# Math Help - 2 Area Volume

1. ## 2 Area Volume

1. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

Set y=y i got x^3. Then intergrate got x^3/3. limit from 0 to 3

2. Volumes of revolution are found by integration just as two-dimensional areas are, only our cross-sections are discs or hollow cylinders (called 'shells') instead of rectangles.

We can find the answer to the first problem by rewriting the function as

$x=y^{1/4}$

and using the Shell Method. To do this, we note that the radius of each shell is $3-y$ and the height $2y^{1/4}$. The volume of each cross section will therefore be

$2y^{1/4}\cdot 2\pi(3-y)\,dy.$

Integrating from $y=0$ to $y=1$, we obtain

$4\pi\int_0^1(3y^{1/4}-y^{5/4})\,dy.$

(Actually, the correct way to rewrite the function is $x=\pm y^{1/4}$, but we accounted for both sides in our integral.)

For the second problem, the volume is infinite. Are you sure it's written correctly?

3. I still dont understand how you do it for number 1.

yes the question is correct..
Originally Posted by Scott H
Volumes of revolution are found by integration just as two-dimensional areas are, only our cross-sections are discs or hollow cylinders (called 'shells') instead of rectangles.

We can find the answer to the first problem by rewriting the function as

$x=y^{1/4}$

and using the Shell Method. To do this, we note that the radius of each shell is $3-y$ and the height $2y^{1/4}$. The volume of each cross section will therefore be

$2y^{1/4}\cdot 2\pi(3-y)\,dy.$

Integrating from $y=0$ to $y=1$, we obtain

$4\pi\int_0^1(3y^{1/4}-y^{5/4})\,dy.$

(Actually, the correct way to rewrite the function is $x=\pm y^{1/4}$, but we accounted for both sides in our integral.)

For the second problem, the volume is infinite. Are you sure it's written correctly?

4. The volume of revolution of the first problem looks kind of like a doughnut (called a 'torus') with a cylindrical hole in the middle. Its axis $y=3$ is parallel to the $x$-axis.

To find the volume, we decompose the region into thinner and thinner concentric shells around the axis $y=3$. To find the volume of one of these shells, we must take into account the height of the shell in the direction of the $x$-axis (here, $2y^{1/4}$) and the circumference of the shell (here, $2\pi(3-y)$, as the circumference decreases when we increase $y$ toward the center). The volume of one shell will therefore be


\begin{aligned}
\mbox{Volume of Shell}&=\mbox{Height}\cdot\mbox{Thickness}\cdot\mb ox{Circumference}\\
&=2y^{1/4}\cdot dy \cdot 2\pi(3-y)\\
&= 4\pi y^{1/4}(3-y)\,dy \\
&= 4\pi (3y^{1/4}-y^{5/4})\,dy.
\end{aligned}

Integrating this from the outside $y=0$ to the inside $y=1$ of the region, we obtain


\begin{aligned}
\mbox{Total Volume} &= \int_0^1 4\pi(3y^{1/4}-y^{5/4})\,dy \\
&= 12\pi\int_0^1 y^{1/4}\,dy - 4\pi\int_0^1 y^{5/4}\,dy.
\end{aligned}

The answer to the second problem is technically $\infty$, as the parabola bounded by $x=3$ is unbounded in the direction of the $y$-axis. If we add the line $y=9$, however, we can find the area by using the Disc Method and integrating along the $y$-axis. Now we're going from $y=0$ to $y=9$, and along the way every disc will have volume


\begin{aligned}
\mbox{Volume of Disc} &= \mbox{Area of Surface}\cdot\mbox{Thickness} \\
&= \pi(\sqrt{y})^2\cdot dy \\
&= \pi y\, dy.
\end{aligned}

The volume of the region would therefore be

$\mbox{Total Volume}=\int_0^9 \pi y\,dy.$

As you can see, finding volumes is just like finding areas under curves, only we are integrating by the volumes of discs and shells rather than by the area of rectangles.

You will not always have to integrate with respect to $y$. Sometimes you can integrate with respect to $x$ and leave the function $y$ as it is.

5. Thanks for explaining and helping me out. I really appreciated.