# areas in polar coordinates

• May 2nd 2010, 07:30 PM
driver327
areas in polar coordinates
I have seen examples where the thetas are preset, but I have trouble with the ones where I have to find them.

Find the area of the regions.

1. Inside the oval limacon r = 4 + 2 cos(theta)

I found the value for theta... -4 = 2cos(theta)

-2 = cos(theta) -> arccos(-2) = theta... but the calc says this is undefined.

• May 2nd 2010, 07:39 PM
skeeter
Quote:

Originally Posted by driver327
I have seen examples where the thetas are preset, but I have trouble with the ones where I have to find them.

Find the area of the regions.

1. Inside the oval limacon r = 4 + 2 cos(theta)

I found the value for theta... -4 = 2cos(theta) ??? r does not equal 0.

-2 = cos(theta) -> arccos(-2) = theta... but the calc says this is undefined.

why are you trying to solve for $\theta$ ?

$A = \int_0^{2\pi} \frac{(4+2\cos{\theta})^2}{2} \, d\theta = \int_0^{\pi} (4+2\cos{\theta})^2 \, d\theta
$
• May 2nd 2010, 08:19 PM
driver327
I did it like that and got the right answer!! my question is: for those types of problems, do I automatically assume that the parameters is 1 full rotation? (from 0 to 2pi or 0 to pi w/o the 1/2).. just trying to understand it
• May 3rd 2010, 08:22 AM
skeeter
Quote:

Originally Posted by driver327
I did it like that and got the right answer!!

not with the posted problem you didn't ... what problem are you talking about?

my question is: for those types of problems, do I automatically assume that the parameters is 1 full rotation? (from 0 to 2pi or 0 to pi w/o the 1/2)

some polar curves require $\theta$ to run from 0 to $2\pi$ to get the full graph ... there are some that require less, like $r = \sin(3\theta)$ ... it only need to run from 0 to $\pi$ to get a complete graph.