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Need help with this.

- Apr 28th 2010, 05:22 PMasdf122345Convert an integral from Cartesian to Polar coordinates and evaluate it.
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Need help with this. - Apr 29th 2010, 08:58 AMAmer
Attachment 16611

$\displaystyle r^2 = x^2 +y^2 $

$\displaystyle y = r\sin \theta $

$\displaystyle x = r \cos \theta $

lets convert the boundaries

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

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

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

$\displaystyle r = 4 \sin \theta $

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

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

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

becomes

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

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

$\displaystyle \int_{0}^{\frac{\pi}{2}} \cos^2\theta (4^3 \sin^4 \theta ) d\theta $ - Apr 29th 2010, 03:12 PMasdf122345
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 PMAmer
$\displaystyle 4^3 \int \cos ^2 \theta (\sin ^4 \theta ) d\theta $

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

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

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

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

$\displaystyle 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 $\displaystyle \sin ^2 2\theta = \frac{1- \cos 4 \theta }{2} $

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

can you continue ? - Apr 30th 2010, 10:46 AMasdf122345
Yea I got it from here. Thanks

- Apr 30th 2010, 01:13 PMasdf122345
I got 4 pi as the answer. Is that correct? Something seems wrong.

- Apr 30th 2010, 01:19 PMshenanigans87
The answer is $\displaystyle 2\pi$

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