous example.

( 12π)

## Surface of revolution

Examples of surfaces of revolution generated by a straight line are cylindrical and conical surfaces depending on whether or not the line is parallel to the axis. A circle that is rotated around any diameter generates a sphere of which it is then a great circle, and if the circle is rotated around an axis that does not intersect the interior of a circle, then it generates a torus which does not intersect itself (a ring torus).

## 8.2: Area of a Surface of Revolution - Mathematics

In this section we are going to look once again at solids of revolution. We first looked at them back in Calculus I when we found the volume of the solid of revolution. In this section we want to find the surface area of this region.

So, for the purposes of the derivation of the formula, let’s look at rotating the continuous function (y = fleft( x ight)) in the interval (left[ ight]) about the (x)-axis. We’ll also need to assume that the derivative is continuous on (left[ ight]). Below is a sketch of a function and the solid of revolution we get by rotating the function about the (x)-axis.

We can derive a formula for the surface area much as we derived the formula for arc length. We’ll start by dividing the interval into (n) equal subintervals of width (Delta x). On each subinterval we will approximate the function with a straight line that agrees with the function at the endpoints of each interval. Here is a sketch of that for our representative function using (n = 4).

Now, rotate the approximations about the (x)-axis and we get the following solid.

The approximation on each interval gives a distinct portion of the solid and to make this clear each portion is colored differently. Each of these portions are called frustums and we know how to find the surface area of frustums.

The surface area of a frustum is given by,

and (l) is the length of the slant of the frustum.

For the frustum on the interval (left[ <<>>,> ight]) we have,

and we know from the previous section that,

Before writing down the formula for the surface area we are going to assume that (Delta x) is “small” and since (fleft( x ight)) is continuous we can then assume that,

So, the surface area of the frustum on the interval (left[ <<>>,> ight]) is approximately,

The surface area of the whole solid is then approximately,

and we can get the exact surface area by taking the limit as (n) goes to infinity.

If we wanted to we could also derive a similar formula for rotating (x = hleft( y ight)) on (left[ ight]) about the (y)-axis. This would give the following formula.

These are not the “standard” formulas however. Notice that the roots in both of these formulas are nothing more than the two (ds)’s we used in the previous section. Also, we will replace (fleft( x ight)) with (y) and (hleft( y ight)) with (x). Doing this gives the following two formulas for the surface area.

#### Surface Area Formulas

There are a couple of things to note about these formulas. First, notice that the variable in the integral itself is always the opposite variable from the one we’re rotating about. Second, we are allowed to use either (ds) in either formula. This means that there are, in some way, four formulas here. We will choose the (ds) based upon which is the most convenient for a given function and problem.

Now let’s work a couple of examples.

The formula that we’ll be using here is,

since we are rotating about the (x)-axis and we’ll use the first (ds) in this case because our function is in the correct form for that (ds) and we won’t gain anything by solving it for (x).

Let’s first get the derivative and the root taken care of.

Here’s the integral for the surface area,

There is a problem however. The (dx) means that we shouldn’t have any (y)’s in the integral. So, before evaluating the integral we’ll need to substitute in for (y) as well.

Previously we made the comment that we could use either (ds) in the surface area formulas. Let’s work an example in which using either (ds) won’t create integrals that are too difficult to evaluate and so we can check both (ds)’s.

Note that we’ve been given the function set up for the first (ds) and limits that work for the second (ds).

Solution 1
This solution will use the first (ds) listed above. We’ll start with the derivative and root.

We’ll also need to get new limits. That isn’t too bad however. All we need to do is plug in the given (y)’s into our equation and solve to get that the range of (x)’s is (1 le x le 8). The integral for the surface area is then,

Note that this time we didn’t need to substitute in for the (x) as we did in the previous example. In this case we picked up a (dx) from the (ds) and so we don’t need to do a substitution for the (x). In fact, if we had substituted for (x) we would have put (y)’s into the integral which would have caused problems.

Solution 2
This time we’ll use the second (ds). So, we’ll first need to solve the equation for (x). We’ll also go ahead and get the derivative and root while we’re at it.

We used the original (y) limits this time because we picked up a (dy) from the (ds). Also note that the presence of the (dy) means that this time, unlike the first solution, we’ll need to substitute in for the (x). Doing that gives,

Note that after the substitution the integral was identical to the first solution and so the work was skipped.

As this example has shown we can use either (ds) to get the surface area. It is important to point out as well that with one (ds) we had to do a substitution for the (x) and with the other we didn’t. This will always work out that way.

Note as well that in the case of the last example it was just as easy to use either (ds). That often won’t be the case. In many examples only one of the (ds) will be convenient to work with so we’ll always need to determine which (ds) is liable to be the easiest to work with before starting the problem.

## Theory of Intense Beams of Charged Particles

### 5.13.1 The Problem Statement

We have seen that using the surface of revolution as a basic stream tube, based on the assumption that Vl, Vψ depend only on l, reduces the problem under consideration to the integration of an ordinary differential equation and, possibly, to the calculation of a quadrature for η. A further task may consist of abandoning the requirement V α = V α (ξ 1 ) and constructing the exact solutions based on studying the group properties of Eqs. (5.225) formulated for a basic surface that is not necessarily a surface of revolution. As such a surface, we can use, as example, any of the surfaces we came across in Section 2 while studying the exact solutions of beam equations (plane, circular cylinder, and cone, as well as helicoid) ( Syrovoy, 1989 ).

## Surface of revolution

Examples of surfaces of revolution generated by a straight line are cylindrical and conical surfaces depending on whether or not the line is parallel to the axis. A circle that is rotated around any diameter generates a sphere of which it is then a great circle, and if the circle is rotated around an axis that does not intersect the interior of a circle, then it generates a torus which does not intersect itself (a ring torus).

A portion of the curve x = 2 + cos z rotated around the z-axis

The sections of the surface of revolution made by planes through the axis are called meridional sections. Any meridional section can be considered to be the generatrix in the plane determined by it and the axis.[2]

The sections of the surface of revolution made by planes that are perpendicular to the axis are circles.

Some special cases of hyperboloids (of either one or two sheets) and elliptic paraboloids are surfaces of revolution. These may be identified as those quadratic surfaces all of whose cross sections perpendicular to the axis are circular.
Area formula

If the curve is described by the parametric functions x(t), y(t), with t ranging over some interval [a,b], and the axis of revolution is the y-axis, then the area Ay is given by the integral

provided that x(t) is never negative between the endpoints a and b. This formula is the calculus equivalent of Pappus's centroid theorem.[3] The quantity

comes from the Pythagorean theorem and represents a small segment of the arc of the curve, as in the arc length formula. The quantity 2πx(t) is the path of (the centroid of) this small segment, as required by Pappus' theorem.

Likewise, when the axis of rotation is the x-axis and provided that y(t) is never negative, the area is given by[4]

If the continuous curve is described by the function y = f(x), a ≤ x ≤ b, then the integral becomes

for revolution around the x-axis, and

for revolution around the y-axis (provided a ≥ 0). These come from the above formula.[5]

For example, the spherical surface with unit radius is generated by the curve y(t) = sin(t), x(t) = cos(t), when t ranges over [0,π]. Its area is therefore

For the case of the spherical curve with radius r, y(x) = √r2 − x2 rotated about the x-axis

A minimal surface of revolution is the surface of revolution of the curve between two given points which minimizes surface area.[6] A basic problem in the calculus of variations is finding the curve between two points that produces this minimal surface of revolution.[6]

There are only two minimal surfaces of revolution (surfaces of revolution which are also minimal surfaces): the plane and the catenoid.[7]
Coordinate expressions

A surface of revolution given by rotating a curve described by y=f(x) around the x-axis may be most simply described in cylindrical coordinates by ( ). In Cartesian coordinates, this yields the parametrization in terms of z and ( heta ) as ( ). If instead we revolve the curve around the y-axis, then the curve is described in cylindrical coordinates by ( ), yielding the expression ( ) in terms of the parameters r and ( heta ).

If x and y are defined in terms of a parameter t, then we obtain a parametrization in terms of t and ( heta ). If x and y are functions of t, then the surface of revolution obtained by revolving the curve around the x-axis is described in cylindrical coordinates by the parametric equation ( ), and the surface of revolution obtained by revolving the curve around the y-axis is described by ( ). In Cartesian coordinates, these (respectively) become ( ) and ( ). The above formulae for surface area then follow by taking the surface integral of the constant function 1 over the surface using these parametrizations.

Geodesics on a surface of revolution

Meridians are always geodesics on a surface of revolution. Other geodesics are governed by Clairaut's relation.[8]
Toroids
Main article: Toroid
A toroid generated from a square

A surface of revolution with a hole in, where the axis of revolution does not intersect the surface, is called a toroid.[9] For example, when a rectangle is rotated around an axis parallel to one of its edges, then a hollow square-section ring is produced. If the revolved figure is a circle, then the object is called a torus.
Applications of surfaces of revolution

The use of surfaces of revolution is essential in many fields in physics and engineering. When certain objects are designed digitally, revolutions like these can be used to determine surface area without the use of measuring the length and radius of the object being designed.

Channel surface, a generalisation of a surface of revolution
Gabriel's Horn
Lemon (geometry), surface of revolution of a circular arc
Liouville surface, another generalization of a surface of revolution
Solid of revolution
Surface integral
Generalized helicoid
Translation surface (differential geometry)

Middlemiss Marks Smart. "15-4. Surfaces of Revolution". Analytic Geometry (3rd ed.). p. 378. LCCN 68015472.
Wilson, W.A. Tracey, J.I. (1925), Analytic Geometry (Revised ed.), D.C. Heath and Co., p. 227
Thomas, George B. "6.7: Area of a Surface of Revolution 6.11: The Theorems of Pappus". Calculus (3rd ed.). pp. 206–209, 217–219. LCCN 69016407.
Singh, R.R. (1993). Engineering Mathematics (6 ed.). Tata McGraw-Hill. p. 6.90. ISBN 0-07-014615-2.
Swokowski, Earl W. (1983), Calculus with analytic geometry (Alternate ed.), Prindle, Weber & Schmidt, p. 617, ISBN 0-87150-341-7
Weisstein, Eric W. "Minimal Surface of Revolution". MathWorld.
Weisstein, Eric W. "Catenoid". MathWorld.
Pressley, Andrew. “Chapter 9 - Geodesics.” Elementary Differential Geometry, 2nd ed., Springer, London, 2012, pp. 227–230.

Weisstein, Eric W. "Surface of Revolution". MathWorld.
"Surface de révolution". Encyclopédie des Formes Mathématiques Remarquables (in French).

## 8.2: Area of a Surface of Revolution - Mathematics

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We compute surface area of a frustrum then use the method of “Slice, Approximate, Integrate” to find areas of surface areas of revolution.

### The area of a frustum

In order to perform the approximation step, we first need to discuss the surface area of a frustrum.

To compute the area of a surface of revolution, we approximate that this area is equal to the sum of areas of basic shapes that we can lay out flat. The argument for this goes way back to the great physicist and mathematician, Archimedes of Alexandria. To follow his argument, we have to begin by computing the area of a ‘lamp shade’ or frustum.

And of course, few things are more interesting than the area of a frustum:

• denote the number of trapezoids,
• denote the length of the top of each trapezoid,
• denote the height of each trapezoid,
• denote the length of the bottom of each trapezoid,

then from geometry, we have that each of the trapezoids, one of which is shown below:

• is the circumference of the top circle,
• is the slant height of the frustum as shown in the above figure, and
• is the circumference of the bottom circle,

and by way of limit laws we find Now, letting

• be the radius of the circle defining the top of the frustum,
• be the slant height of the frustum, and
• be the radius of the circle defining the base of the frustum,

### The area of a surface of revolution

Let’s consider a function with a continuous derivative, and form a surface of revolution formed by this curve by rotating the portion of the curve from to about the -axis:

We can find a formula that gives the surface area of this surface of revolution using the procedure of “Slice, Approximate, Integrate”!

Step 1: Slice Since we have the curve to be revolved expressed as a function of , we choose to slice with respect to :

Step 2: Approximate We have seen how to find the surface area of a frustrum, so we should thus approximate each slice as a frustrum.

Thus the surface area, of this frustum is: Note that there is a value between and such that , so we write:

and can find the total approximate surface area by using frustra by adding together all of the surface areas:

Step 3: Integrate The formula above has good conceptual meaning, it does not readily pass to an integral quite yet! and have seen that we can express free in terms of either or , which allows us to express the infinitesimal by:

Note also that as the slice widths shrink, the value above approaches the distance that the corresponding slice is away from the axis of rotation.

To make sure that we emphasize this freedom in expressing as well as the inherent geometric results we used to build the surface area, we write:

where the radius is the distance from the axis of rotation to the slice and is the slant height of the slice.

To compute this surface area, we first choose to express either:

We then have to express the distance in terms of the variable of integration. This will always be a vertical or horizontal distance, which can be computed just as we have been doing in previous sections!

Using the remark, and letting and we can therefore write:

where the radius is the distance from the axis of rotation to the slice at .

An important concept to note is that the slice is located at a point on the curve. The choice of variable of integration may require that we express either or in terms of the other by using the equation that describes the curve. We will see this in the following examples.

We begin by considering looking at a picture,

As a brief aside, note that this slice gives rise to the following frustrum when revolved about the -axis:

Let’s first set up the integral with respect to . In order to do this, we choose:

Note that here is a vertical distance that must be expressed in terms of . Since the slice is located at a point on the curve :

Note that here is a vertical distance that must be expressed in terms of . Since the slice is located at a point on the curve :

Now, since we have , write with me: So

This integral can be computed using the substitution . Working out the details (which you should do on your own), gives:

As our final example, we will compute the surface area of the sphere.

### Final thoughts

The key formulas in this section are:

We are free to choose the variable of integration here since we can express in terms of either or easily. Once this choice of variable has been determined, we need to express the radius of the infinitesimal frustrum and the limits for the integral in terms of the variable of integration.

This radius is the distance from the axis of rotation to the slice at , which is either a horizontal or vertical distance. We just need to make sure that we express it in terms of the variable of integration appropriately.

Many of the integrals that arise in the context of these problems can be difficult. Careful differentiation and algebra, as well as a good grasp of integration techniques can be vital when finding surface areas. As usual, this can be challenging and practice is the key here.

“Math is not a spectator sport. It’s not a body of knowledge. It’s not symbols on a page. It’s something you play with, something you do” - Keith Devlin

## What's going on?

So how can it be that we have a finite volume, but an infinite surface area?

One way of looking at it is to recall how a volume of solid of revolution is actually calculated using integral calculus. We are finding the sum of the volumes of an infinite number of disks, radius y (for different values of x). The disks have area

and infinitesimal height dx. So each disk has volume

dx" />

The sum of an infinite number of such disks will converge, since this general sum converges:

### The dispute

The paradox about an object having finite volume but infinite surface area caused a lot of dispute regarding the nature of infinity among mathematicians of the 17th century, including Galileo and Wallis. Such paradoxes are a great way to get us thinking!

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Posted in Mathematics category - 22 Mar 2018 [Permalink]

### 6 Comments on “The object with finite volume but infinite surface area”

I was wondering, though, with regard to the surface area of the solid of revolution, as

is

When the derivative is squared, it's

The end result would be the same, though, as above as x approaches infinity, with the expression in the integrand being,

Good evening
A very nice problem with a great answer.
I would like to add or I must say that I want comments on these lines particularly which I am going to write. If we see this problem practically then
Actually the volume we get after integrating is tending to that volume as length is infinitesimally large. So it will hardly make any difference in volume whereas it's surface area will keep on increasing.

@Christian: Thanks for pointing out the error, which I have fixed in the post. I didn't question myself as the end result was the same, as you said!

So what would happen if someone were to hold the Horn vertically and fill it up with π litres of paint? Would the inside surface of the Horn get covered in paint entirely?

@Jan: Excellent question! Counter-intuitive it may be, but I suspect not. Anyone else like to weigh in?

Great article about the object with finite volume but infinite surface area.

## 8.2: Area of a Surface of Revolution - Mathematics

Consider a plane y=f(x) in the x-y plane between ordinates x=a and x=b. If a certain portion of this curve is revolved about an axis, a solid of revolution is generated.

1. Cartesian Form:
• Area of solid formed by revolving the arc of curve about x-axis is-
• Area of revolution by revolving the curve about y axis is-

• About the x-axis: initial line

Here replace r by f(θ)

Here replace r by f(θ)
• Limits for x: x = a to x = b

Here PM is in terms of x.
• Limits for y: y = c to y = d

Here PM is in terms of y.

Example:
Find the area of the solid of revolution generated by revolving the parabola about the x-axis.
Explanation:
Now we are given with the Cartesian form of the equation of parabola and the parabola has been rotated about the x-axis. Hence we use the formula for revolving Cartesian form about x-axis which is:

Here . Now we need to calculate dy/dx

Differentiating w.r.t x we get:

Using

Now we are provided with limits of x as x=0 to x=3. Plugging our calculated values in the above formula we get:

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## Lateral & Surface Areas, Volumes

The lateral area of a regular pyramid or right cone is similar to that of prisms, but since each face is a triangle (or triangle-like), there is a factor of one half. The lateral area is thus half the slant height times the perimeter. The slant height is the distance from the vertex to the edge of the base where it is halfway between the base's vertices. If the pyramid is irregular and certainly if the cone is oblique, the surface area might not be calculatable using elementary techniques (which is a fancy way to say you may need calculus). It depends on if you can obtain the altitude (slant height) of each triangular face.

 Surface Area = Lateral Area + n × Bases n = 2 for prisms/cylinders n = 1 for pyramids/cones n = 0 for spheres.

The surface area of a pyramid or cone is the lateral area plus the area of the single base.

The surface area of a sphere is equal to 4 r 2 . Analogous to the unit circle is the unit sphere. Similarly, just as there are 2 radians of angle in one revolution, there are 4 steradians of solid angle in all directions.

Example: Consider a right pyramid A-BCDE with vertex A and square base BCDE of length 20" on each side and a slant height of 26". What are its lateral and surface areas?

Answer: We don't need the height for this calculation, but we will calculate it anyway to stress the difference between slant height and height. The slant height is the hypotenuse of a right triangle where the height is one leg and 20"/2 = 10" is the other leg. Thus 10 2 + h 2 = 26 2 or 100 + h 2 = 676. Thus h 2 = 576 or h = 24". The lateral surfaces are all triangles with a base of 20" and a height (the slant height) of 26". There are four of them. Thus the lateral area is 4×½吐"吖" = 1040 in 2 . The base is 20" square or 400 in 2 . Thus the [total] surface area is 1440 in 2 .

Understanding surface area may be clearer if you refer back to the net associated with the object. At left is a net for a cube and at right a portion of a net for a sphere. Each of these portions of a sphere is called a gore .

Now is a good time to review something learned in algebra, namely ( x + y ) 2 = x 2 + xy + xy + y 2 = x 2 + 2 xy + y 2 . The diagram at the right should clarify this further, help you remember the FOIL method , as well as give a physical basis for this relationship. (Remember also, the square root of ( x 2 + y 2 ) does NOT equal x + y .) Consider extending the FOIL method first into trinomials: ( a + b + c )( d + e + f ) = ad + ae + af + bd + be + bf + cd + ce + cf . The distributive property is another way to consider this situation. Here the box method is useful.

d e f
b bd be bf
c cd ce cf

Now extend the method into three dimensions to find: V = ( a + b )( c + d )( e + f ) = ace + acf + ade + adf + bce + bcf + bde + bdf . This would be useful in finding the volume, which is why it is difficult to display in two dimensions.

• Every polyhedral region has an unique volume , dependent only on your unit cube .
• A box has a volume of length × width × height ( V = lwh ).
• Congruent figures have equivalent volume.
• Total volume is the sum of all nonoverlapping regions.

By knowing the volume, one can determine the dimensions of a polyhedron. Specifically for a cube with edge s and volume s 3 , given a cube with volume 1000 cubic centimeters (1 liter), you can take the cube root to determine each side had length 10 centimeters or about 3.937 inches. Since one gallon is 231 cubic inches, it is thus about 3.785 liters. Other unit conversions can be expected and are summarized in Numbers lesson 9. Cube roots and volume are at the heart of an ancient impossible geometric construction from antiquity, the Delian Cube Doubling problem. Another important concept is that if you double the dimensions of a cube, the volume goes up by a factor of 8=2 3 , just like area went up by a factor of 4=2 2 . This is a problem commonly encountered when converting from cubic feet into cubic yards!

Example: Suppose you wish to pour concrete 4" deep in your driveway which is 90' long and 9' wide.

Answer: You quickly discover that there are 90࡯঩=270 feet 3 . However, there are only 10 yards 3 since each yard is 3 feet and 3 3 =27.

Calculating in the "native unit" of yards: 30ࡩয can help prevent such an error. By "native" we mean here that the final results are expected in cubic yards. If the initial units are converted to yards fewer mistakes will be made. It is EXTREMELY common to erroneously divide by 3 or 9 and not 27 when converting cubic feet into cubic yards.

Example: Suppose you wish to find the volume of the square based right pyramid A-BCDE given in an earlier example with slant height 26" and base 20" on each side.

Answer: The height is 24" as calculated in the previous example. Thus the volume is (1/3)× B × h = (1/3) × 20" × 20" × 24" = 3200 in 3 .

Volume Formulae

 Prism or cylinder: V = base area × height Pyramid or cone: V = (1/3) × base area × height Sphere: V = (4/3) × (radius) 3

Typically these formula are written as V = Bh (prism or cylinder), V =(1/3) Bh (pyramid or cone), or V =(4/3) r 3 (sphere). Note how a big B is used to signify that this is a two dimensional base or area and not the same (linear) b we use in triangles.

Oblique Prisms and cylinders have the same volume as a right prism or cylinder with the same height and base area. Think of a stack of paper whose top has been pushed to one side. The stack is no longer vertical. However, the volume of paper hasn't changed. In the formula for finding the volume of an oblique prism please note that the height is the perpendicular segment between the top and bottom bases. When you learn calculus you will discover the surface area of a sphere to be the derivative with respect to r of the sphere's volume formula. A similar thing happens between area of a circle and its circumference. This may be happenstance or there may be a deep reason which I'd like to know.

Example: A favorite volume/surface area problem is as follows. A swimming pool is 24' long, 20' wide, 3' deep at the shallow end, and 10' deep at the deep end. The floor slopes evenly. What is the inside surface of the swimming pool and what is the volume (in gallons)?

Answer: The swimming pool is a trapezoidal prism. The floor is 25 feet long since 10' - 3' = 7' and 7 2 +24 2 = 49 + 576 = 625 = 25 2 . The surface area is the sum of 5 surfaces: 2 congruent trapezoidal sides (½(3+10)㩌), 2 rectangular ends (3㩈 + 10㩈), and the bottom (20㩍). This is 2𤚤 + 60 + 200 + 500 = 1072 feet 2 . The volume is Base × height, where Base is the area of one side (½(3+10)㩌), and the height is the width of the pool (20). Thus the volume is 3120 feet 3 or 23339 gallons (multiply by 12 3 cubic inches per cubic foot and divide by 231 cubic inches per gallon).

A gedanken experiment (thought experiment) used to justify the volume formula for a sphere is as follows. First, remember the circle area activity where we cut the circle into 16 wedges, then rearranged the wedges into a r × r parallelogram. Along the same lines, cut a sphere into pyramids. The total area of the bases of these pyramids is 4 r 2 . The height of each is r . Hence the formula is derived. Along the same lines, some have suggested remembering the 1/3 in conal volume formulae by correlating it to the analoguous two dimensional trianglular area formula which has a 1/2 in it.

Given two solids included between parallel planes. If every plane cross section parallel to the given planes has the same area in both solids, then the volumes of the solids are equal. This is know as Cavalieri's Principle .

The Greek Archimedes is one of the three greatest mathematicians of all time. Among his important discoveries is the relationship between the volumes of the cone, sphere, and cylinder. In fact, this discovery was so much his favorite that he requested it to be inscribed on his tombstone. Specifically, consider a sphere of radius r , two cones each with the same radius and height ( r ), and a cylinder with the same radius and height (2 r ). The cylinder will contain either the two cones or the sphere. Their volumes can easily be seen to be (4/3) r 3 , 2(1/3) r 3 , and 2 r 3 . Thus the cones plus the sphere equals the cylinder exactly. (Actually, Archimedes is more commonly credited with showing the sphere's volume to be 2/3's that of the cylinder.) See the corresponding diagrams in the textbook related to the proof of Cavalieri's Principle.

Example: Question 10.2#24 in our text asked the students about cones made from circles (radius 4") with central angles of 45°, 60°, and 120° removed (which got taped to the board amid Madonna jokes). Perform the following. Find the volume of each cone. Find the central angle which maximizes volume. Find the central angle which maximized volume to surface ratio.

Answer: For bonus points hand in your solution by the time the chapter reviews are due.

## CALC 2: Area of a Surface of Revolution: x=y+y^3 from 0 to 4

Hi, so I'm having trouble with one of my online calc homework..

Consider the following.
x=y+y^3 from 0 to 4
(a) Set up an integral for the area of the surface obtained by rotating the curve about the x-axis and the y-axis.
(i) the x-axis, the answer is S= 2piy(sqrt((3y^2+1)^2)+1)dy
(ii) the y-axis, the answer is S= 2pi(y^3+y)sqrt((3y^2+1)^2+1)dy

(b) Use the numerical integration capability of a calculator to evaluate the surface areas correct to four decimal places.
(i) the x-axis, the answer is 1258.6212
(ii) the y-axis, I'm really stuck on how to take the integral. I used wolfram to show how they did it, but I don't understand how they got it! I would really like to understand how to do this problem!

#### Subhotosh Khan

##### Super Moderator

Hi, so I'm having trouble with one of my online calc homework..

Consider the following.
x=y+y^3 from 0 to 4
(a) Set up an integral for the area of the surface obtained by rotating the curve about the x-axis and the y-axis.
(i) the x-axis, the answer is S= 2piy(sqrt((3y^2+1)^2)+1)dy
(ii) the y-axis, the answer is S= 2pi(y^3+y)sqrt((3y^2+1)^2+1)dy

(b) Use the numerical integration capability of a calculator to evaluate the surface areas correct to four decimal places.
(i) the x-axis, the answer is 1258.6212
(ii) the y-axis, I'm really stuck on how to take the integral. I used wolfram to show how they did it, but I don't understand how they got it! I would really like to understand how to do this problem!

Please share your work with us . even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.


text:
Home > CC2MN > Chapter cc310 > Lesson cc310.1.4 > Problem10-56

10-56.

If the volume of a sphere is 113.04 ft3, write and solve and equation to find the radius. Homework Help ✎

$\text{The\: formula\: to\: find\: the\: volume\: of\: a\: sphere\: is \:} V = \frac{4}{3}\pi\:r^3.$

How would you undo an exponent?

$113.04\; =\; \frac{4}{3}\pi {\it r}^3$

$\frac{\left(113.04\right)\left(3\right)}{4\pi }={\it r}^3$

$26.99\; \approx\; {\it r}^3$

$\sqrt[3]{26.99}\approx\; \it r$

$3 \approx \it r$

format: plain


text:
Tag Info

46

It did, with the third rocket stage. Instead of just becoming another object in solar orbit like the previous Apollo third stages, this time the third stage was sent into the Moon for a crash landing whose impact would be recorded on the seismometer installed by Apollo 12. This test went off without a hitch and successfully returned data from the ...

40

According to the link here, In meteorological rockets, the temperature sensor is not measured as the rocket ascends but a payload containing the sensor is ejected from a high altitude and as it parachutes down measurements are beamed back.

35

To start with a summary: (1) the extra delta-V needed to halve the journey time is not as much as you think and (2) the SpaceX mission plan is, effectively, a really, really huge launch from Earth. To expand on the first point, we need to look a bit more closely at how interplanetary travel actually works. There are lots of discussions on stack exchange (...

30

In addition to crashing the Saturn V's S-IVB into the moon to collect seismic data from sensors installed by the crews of Apollo 11 and 12, several life sciences experiments were performed on the crew before and after Apollo 13. While all of the inflight experiments were canceled, researchers still managed to collect data on the cardiovascular's response to ...

27

Watney did have trouble controlling himself with the "iron man" suit puncture trick; it's plausible that he could quickly get the hang of positioning his hand to thrust through his center of mass (though that all happened fast enough that I didn't see if it was so depicted). Mastering that technique in the few seconds he had available was pretty unlikely, ...

25

The most inaccurate "artistic license" thing was the Martian storm - Andy Weir admitted that he misrepresented it for dramatic effect. The most blatant error though was in growing the potatoes. In particular, the matter of lighting. For the depicted indoor growth, Mark would need 60 kilowatt worth of LED growth lights, and a proportionally sized solar farm. ...

22

Excellent question! That's probably ground velocity or more likely velocity in an ECEF (earth-centered, earth-fixed), also known as ECR (earth-centered rotational) frame that they are displaying, referenced to the rotating frame of the Earth. A hint would be that it is zero before lift-off even though the Earth is rotating. Wikipedia gives the value of $3.... 16 The most obvious thing in my opinion was the orientation of Hermes in the scene right before they receive the hidden transmission with instructions for the Rich Purnell manoeuvre. They had just discussed the fact that Hermes is currently slowing down in its approach to Earth, which would mean the vehicle is pointing retrograde. However, when it cuts to ... 15 Suborbital sounding rockets cost about 1/100 as much as an orbital launcher. Black Brant XII, one of the most advanced sounding rockets in use, can take 100-400kg payload into space for \$600K. For a Falcon-9-based sounding rocket approach to make any sense, you'd have to describe a single payload that delivered as much science as 100 individual 100kg ...

15

The Voyager probes are currently further away from earth than any object before. Knowing their position alone gives interesting scientific data, because it tells us their speed and direction, which gives us information about the forces which act on objects so far outside of the solar system (microgravity, acceleration through solar winds, friction with ...

13

Here are some Apollo specific innovations: microchip, cordless tools, joystick, CAT scans, technology in MRI machines, modern shoe designs, freeze dried food, vacuum sealed packages, dampening material, retro-reflector (detects chemical leaks), water purification, silicon based storage of records, fly-by-wire, ground water cleaning, large fabric roofs used ...

12

Here is an answer from over on aviation.stackexchange.com which addresses this issue. I've quoted a portion of it below, but see that link for more info. As you measure temperature moving at high velocities, your outside thermometer will measure a higher temp than what is actually outside (what a non-moving thermometer would get). That's because as ...

8

The two Voyager spacecraft are now conducting the Voyager Interstellar Mission: The mission objective of the Voyager Interstellar Mission (VIM) is to extend the NASA exploration of the solar system beyond the neighborhood of the outer planets to the outer limits of the Sun's sphere of influence, and possibly beyond. This extended mission is ...

7

NASA actually have looked into mining the gas giants, http://mdcampbell.com/TM-2006-214122AtmosphericMining.pdf, this paper outlines some the methods they might use, whether they will or not is a different question. They were considering mining methods, of Uranus due to its relatively low wind speeds (compared to Jupiter, Saturn and Neptune, such as ...

6

Combustion analysis computer programs and specifically exact reacting gas chemical kinetics programs were developed on the Apollo program to analyze and optimize rocket nozzle designs for efficiency improvements. Adaptation of these programs to other than rocket fuels provided the ability to analyze the mechanisms of NOx formation in fuel/air combustion ...

6

There are studies into mining in space, but not for the gas giants. The outer planets are massive. Very massive. This means they have very big gravity wells. Lifting anything from Jupiter is going to be extremely energy-intensive, getting it back from Jupiter to Earth is too. So, currently, this is the realm of science-fiction. As for the second part ...

6

Exactly at the center-of-mass of the Earth there is no net gravitational force. Gravity comes from mass, and all the mass of the Earth is evenly distributed around the center-of-mass (by definition), so all the forces cancel out. As you get deeper and deeper, the net gravitational force falls -- linearly if density is constant. In practice, of course, you ...

6

The important thing to note here is that to say "it takes 8-9 years" doesn't make sense without specifying who it applies to. When relativistic effects start to apply, it's not the same in all reference frames. Let's take your Proxima Centauri example. At half the speed of light, the spacecraft reaches Proxima Centauri in 8.5 years, and if it immediately ...

5

With 2 mm/s^2 acceleration, Hermes' 124 day trip from Low Earth Orbit to Low Mars Orbit is impossible. "Low earth orbit?" a Weir defender might object, "It's all hyperbolic fly bys." Which is wrong, of course. The hyperbolic rendezvous were extraordinary maneuvers made under unusual circumstances (Watney needing rescue). The 124 day earth to Mars trip ...

5

In this answer, I'm going to assume the tunnel is a perfect vacuum and there's no intense heat or pressure. This is a thought experiment. If you and John fall at the same time, you guys will meet approximately 20 minutes after you jump into the tunnel. Depending on how wide the tunnel is, you guys will most likely collide at a velocity of around 11 km/s. ...

4

I know that it has been theorized about harvesting isotopes of Helium or Hydrogen space, including studies for / by NASA. In fact, NASA did study mining celestial bodies, including other planets, and still does. The study report "Space Resources" (NASA SP-509, 1992) has become a classic. It can be found here http://www.nss.org/settlement/spaceresources/...

4

Advancement in science comes from repeated (and repeatable) observations that are systematically processed, hypotheses that get tested on experimental data, and conclusions that get peer review. One high-altitude balloon with low-cost off-the-shelf sensor payload will add nothing to existing space/earth sciences if you do not do the following: read ...

4

Without its main thruster, orbit maneuver engine (OME), that's been completely dismissed as possible to restart (they did try but only got a small fraction of thrust out of it, about 10%) and have even dumped oxidizer (Mixed Oxides of Nitrogen, or MON) to save on spacecraft's mass as reaction control system (RCS) thrusters don't need it, extremely low (300 ...

4

As time goes by, the technological backbone of any system becomes dated and outmoded. For example, several of the space shuttle systems were still using items designed in the Apollo days. These can be refit to use newer technology (i.e. using laptops to supplement shuttle computers, new glass cockpit technology, etc.), but there always comes a point where ...

4

No evidence for identifiable scientific product(s) later used by the ISS and/or other manned spaceflight. However, researchers of biosphere two found the original crew's lowered blood pressure, blood cholesterol, and enhancement of there immune systems to be caused by Roy Walford's low calorie restriction diet (Walford, R., Mock D, Verdery R, MacCallum T, ...

4

One addition that I struggle with was the rapid way he was able to use poop and Martian regolith to create soil to grow the potatoes. In the book at least (didn't see the movie). I can't anywhere find data on how long it takes to make organic soil for plants; sure hydroponics are great, but that's not what Mr. Whatney did. Soil takes a long time to create ...

4

I think it's just barely possible that 1850s technology could achieve a crewed suborbital launch above the Kármán line, similar to the first crewed Mercury missions, but that anything beyond that would not be possible. Without modern rocket engines, the best propulsion option is a black powder rocket. This has a specific impulse of about 80 seconds (a ...

3

I liked the film and found it quite realistic overall, but what I found very unrealistic was his repair of the hab using plastic foil and duct tape. It has already been mentioned that the martian atmosphere is very thin. Inside the hab the air is breathable, so it's pressure must be substantially higher than that of the martian atmosphere. If the pressure ...

3

I've been searching around, and found some piecemealed data. Here's what I can find. Some very limited data is available from the Planetary Data Archives, mostly in the form of cooperative missions with NASA I found some data on this website, from the Space Research Institute Some of it appears to be hosted on a private network similar to the WW2, known as ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# 39+ Sliding Friction Definition Physics Picture

Sliding Friction Definition Physics. If the limiting kinetic friction force is $\mathrm{f}_{\mathrm{k}}$ and the normal reaction is n. It is also called as kinetic friction.

Several friction models have been formulated in terms of friction coefficients as a function of the relative sliding velocity, displacement, and quiescent time [4] [5] [6] [7] [8] [9] [10] [11. Limiting friction do not depend on area of contact surfaces but depends on their nature, i.e. Viscous force is the equivalence of friction in fluids that resists the relative motion of a body through the fluid.

### Sliding Friction (Also Called Kinetic Friction) Is A Contact Force That Resists The Sliding Motion Of Two Objects Or An Object And A Surface.

You can see why the sol of your shoes is created in uneven surface. So, this cannot be truly considered as sliding friction. Love physics, study a lot about friction facts.

### The Sliding Friction Is Friction Which Comes Into Play When Surface Of One Object Moves Relative To Surface Of Another Object.

The definition of static sliding friction is the same as sliding friction, except that the force applied is unable to make the object move. The force of static friction is a force between two surfaces that prevents those surfaces from sliding or slipping across each other. If one body begins to slides over the other, then the friction between the surfaces is called sliding friction.

### Limiting Friction, F S(Max) = Μ L R.

It is the maximum value of static friction when body is at the verge of starting motion. Viscous force is the equivalence of friction in fluids that resists the relative motion of a body through the fluid. Where, μ l = coefficient of limiting friction and r = normal reaction.

### The Usual Formula For That Is Just:

Sliding friction is almost always less than that of static friction; Several friction models have been formulated in terms of friction coefficients as a function of the relative sliding velocity, displacement, and quiescent time [4] [5] [6] [7] [8] [9] [10] [11. For example, suppose we want to push a washing machine across the floor, it takes some time to push it, till that time static sliding friction is opposing the movement.

### Kinetic Friction, Also Known As Dynamic Friction Or Sliding Friction, Occurs When Two Objects Are Moving Relative To Each Other And Rub Together (Like A Sled On The Ground).

If the limiting kinetic friction force is $\mathrm{f}_{\mathrm{k}}$ and the normal reaction is n. Another common force is friction. Friction is a force that opposes motion.


text:
# Integration by substitution Calculator

Go!
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### Difficult Problems

1

Example

$\int\sqrt{5x-3}dx$
2

Solve the integral $\int\sqrt{5x-3}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=5x-3 \\ du=5dx\end{matrix}$
3

Isolate $dx$ in the previous equation

$\frac{du}{5}=dx$
4

Substituting $u$ and $dx$ in the integral

$\int\frac{\sqrt{u}}{5}du$
5

Take the constant out of the integral

$\frac{1}{5}\int\sqrt{u}du$
6

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{2}{15}\sqrt{u^{3}}$
7

Substitute $u$ back for it's value, $5x-3$

$\frac{2}{15}\sqrt{\left(5x-3\right)^{3}}$
8

$\frac{2}{15}\sqrt{\left(5x-3\right)^{3}}+C_0$


text:
# cli¶

A python command line interpreter which has all the Gaffer modules available to it.

Caution

This application is deprecated. Use gaffer env python instead.

## -help¶

Prints names and descriptions of each parameter rather than running the application.

## -threads¶

The maximum number of threads used for computation. The default value of zero causes the number of threads to be chosen automatically based on the available hardware.

## -profileFileName¶

If this is specified, then the application is run using the cProfile profiling module, and the results saved to the file for later examination.


text:
# [time 495] Re: [time 494] Observation models

Stephen P. King (stephenk1@home.com)
Mon, 26 Jul 1999 11:46:06 -0400

Dear Matti,

Focusing on two notions...

Matti Pitkanen wrote:

snip

> [MP]
> > By the way, motion in zero modes occurs as hopping quantum jump by
> quantum
> > jump:
> >
> > x1-->x2--x3---
> >
> > Just like Brownian motion. In fact, I believe that Brownian motion
> > indeed corresponds to this kind of motion. The possibility to
> > Brownian motion to move faster than light would not be a
> > result of nonrelativistic approximation but real physical
> > effect.
[SPK]
> Yes! WE need to look at this very carefully! I say that the
> forward
> jumps ... -->x1 --> x2--> ...
> are dual to "backward jumps" ... i2 <-- i1 <-- ... of logical
> implication (following Pratt). It looks like this:
>
> ... -->x1 --> x2--> ... Time
> | |
> ... |g |h ...
> | |
> ... <--i2 <-- i1<-- ... Logic
>
> were g and h are infomorphic bisimulations...
>
> [MP] The problem is that logical causation and temporal causation
> seem to have different directions unless one replaces A--> B with
> not A <--- not B.

It is logical entailment or implication that "goes backwards" with
respect to time. This is discussed at length in Pratt's paper ratmech.ps
If we take the negation: "not A <--- not B", we are just reversing the
respective arrows. Umm, only in the strict binary boolean case is ~~A =
A, so it gets a bit complicated when we are using a Chu_[0,1] space to
represent the interactions of a given pair of LSs, since fuzzification
allows for ~A \intersect A =/= 0 under certain conditions!

> I would take only the upper part of the diagram:
>
> ... -->x1 --> x2--> ... Times for generation of subselves x1, x2
>
> and denote selves by x1, x2,.... Since x2 is subself of x1
> implication x1-->x2 is realized as set-theoretical inclusion.
> Subself is implied by self and in correct temporal order!

Yes, but I am trying to discuss how the ordering by inclusion is
constructed. If we just assume that the ordering is a priori, then we do
not have to deal with the question as to why the selves (or more
generally, observations) are ordered as they are. It is this assumption
that exists in the classical model of space-time and is in severe
contradiction with facts, such as those illustrated in the EPR and
delayed choice situations.
We can just say that temporal orderings are given by the MNP of the
quantum jumps, but I am trying to understand the details better. I think
of this as asking what decides the winner in a tournament; while we can
not say with certainty 1 who will win. All we can do is to set up
"pay-off matrixes", etc.

> There are several interpretations depending on the nature
> of quantum jump: interpretation as logical causation
> or as genuine set theoretic inclusions: selves as representation
> for objects of figure. Entire figure as big self, background and
> objects of figure as subselves containing smaller objects
> as subsubselves.... Or tree like
> structure of linguistic expression...
>
snip

> snip
> [MP]
> > > Actually the measurement can be said to be disjoint in well
> > > defined sense. Separate selves mutually unentangled.
> [SPK]
> > > Yes! This is why separate selves have separate indentities! I say that
> > > experiences are "almost disjoint" since they differ in at least the
> > > "point of identity". The notion of a fixed point under a given set of
> > > transformations is what I considered before I has educated by Peter's
> > > work. The Self, within Peter's formalism, is a greatest fixed point, I
> > > think. Umm, I need to review his paper:
> > > http://www.cs.brown.edu/~pw/papers/math1.ps
> [MP]
> > Quantum entanglement is what destroys selves. For instance, could
> > it be that the separate notes in chord correspond to
> > frequency-selves which have entangled to form chord-self?
> > This would explain why we do not hear separate notes but entire
> > chord.
>
> This is why we can not do a "Vulcan Mind Meld"! It would be
> impossible to unentangle the two minds!
>
> [MP] There are objections agains simple rule that unentangled
> subselves are experienced as average subself by self
> containing them. Probably the subselves must have large
> enough mutual resemblance before they are averaged. Perhaps
> they must have same p-adic prime, belong to same spacetime sheet:
> something like that. Must think about.

I think that a situation is required such that a "large enough"
sampling of the possible actions of the selves can be had so that a
winner can be reasonably predicted. We see an example of this in
political elections in a democratic country. The individual electorate
(representing the individual subselves) vote and a sample of these are
polled by the media to form a statistical sample of the whole and this
sample is used to predict the outcome "sooner than the actual counting
of the ballots can occur". I think that the situation involved with
"selves" is a continuous voting/polling situation since the "elections"
occur at every quantum jump!
The p-adic prime looks to me to well model the size of the sample
population!

snip
[SPK]
> > > How is the information content of this localization given? This is an
> > > infinite NP-Complete problem that requires ETERNITY to compute! It can
> > > not be given a priori!
> [MP]
> > TGD generously provides this eternity! The quantum computation defined
> > by informational time development operator lasts infinite amount
> > of lightcone proper time (not our psychological time which
> > is quite different thing).
>
> Yes! :-) I just try to think about how each observation involves
> its own lightcone of causality...
>
> [MP] In TGD framework causality is not involved at level of experiencing.
> It is involved at the dynamics determining spacetimes as
> absolute minima of Kahler action: that is at the level of configuration
> space geometry. In your approach situation is obviously different.

I say that since the determination of the particular experience is not
modelable locally, as it involves an entire light-like hypersurface, the
causality not considered as an experience itself. Only the results of
the determination enter into the "picture". So I agree this you here. My
only question is in the nature of the "absoluteness" of the minima of
the Kaehler action. Umm, what are the quantities involved in the Kaehler
action? Is there an associated Lagrangian or Hamiltonian? What
"information measures" are related to it?

snip

> [MP]
> > I understand what you are saying. This is probably the only possible
> > consistent philosophy in the framework in which one does not introduce
> > quantum jump and replace single objective reality with all
> > possible objective realities allowed by physical theory.
>
> Huh, how do you conclude that?
>
> [MP] Well, I don't know! Somehow I interpreted the idea about
> several almost non-intersecting spacetimes as geometric counterpart
> for giving up the notion of single objective reality, which
> is quantum concept.

Since we have many observers, it is natural to consider that there are
many observational experiences to be considered. The notion of a single
objective reality only makes sense IFF the class of observables is
strictly sharp (binary certainty). The problem I see is that only a
prediction can be made up to the accuracy allowed by the p-ary cut-off
(\epsilon of accuracy). So the smearing of "reality" that QM predict is
no surprise!

snip

> [MP]
> > I think that a lot of confusion is created by different views about
> > imbedding space. I do not regard its points as reprsentations of events:
> > it is quite too simple thing for that.
>
> Indeed! But, it is helpful as a concept to start out with...
>
> [MP]
> OK. It is certainly what Einstein believed. Classical geometric
> structures in spacetime would characterize entire physics.

Sure! Invariance structures are defined by geometry, so a physics = a
geometry.

Next I would like to better understand the problem you have with
Frieden's work...

Onward!

Stephen

This archive was generated by hypermail 2.0b3 on Sun Oct 17 1999 - 22:36:57 JST


text:
# how can i prove that square root of n is space constructible

I know that square-root of n is space-constructible. I can't prove it by the space-constructible definition. How can I show that only $\sqrt{n}$ space is used?

-
I don't know what space-constructible means. I know $\sqrt n$ is "Euclid-constructible," that is, I can show you how to construct it with ruler and compass, if that's what you want. –  Gerry Myerson Jan 28 '12 at 22:02
@Gerry: The tag says "complexity," so... –  anon Jan 28 '12 at 22:10
Thanks. space-constructible means that it is possible to construct in o(n) space from 1^n to the binary representation of f(n). in our case f(n) is sqrt. Maybe the Euclid-constructible can help. –  Eyal Golan Jan 28 '12 at 22:17
@anon, my comment was intended to extract some useful information from OP. I think it succeeded. –  Gerry Myerson Jan 28 '12 at 23:11
You don't need anything nearly so complicated as Newton - keep in mind that there are no time bounds on the construction. Just test all the numbers $t$ from 1 forward until you find one with $t^2\gt n$. This is easy to do in $O(\log n)$ space just by using binary. –  Steven Stadnicki Jan 28 '12 at 23:36


text:
# Triple Integral

Given the triple integral $$\int\int\int_{G}$$ xyz dV
Where G is the region bounded by x=1, y=x, y=0, z=0, z=2.
How do I evaluate it.

daniel_i_l
Gold Member
First of all, is there any other constraint on x? I'll assume that x>=0.
Do you know how to change triple integrals into single-variable integrals?
The general idea is to first evaluate the integral while pretending that 2 of the variables are constant. Then you use that result to integrate over the other variables.

In this case it might be easiest to first evaluate the integral in the triangle 0<=x<=1 and
0<=y<=x while assuming that z is constant. Then integrate that result with z as a variable from 0 to 2. The triangle can be evaluated in a similar way. In other words:
$$I = \int^{2}_{0}(\int^{1}_{0}(\int^{1-x}_{0} xyz dy)dx)dz$$

Cheers. I got an answer of 1/3


text:
# linear programming feasiblity checking

Is there any sufficient and necessary condition to check the feasibility of the linear constraints

$Ax=b, x\geq 0$

without solving an LP with a constant objective function? $x$ is the variable and $A\in\mathbb{R}^{m\times n}$ is given and has full row rank. $b$ is also given.

Farka's lemma could be an option. But it is not what I'm looking for since if has the flavor of "if a solution exists for a set of equalities and inequalities" but does not provide a direct way to determine whether the solution exists.

Is there any more direct method to do the feasibility checking, such as by merely looking at the rank of augmented matrix and coefficient matrix similar to what Rouché–Capelli theorem does?

There are methods called "preprocessing" or "presolve", that will take a problem with constraints $b_{l} \leq Ax \leq b_{u}$ and $l \leq x \leq u$ as input. Using tests that are usually linear time, you can check sufficient conditions for infeasibility (but not necessary conditions). These methods are commonly used to transform a problem to reduce the runtime of standard algorithms for linear programming. You should look at a 1994 paper by Savelsbergh, a paper by Andersen and Andersen, and a survey paper by Ashutosh Mahajan. These papers all cover more general transformations that are also of interest; the Mahajan paper probably has the most modern and readable notation.


text:
# Samacheer Kalvi 10th Maths Model Question Paper 1 English Medium

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 1 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 1 English Medium

Instructions

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III and IV are to be attempted separately.
• Question numbers 1 to 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
• Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
• Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
• Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 100

PART – I

I. Choose the correct answer. Answer all the questions. [14 × 1 = 14]

Question 1.
If {(a, 8),(6, b)} represents an identity function, then the value of a and b are respectively ……………. .
(1) (8,6)
(2) (8,8)
(3) (6,8)
(4) (6,6) .
(1) (8,6)

Question 2.
If the H.C.F of 65 and 117 is expressible in the form of 65m -117, then the value of m is ……………. .
(1) 4
(2) 2
(3) 1
(4) 3
(2) 2

Question 3.
The next term ol the sequence $$\frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}$$ is …………… .
(1) $$\frac{1}{24}$$
(2) $$\frac{1}{27}$$
(3) $$\frac{2}{3}$$
(4) $$\frac{1}{81}$$
(2) $$\frac{1}{27}$$

Question 4.
Which of the following should be added to make x4 + 64 a perfect square ……………… .
(1) 4x2
(2) 16x2
(3) 8x2
(4) -8x2
(2) 16x2

Question 5.
Find the matrix x if 2x + $$\left[ \begin{matrix} 1 & 3 \\ 5 & 7 \end{matrix} \right] =\left[ \begin{matrix} 5 & 7 \\ 9 & 5 \end{matrix} \right]$$ ……………. .
(1) $$\left[ \begin{matrix} -2 & -2 \\ 2 & -1 \end{matrix} \right]$$
(2) $$\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right]$$
(3) $$\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix} \right]$$
(4) $$\left[ \begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix} \right]$$
(2) $$\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right]$$

Question 6.
Two poles of heights 6 m and 11 m stand vertically on a plane ground if the distance between their feet is 12m, what is the distance between their tops?
(1) 13 m
(2) 14 m
(3) 15 m
(4) 12.8 m
(1) 13 m

Question 7.
If (5, 7), (3,p) and (6, 6) are collinear then the value of p is ……………… .
(1) 3
(2) 6
(3) 9
(4) 12
(3) 9

Question 8.
The value of sin2θ + $$\frac{1}{1+\tan ^{2} \theta}$$ is equal to …………….. .
(1) tan2θ
(2) 1
(3) cot2θ
(4) 0
(2) 1

Question 9.
If the radius of the base of a cone is tripled and the height is doubled then the volume is ……………… .
(4) unchanged

Question 10.
Standard deviation of a collection of data is 2√2. If each value is multiplied by 3, then the standard deviation of the new data is
……………… .
(1) √l2
(2) 4√2
(3) 6√2
(4) 9√2
(3) 6√2

Question 11.
A page is selected at random from a book. The probability that the digit at unit place of the page number chosen’ is less than 7 is …………. .
(1) $$\frac{3}{10}$$
(2) $$\frac{7}{10}$$
(3) $$\frac{3}{9}$$
(4) $$\frac{7}{9}$$
(2) $$\frac{7}{10}$$

Question 12.
If f = {(6, 3), (8, 9), (5, 3), (-1, 6)} then the pre-images of 3 are ……………… .
(1) 5 and -1
(2) 6 and 8
(3) 8 and -1
(4) 6 and 5
(4) 6 and 5

Question 13.
If a and p are the zeros 6f the polynomial P(x) = 4x2 + 3x + 7 then $$\frac{1}{\alpha}+\frac{1}{\beta}$$ is equal to …………….. .
(1) $$\frac{7}{3}$$
(2) $$\frac{-7}{3}$$
(3) $$\frac{3}{7}$$
(4) $$\frac{-3}{7}$$
(4) $$\frac{-3}{7}$$

Question 14.
The probability that a-leap year will have 53 Fridays or 53 Saturdays is …………….. .
(1) $$\frac{2}{7}$$
(2) $$\frac{1}{7}$$
(3) $$\frac{4}{7}$$
(4) $$\frac{3}{7}$$
(4) $$\frac{3}{7}$$

PART – II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
A relation “f” is defined by f(x) = x2 – 2 where x ∈ {-2, -1, 0, 3}
(i) List the elements of f
(ii) If f a function?
f(x) = x2 – 2 where x ∈ {-2,-1, 0, 3}
(i) f(-2) = (-2)2 – 2 = 2 ; f(-1) = (-1)2 – 2 = -1
f(0) = (0)2 – 2 = -2 ; f(3) = (3)2 – 2 = 7
Therefore, f = {(-2,2), (-1,-1), (0,-2), (3,7)}

(ii) We note that each element in the domain of f has a unique image.
Therefore f is a function.

Question 16.
Let f(x) = x2 – 1. Find fofof
fofof = fof [(x)]
= fof (x2 – 1)
= f (x2 – 1)2 – 1
= f (x4 – 2x2 + 1 – 1)
= f (x4 – 2x2)
fofof = (x4 – 2x2)2 – 1

Question 17.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Find the L.C.M of 35, 56, and 91
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
L.C.M = 23 × 5 × 7 × 13
= 3640
Since it leaves remainder 7
The required number = 3640 + 7 = 3647
∴ The smallest number is 3647

Question 18.
In a G.P. 729, 243, 81,… find t7.
The G.P. is 729,243,81, …………..
a = 729; r = $$\frac{t_{2}}{t_{1}}=\frac{243}{729}=\frac{81}{243}=\frac{27}{81}=\frac{3}{9}=\frac{1}{3}$$
tn = a rn – 1
t7 = 729 $$\left(\frac{1}{3}\right)^{7-1}$$
= $$729\left(\frac{1}{3}\right)^{6}=729 \times \frac{1}{729}=1$$
∴ The 7th term is 1

Question 19.
Find the excluded values of the expression $$\frac{t}{t^{2}-5 t+6}$$

The expression $$\frac{t}{t^{2}-5 t+6}$$ is undefined
when t2 – 5t + 6 = 0
(t – 3) (t – 2) = 0
t – 3 = 0 or t – 2 = 0
t = 3 or t = 2
The excluded values are 2 and 3

Question 20.
Find the square root of 9x2 – 24xy + 30xz – 40yz + 25z2 + 16y2
$$\sqrt{9 x^{2}-24 x y+30 x z-40 y z+25 z^{2}+16 y^{2}}$$
$$\sqrt{(3 x)^{2}+(4 y)^{2}+(5 z)^{2}-2(3 x)(4 y)-2(4 y)(5 z)+2(3 x)(5 z)}$$
$$\sqrt{(3 x-4 y+5 z)^{2}}$$ [using (a – b + c)2 = a2 + b2 + c2 -2ab -2bc + 2ac]
= |3x – 4y + 5z|

Question 21.
construct a 3 x 3 matrix whose elements are given by aij = $$\frac{(i+j)^{3}}{3}$$
aij = $$\frac{(i+j)^{3}}{3}$$

Question 22.
If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm2 and the area of ∆DEF is 16 cm2 and BC = 2.1 cm. Find the length of EF.
Given ∆ABC ~ ∆DEF

$$\frac{9}{16}=\frac{(2.1)^{2}}{\mathrm{EF}^{2}}$$
$$\left(\frac{3}{4}\right)^{2}=\left(\frac{2.1}{\mathrm{EF}}\right)^{2}$$
$$\frac{3}{4}=\frac{2.1}{\mathrm{EF}}$$
EF = $$\frac{4 \times 2.1}{3}$$ = 2.8 cm
Length of EF = 2.8 cm

Question 23.
If the area of the triangle formed by the vertices (p,p), (5,6), (5, -2) is 32 sq.units. Find the value of ‘p’.
Let the vertices be A (p,p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
$$\frac{1}{2}$$[x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3] = 32

$$\frac{1}{2}$$[(6p – 10 + 5p) – (5p + 30 – 2p)] = 32
$$\frac{1}{2}$$[11p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = $$\frac{104}{8}$$ = 13
The value of p = 13

Question 24.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 √3 m.
Height of the tower (AC) = 10 √3 m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ

In the right ∆ ABC, tan θ = $$\frac{A C}{A B}$$
$$=\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}$$
tan θ = $$\frac{1}{\sqrt{3}}$$ = tan 30°
∴ Angle of inclination is 30°

Question 25.
Find the diameter of a sphere whose surface area is 154m2.
Let r be the radius of the sphere.
Given that, surface area of sphere = 154 m2
4πr2 = 154
4 × $$\frac{22}{7}$$ × r2 = 154
gives r2 = $$154 \times \frac{1}{4} \times \frac{7}{22}$$
hence, r2 = $$\frac{49}{4}$$ we get r = $$\frac{7}{2}$$
Therefore, diameter is 7m

Question 26.
Find the standard deviation and the variance of first 23 natural numbers.
Standard deviation of first “n” natural numbers = $$\sqrt{\frac{n^{2}-1}{12}}$$
Standard deviation of first 23 natural numbers = $$\sqrt{\frac{23^{2}-1}{12}}$$
= $$\sqrt{\frac{529 – 1}{12}}$$
$$=\sqrt{\frac{528}{12}}$$
= √44 (variance)
Variance = 44 and standard deviation = 6.63

Question 27.
Form the quadratic equation whose roots are 3 + √7 ; 3 – √7
Sum of the roots = 3 + √7 + 3 – √7
= 6
Product of the roots = (3 + √7)(3 – √7)
= 9 – 7 = 2 .
The required equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (6) x + 2 = 0
x2 – 6x + 2 = 0

Question 28.
Two coins are tossed together. What is the probability of getting at most one head?
Sample space = {HH, HT, TH, TT}
n(S) = 4
Let A be the event of getting atmost one head A = {HT, TH, TT}
n(A) = 3
P(A) = $$\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}$$
The probability is $$\frac{3}{4}$$

PART – III

III. Answer any ten questions. Question No 42 is compulsory. [10 × 5 = 50]

Question 29.

f(x) = 6x + 1 ; x = {-5, -4, -3, -2, -1, 0, 1}
f(x) = 5x2 – 1 ; x = {2,3,4,5}
f(x) = 3x – 4; x = {6,7,8,9}

(i) f(-3) + f(2)
f(x) = 6x+1
f(3) = 6(-3) + 1 = -18 + 1 = -17
f(x) = 5x2 – 1
f(2) = 5(2)2 – 1 = 20 – 1 = 19
f(-3) + f(2) = -17 + 19 = 2

(ii) f(7) – f(1)
f(x) = 3x – 4
f(7) = 3(7) – 4= 21 – 4 = 17
f(x) = 6x + 1
f(1) = 6(1)+ 1 = 6 + 1 = 7
f(7) – f(1) = 17 – 7
= 10

(iii) 2f(4) + f(8)
f(x) = 5x2 – 1
f(4) = 5(4)2 – 1 = 15(16) – 1
= 80 – 1 = 79
f(x) = 3x – 4
f(8) = 3(8) – 4 = 24 – 4 = 20
2f(4) + f(8) = 2(79) + 20
= 158 + 20 .
= 178

(iv) $$\frac{2 f(-2)-f(6)}{f(4)+f(-2)}$$
f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11
f(x) = 3x – 4
f(6) = 3(6) – 4 = 18 – 4 = 14
f(x) = 5x2 – 1
f(4) = 5(4)2 – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 6x+1
f(-2) = 6(-2) + 1 = -12 + 1 = -11
$$\frac{2 f(-2)-f(6)}{f(4)+f(-2)}$$ = $$\frac{2(-11)-14}{79-11}$$
= $$\frac{-22-14}{68}=\frac{-36}{68}=\frac{-9}{17}$$

Question 30.
If f(x) = 2x + 3, g(x) = 1 – 2x and h(x) = 3x Prove that fo(goh) = (fog)oh
f(x) = 2x + 3, g(x) = 1 – 2x, h(x) = 3x
Now, (fog)(x) = f(g(x)) = f(1 – 2x) = 2(1 – 2x) + 3 = 5 – 4x
Then, (fog)oh(x) = (fog) (h(x)) = (fog)(3x) = 5 – 4(3x) = 5 – 12x ….(1)
(goh)(x) = g(h(x)) = g(3x) = 1 – 2(3x) = 1 – 6x
So, fo (goh)(x) = f(1 – 6x) = 2(1 – 6x) + 3 = 5 – 12x …….. (2)
From (1) and (2), we get (fog)oh = fo(goh)

Question 31.
If S1 , S2 , S3, ……. Smare the sums of n terms of m A.P.,s whose first terms are 1,2, 3, …….. m and whose common differences are 1,3,5,….(2m – 1) respectively, then show that (S1 + S2 + S3 + …….. + Sm) = $$\frac { 1 }{ 2 }$$ mn(mn + 1)
First terms of an A.P and 2, 3 ……….. m
The common difference are 1, 3, 5 , ………….. (2m – 1)
Sn = $$\frac { n }{ 2 }$$[2a+(n – 1)d]
S1 = $$\frac { n }{ 2 }$$[2+(n – 1)(1)]
= $$\frac { n }{ 2 }$$[2 + n – 1]
S1 = $$\frac { n }{ 2 }$$[n + 1] …….. (1)
S2 = $$\frac { n }{ 2 }$$[2(2) + (n – 1)3]
= $$\frac { n }{ 2 }$$[4 + 3n – 3]
S2 = $$\frac { n }{ 2 }$$(3n – 2) ……. (2)
S3 = $$\frac { n }{ 2 }$$[2(3) + (n – 1)5]
= $$\frac { n }{ 2 }$$[6 + 5n – 5}
= $$\frac { n }{ 2 }$$[5n + 1] …….. (3)
Sm = $$\frac { n }{ 2 }$$[2m + (n – 1)(2m – 1)]
= $$\frac { n }{ 2 }$$[2m + 2mn – n – m – 2m + 1]
= $$\frac { n }{ 2 }$$[n(2m – 1) + 1]
By adding (1) (2) (3) we get
S1 + S2 + S3 + ……….. + Sm = $$\frac { n }{ 2 }$$(n + 1) + $$\frac { n }{ 2 }$$(3n + 1) + $$\frac { n }{ 2 }$$(5n + 1)+ …………+ $$\frac { n }{ 2 }$$[n(2m – 1) + 1]
= $$\frac { n }{ 2 }$$[n + 1 + 3n + 1 + 5n + 1 …………… + n(2m – 1) + m]
= $$\frac { n }{ 2 }$$[n + 3n + 5n + ………….. + n(2m – 1) + m]
= $$\frac { n }{ 2 }$$[n$$(\frac { m }{ 2 })$$(2m) + m]
= $$\frac { n }{ 2 }$$[nm2 + m]
S1 + S2 + S3 + ……… + Sm = $$\frac { mn }{ 2 }$$[nm + 1]

Hint:
1 + 3 + 5 + …….. + 2m – 1
Sn = $$\frac { n }{ 2 }$$(a + 1)
= $$\frac { m }{ 2 }$$ (1 + 2m – 1)
= $$\frac { m }{ 2 }$$ (2m)

Question 32.
Find the sum to n terms of the series 0.4 + 0.44 + 0.444 + ………. to n terms

Question 33.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years .
By the given first condition.
$$\frac{x+y+z}{3}=53$$
x + y + z = 159 ….(1)
By the given 2nd condition.
$$\frac{1}{2} z+\frac{1}{3} y+\frac{1}{4} x=65$$
Multiply by 12
6z + 4y + 3x = 780 .
3x + 4y + 6z = 780 ….(2)
By the given 3rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
– 4x + z = – 16 + 4
4x – z = 12 ….(3)

Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
– z = 12 – 96
z = 84

Substitute the value of x = 24 and
z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51

Vani age = 24 years
Vani’s father age = 51 years
Vani grand father age = 84 years

Question 34.
Find the values of a and b if the following polynomials are perfect squares. ax4 + bx3 + 361x2 + 220.x + 100
Re-arrange the order we get .
100 + 220x + 361x2 + by3 + ax4

Since it is a perfect square
b – 264 = 0
b = 264

a – 144 = 0
a = 144

∴ The value of a = 144 and b = 264

Question 35.
State and prove Pythagoras Theorem
Statement
In a right angle triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Proof
Given : In ∆ABC, ∠A = 90°.
To prove : AB2 + AC2 = BC2
Construction : Draw AD ⊥ BC

Adding (1) and (2) we get
AB2 + AC2 = BC × BD + BC × DC
= BC (BD + DC) = BC × BC
AB2 + AC2 = BC2
Hence the theorem is proved.

Question 36.
Find the equation of the median and altitude of AABC through A where the vertices are A(6, 2), B(-5, -1) and C(1,9).
(i) To find median

Mid point of BC(D) = $$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$$
$$=\left(\frac{-5+1}{2}, \frac{-1+9}{2}\right)$$
$$=\left(\frac{-4}{2}, \frac{8}{2}\right)$$ = (-2, 4)
Equation of the median AD is
$$\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}$$
$$\frac{y-2}{4-2}=\frac{x-6}{-2-6}$$
$$\frac{y-2}{2}=\frac{x-6}{-8}$$
= 2(x – 6) = 8(y – 2)
2x – 12 = -8y + 16
2x + 8y – 28 = 0
(÷by 2) x + 4y – 14 = 0

(ii) To find the equation of the altitude

Slope of BC = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
$$=\frac{9+1}{1+5}$$
$$=\frac{10}{6}=\frac{5}{3}$$
Slope of the altitude = $$-\frac{3}{5}$$
Equation of the altitude AD is
y – y1 = m(x – x1)
y – 2 = $$-\frac{3}{5}$$(x – 6)
-3 (x – 6) = 5 (y – 2)
-3x + 18 = 5y – 10
-3x – 5y + 18 + 10 = 0
-3x – 5y + 28 =0
3x + 5y – 28 = 0
∴ Equation of the median is x + 4y – 14 = 0
Equation of the altitude is 3x + 57 – 28 = 0

Question 37.
Two ships are sailing in the sea on either sides of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the ships are 30° and 45° respectively. If the lighthouse is 200 m high, find the distance between the two ships. (√3 = 1.732 )
Let AB be the lighthouse. Let C and D be the positions of the two ships.
Then, AB = 200 m
∠ACB = 30°, ∠ADB = 45°
In right triangle BAC, tan 30° = $$\frac{A B}{A C}$$

$$\frac{1}{\sqrt{3}}=\frac{200}{\mathrm{AC}}$$ gives AC = 200√3
In right triangle BAD, tan 45° = $$\frac{A B}{A D}[latex] 1 = [latex]\frac{200}{\mathrm{AD}}$$ gives AD = 200
Now, CD = AC + AD = 200√3 + 200 [by (1) and (2)]
CD = 200 (√3 + 1) = 200 × 2.732 = 546.4
Distance between two ships is 546.4 m

Question 38.
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Radius of the cylinder = $$\frac{4.5}{2} \mathrm{cm}$$
Height of the cylinder = 10 cm
Volume of the cylinder = π r2 h cu. units

Radius of the coin (r) .= $$\frac{1.5}{2} \mathrm{cm}$$
Thickness of the coin (h) = 2 mm = $$\frac{2}{10} \mathrm{cm}$$
Volume of one coin = π r2 h cu. units

Number of coins = 450

Question 39.
A box contains cards numbered 3, 5, 7, 9, … 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
Sample space = {3, 5, 7,9 ,… ,35, 37} $$\left[\frac{37-1}{2}=\frac{36}{2}=18\right]$$
n (S) = 18
Let A be the event of getting a multiple of 7
A = {7,21,35}
n(A) = 3
p(A) = $$\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{18}$$
‘ Let B be the event of getting a prime number
B = {3,5, 7, 11, 13, 17, 19, 23.29,31.37}
n (B) = 11
p(B) = $$\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{11}{18}$$
A∩B = {7}
n(A ∩ B) = 1
P(A ∩ B) = $$\frac{n(\mathrm{A} \cap \mathrm{B})}{n(\mathrm{S})}=\frac{1}{18}$$
p(A∪B) = P(A) + P(B) – P(A ∩ B)
$$=\frac{3}{18}+\frac{11}{18}-\frac{1}{18}=\frac{3+11-1}{18}=\frac{13}{18}$$
Probability of getting a muliple of 7 or a priime number = $$\frac{13}{18}$$

Question 40.
Find X and Y if X – Y = $$\left[ \begin{matrix} 2 & 1 \\ 4 & 3 \\ 0 & 6 \end{matrix} \right]$$ and X + Y = $$\left[ \begin{matrix} 12 & 13 \\ 6 & 5 \\ 4 & 8 \end{matrix} \right]$$

Question 41.
A lead pencil is in the shape of a right circular cylinder. The pencil is 28 cm long and its radius is 3 in. If the led is of radius one mm (1 mm) then find the volume of the wood used in the pencil.
Length of a pencil (h) = 28 cm
Radius of a pencil (R) = 3mm $$\left(\frac{3}{10} \mathrm{cm}\right)$$
Radius of a led (r) = 1 mm $$\left(\frac{1}{10} \mathrm{cm}\right)$$
Volume of the wood used = Volume of the pencil – Volume of the led
= πh(R2 – r2cu. unit

Volume of the wood used in the pencil = 7.04 cm3

Question 42.
The following table shows the marks obtained by 48 students in a quiz competition in Mathematics. Calculate the variance and standard deviation.

Let us form the following table using the given data.

PART – IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) Draw a circle of radius 4.5 cm. Take a point on the circle. Draw the tangent at that point using the alternate segment theorem. Rough Diagram
Radius of the circle = 4.5 cm

Steps of construction:

1. With O as centre, draw a circle of radius 4.5 cm.
2. Take a point L on the circle. Through L draw any chord LM.
3. Take a point M distinct from L and N on the circle, so that L, M, N are in anti-clockwise direction. Join LN and NM.
4. Through “L” draw tangent TT’ such that ∠TLM = ∠MNL.
5. TT’ is the required tangent.

[OR]

(b)
Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.

Steps of construction

1. Draw a line segment RQ = 5 cm.
2. At R draw RE such that ∠QRE = 40°.
3. At R. draw RF such that ∠ERF = 90°.
4. Draw the perpendicular bisector to RQ, which intersects RF at O and RQ at G.
5. With O as centre and OP as radius draw a circle.
6. From G mark arcs of radius 4.4 cm on the circle. Mark them as P and S.
7. Join PR and PQ. Then ∆PQR is the required triangle.
8. From P draw’ a line PN which is perpendicular to RQ it meets at N.
9. Measure the altitude PN.
PN = 2.2 cm.

Question 44.
(a) Draw the graph of y = x2 – 4x + 3 and use it to solve x2 – 6x + 9 = 0.
Step 1: Draw the graph of y = x2 – 4x + 3 by preparing the table of values as below.

Step 2: To solve x2 – 6x + 9 = 0, subtract x2 – 6x + 9 = 0 from y = – x2 – 4x + 3

The equation y = 2x – 6 represent a straight line. Draw the graph of y = 2x – 6 form in the table of values as below.

The line y = 2x – 6 intersect y = x2 – 4x + 3 only at one point.

Step 3: Mark the point of intersection of the curve y = x2 – 4.r 3 and y = 2x – 6 that is (3,0).
Therefore, the x coordinate 3 is the only solution for the equation x2 – 6x + 9 = 0.

[OR]

(b) Draw the Graph of y = x2 – 4x + 4


text:
## Wednesday, May 07, 2008

### Confidence is back

The figure shows the movement of VIX and the S&P 500 (adjusted for fluctuations of the USD) in recent weeks. The vertical yellow line is 17th of March. Click on the figure to see it more clearly. We see a sharp outburst of fear and then it rapidly subsided. I'm feeling fairly pleased with myself about an article that I wrote on 18th afternoon in Indian time (i.e. early morning of 18th in the US); a "long S&P + long dollar + short VIX" position from that date would have worked quite nicely. Also see the associated blog entry, and this synthesis of analysis of the disturbances of 2007 and 2008.

Using a dataset from 1/1990 till yesterday, the overall average value for VIX is 19.07. We are now at 18.21. So I guess we're pretty much back to the unconditional mean.

Please note: Comments are moderated. Only civilised conversation is permitted on this blog. Criticising me is perfectly okay; uncivilised language is not. I delete any comment which is spam, has personal attacks against anyone, or uses foul language. I delete any comment which does not contribute to the intellectual discussion about the blog article in question.

Please note: LaTeX mathematics works. This means that if you want to say $10 you have to say \$10.


text:
Example: Neo-Riemannian Orbit

Suppose we want to generate chord progressions based on Neo-Riemannian Theory, which gives a theoretical framework to navigate through all possible major- and minor triads by applying certain transformations. To quote from Wikipedia about Neo-Riemannian Theory:

The principal transformations of neo-Riemannian triadic theory connect triads of different species (major and minor), and are their own inverses (a second application undoes the first). These transformations are purely harmonic, and do not need any particular voice leading between chords: all instances of motion from a C major to a C minor triad represent the same neo-Riemannian transformation, no matter how the voices are distributed in register.

The three transformations move one of the three notes of the triad to produce a different triad:

• The P transformation exchanges a triad for its Parallel. In a Major Triad move the third down a semitone (C major to C minor), in a Minor Triad move the third up a semitone (C minor to C major)
• The R transformation exchanges a triad for its Relative. In a Major Triad move the fifth up a tone (C major to A minor), in a Minor Triad move the root down a tone (A minor to C major)
• The L transformation exchanges a triad for its Leading-Tone Exchange. In a Major Triad the root moves down by a semitone (C major to E minor), in a Minor Triad the fifth moves up by a semitone (E minor to C major)

The transformations P, R and L

We will have to implement the transformations P, R and L. Let's keep them in a dictionary called riemann, where we put each transformation associated with it's symbol, like \L.

var riemann = (); // Dictionary <symbol> -> <neo-riemannian-transformation>

The P transformation exchanges a triad for its Parallel. In a Major Triad move the third down a semitone (C major to C minor), in a Minor Triad move the third up a semitone (C minor to C major).

Even if we don't have yet the means to decide programmatically if a given triad is a major or minor one, or how to raise or lower particular notes in it, we can start expressing our intend right away by writing code that calls not-yet-available functions, but with names that we like to use. For example, the description for the transformation \P above could be expressed like this:

riemann[\P] = { |triad|  // [a, b, c]
case
// In a Major Triad move the third down a semitone (C major to C minor),
// in a Minor Triad move the third up a semitone (C minor to C major)
};

We use the case {}{}…{}{} construct to differentiate between different cases (major, minor and catch-all case (for detecting errors)) and call functions isMajor, isMinor and moveThird, which we will need to implement seperately. Nevertheless, our intend has been clearly stated and readers of our code will understand.

Before implementing the missing functions, let's write the other two transformations in a similar manner.

The R transformation exchanges a triad for its Relative. In a Major Triad move the fifth up a tone (C major to A minor), in a Minor Triad move the root down a tone (A minor to C major)

riemann[\R] = { |triad| // [a, b, c]
case
// In a Major Triad move the fifth up a tone (C major to A minor),
// in a Minor Triad move the root down a tone (A minor to C major)
};

The L transformation exchanges a triad for its Leading-Tone Exchange. In a Major Triad the root moves down by a semitone (C major to E minor), in a Minor Triad the fifth moves up by a semitone (E minor to C major)

riemann[\L] = { |triad| // [a, b, c]
case
// In a Major Triad the root moves down by a semitone (C major to E minor),
// in a Minor Triad the fifth moves up by a semitone (E minor to C major)
};

Helper functions

Finally, we will now implement the missing functions and put them before what we have written so far. First, we need function to distinguish Major and Minor chords:

var isMajor = { |triad|

// Ex.:	triad == [0, 4, 7] (C Major)
//    ⇒	f == 4 and s == 3

// Check all possible inversions of a major chord by looking at the
// intervals:
case
{(f==4 && s==3) || (f==3 && s==5) || (f==5 && s==4)} { true  } // major!
{(f==3 && s==4) || (f==5 && s==3) || (f==4 && s==5)} { false } // minor!
// Other interval combinations mean invalid triad, help debugging it:
{ true }{ triad.debug("[isMajor] unknown modus"); false } // error!
};

case
{(f==3 && s==4) || (f==5 && s==3) || (f==4 && s==5)} { true  } // minor!
{(f==4 && s==3) || (f==3 && s==5) || (f==5 && s==4)} { false } // major!
{ true }{ triad.debug("[isMinor] unknown modus"); false } // error!
};

... and functions manipulating chords in specific ways:

var moveRoot = { |triad, d|
// triad = [a, b, c]

case
// root position: fifth between a and c, root == a
{ c-a == 7 }{ triad + [ d, 0, 0 ] }
// first inversion: forth between b and c, root == c
{ c-b == 5 }{ triad + [ 0, 0, d ] }
// second inversion: forth between a and b, root == b
{ b-a == 5 }{ triad + [ 0, d, 0 ] }
// Couldn't identify the root! Help debugging it.
{ true }{ triad.debug("unknown root") }
};

var moveThird = { |triad, d|
case
// root position: fifth between a and c, third == b
{ c-a == 7 }{ triad + [ 0, d, 0] }
// first inversion: forth between b and c, third == a
{ c-b == 5 }{ triad + [ d, 0, 0] }
// second inversion: forth between a and b, third == c
{ b-a == 5 }{ triad + [ 0, 0, d] }
{ true }{ triad.debug("unknown third") }
};

var moveFifth = { |triad, d|
case
// root position: fifth between a and c, fifth == c
{ c-a == 7 }{ triad + [ 0, 0, d] }
// first inversion: forth between b and c, fifth == b
{ c-b == 5 }{ triad + [ 0, d, 0] }
// second inversion: forth between a and b, fifth == a
{ b-a == 5 }{ triad + [ d, 0, 0] }
{ true }{ triad.debug("unknown fifth") }
};

For printing chord names:

var noteNames = ["C", "C♯", "D", "D♯", "E", "F", "F♯", "G", "G♯", "A", "A♯", "B"];

case
{ c-a == 7 }{ a }
{ b-a == 5 }{ b }
{ c-b == 5 }{ c }
{ true }{ triad.debug("[getRoot] unknown root") }
};

var name = noteNames[root];

case
{ isMajor.(triad) } { name ++ " Major" }
{ isMinor.(triad) } { name.toLower ++ " Minor" }
{ true }{ triad.debug("[chordName] unknown modus") }
};

Secondary transformations

The N (or Nebenverwandt) relation exchanges a major triad for its minor subdominant, and a minor triad for its major dominant (C major and F minor). The "N" transformation can be obtained by applying R, L, and P successively.

riemann[\N] = riemann[\R]<>riemann[\L]<>riemann[\P];

S (or Slide) relation exchanges two triads that share a third (C major and C♯ minor); it can be obtained by applying L, P, and R successively in that order:

riemann[\S] = riemann[\L]<>riemann[\P]<>riemann[\R];

The H relation (LPL) exchanges a triad for its hexatonic pole (C major and A♭ minor):

riemann[\H] = riemann[\L]<>riemann[\P]<>riemann[\L];

The combination PRL has no name on wikipedia, transforming C major to D major. Let's call it U.

riemann[\U] = riemann[\P]<>riemann[\R]<>riemann[\L];

An orbit routine

neoOrbit is a function that expects a chord and a stream of neo-riemannian transformations (as symbols) and returns a routine that yields the progression of chords by applying the transformations consecutively.

~neoOrbit = {
arg chord = [0, 4, 7],  stream = Pseq([\P, \L, \R], inf).asStream;

Routine {
var tr;
loop{
chordName.(chord).debug("chord");

// Eject the current chord to the consumer of this
// routine, when called .next() on us
chord.yield;

// Get the next transformation (symbol) from the stream
tr = stream.next().debug("trans");

// Apply the transformation
chord = riemann[tr].(chord);

// Stay within the 0-11 range
chord = chord%12;

// After applying %12 the array might be out of order
chord.sort;
};
};
};

Following the orbit

Let's put everything together and generate a chord progression, starting at D-Major and alternating between the transformations R and L:

(
Pbind(
\scale, Scale.chromatic,
\degree, ~neoOrbit.value(chord: [0, 4, 7]+2, stream: Pseq([\R, \L], inf).asStream),
\dur, 0.5,
).play;
)

If everything worked well, you should hear a chord progression, starting at D-Major and moving through the whole Tonnetz, i.e. orbiting through all 24 possible chords. The output in the Post window should look like this:

chord: D Major
trans: R
chord: b minor
trans: L
chord: G Major
trans: R
chord: e minor
trans: L
chord: C Major
trans: R
chord: a minor
trans: L
chord: F Major
trans: R
chord: d minor
trans: L
chord: A♯ Major
trans: R
chord: g minor
trans: L
chord: D♯ Major
⋮

Have a listen to the first few seconds of the progression: