# Area of a region enclosed by polar equation

• July 27th 2008, 06:53 PM
auslmar
Area of a region enclosed by polar equation
I have this problem that reads "Find the area enclosed by $r^2$ = $4cos(\theta)
$
."

so, what I ended up with

2 $
\int_a^b$
$cos(\theta)$ $d\theta$

I may have messed up in finding the correct integral, but I think the problem is that I don't know what to choose for $a$ or $b$. Can anyone help explain how I would go about choosing proper values for $a$ and $b$?

P.S. I seem to be having trouble with the LaTex, is anyone else having trouble?
• July 27th 2008, 07:00 PM
Chris L T521
Quote:

Originally Posted by auslmar
I have this problem that reads "Find the area enclosed by $r^2$ = $4cos(\theta)
$
."

so, what I ended up with

2 $
\int_a^b$
$cos(\theta)$ $d\theta$

I may have messed up in finding the correct integral, but I think the problem is that I don't know what to choose for $a$ or $b$. Can anyone help explain how I would go about choosing proper values for $a$ and $b$?

P.S. I seem to be having trouble with the LaTex, is anyone else having trouble?

Have you tried to graph the Lemniscate? The limits are a little tricky, but its not that complex...take into account that $4$ $\cos\theta$ can't be negative...

Just wondering, what did you get for your limits of integration? You're integral set up is correct.

--Chris
• July 27th 2008, 07:34 PM
Chris L T521
Quote:

Originally Posted by auslmar
I have this problem that reads "Find the area enclosed by $r^2$ = $4cos(\theta)
$
."

so, what I ended up with

2 $
\int_a^b$
$cos(\theta)$ $d\theta$

I may have messed up in finding the correct integral, but I think the problem is that I don't know what to choose for $a$ or $b$. Can anyone help explain how I would go about choosing proper values for $a$ and $b$?

P.S. I seem to be having trouble with the LaTex, is anyone else having trouble?

http://img.photobucket.com/albums/v4...e2143a3-19.jpg

Does this make sense?

--Chris
• July 27th 2008, 07:37 PM
Soroban
Hello, auslmar!

By the way, this is not a lemniscate.
. . It could be called "the square root of a circle."

Quote:

I have this problem that reads: Find the area enclosed by $r^2 = 4$ $\cos\theta$

so, what I ended up with: .2 $
\int_a^b$
$cos(\theta)$ $d\theta$

I may have messed up in finding the correct integral.

Set r = 0 and solve for θ.

. . 4 cos θ .= .0 . . θ .= .±½π

and there are our limits . . .

• July 27th 2008, 08:01 PM
Chris L T521
Quote:

Originally Posted by Soroban
Hello, auslmar!

By the way, this is not a lemniscate.
. . It could be called "the square root of a circle."

Um...

http://img.photobucket.com/albums/v4...e2143a3-20.jpg

http://img.photobucket.com/albums/v4...Lemniscate.jpg

It might not be a lemniscate per se, but it has the form of one.

--Chris