# Convert an integral from Cartesian to Polar coordinates and evaluate it.

• Apr 28th 2010, 05:22 PM
asdf122345
Convert an integral from Cartesian to Polar coordinates and evaluate it.
• Apr 29th 2010, 08:58 AM
Amer
Quote:

Originally Posted by asdf122345

Attachment 16611

$r^2 = x^2 +y^2$
$y = r\sin \theta$
$x = r \cos \theta$
lets convert the boundaries

$x = \sqrt{4y-y^2 }$

$x^2 = 4y -y^2 \Rightarrow x^2+y^2 = 4y$

$r^2 = 4r\sin \theta$

$r = 4 \sin \theta$

r changes from 0 to $4\sin \theta$

and theta from 0 to pi/2 the first quarter so

$\int_{0}^{4}\int_{0}^{\sqrt{4y-y^2}} x^2 \; dx dy$

becomes

$\int_{0}^{\frac{\pi}{2}} \int_{0}^{4\sin \theta} r^2 \cos^2 \theta (r) dr d\theta$

$\int_{0}^{\frac{\pi}{2}} \cos^2 \theta \left(\frac{r^4}{4}\right)\mid^{4 \sin \theta}_{0} d\theta$

$\int_{0}^{\frac{\pi}{2}} \cos^2\theta (4^3 \sin^4 \theta ) d\theta$
• Apr 29th 2010, 03:12 PM
asdf122345
Is it possible to integrate where you left off? I can't figure out how to integrate that. Or does it get messy from integrating from that.
• Apr 29th 2010, 10:07 PM
Amer
$4^3 \int \cos ^2 \theta (\sin ^4 \theta ) d\theta$

$\sin \theta \cos \theta = \frac{\sin 2\theta }{2}$

$\sin ^2 \theta = \frac{1-\cos 2\theta}{2}$

$4^3 \int \cos ^2 \theta \sin ^2 \theta ( \sin ^2\theta ) d \theta$

$4^3 \int \frac{\sin ^2 2\theta}{4} \left(\frac{1- \cos 2\theta }{2}\right) d\theta$

$4^3 \left(\int \frac{\sin ^2 2\theta}{4} d\theta - \int \frac{\sin ^2 2 \theta \cos 2\theta}{8} d\theta\right)$

first integral convert $\sin ^2 2\theta = \frac{1- \cos 4 \theta }{2}$

second one sub $u = \sin 2\theta$

can you continue ?
• Apr 30th 2010, 10:46 AM
asdf122345
Yea I got it from here. Thanks
• Apr 30th 2010, 01:13 PM
asdf122345
I got 4 pi as the answer. Is that correct? Something seems wrong.
• Apr 30th 2010, 01:19 PM
shenanigans87
The answer is $2\pi$

So you may have just made a small error by a factor of 2.