# area; polar coordinates

• Nov 13th 2009, 02:56 AM
billym
area; polar coordinates
I have $\int\int_Rx^3+xy^2dxdy$,

$R$ = second quadrant of $1/4<=x^2+y^2<=1$

I changed $x^3+xy^2$ to polar coordinates:

$r^3cos^3(\theta)+r cos(\theta)r^2sin^2(\theta)=r^3cos(\theta)$

Do I evaluate the integral like this?

$\int_{\pi/2}^{\pi}\int_{1/2}^{1}r^3cos(\theta)drd\theta$
• Nov 13th 2009, 03:00 AM
jeneverboy
I think your boundaries are correct.
• Nov 13th 2009, 03:57 AM
galactus
Don't forget about that extra r when using polar integration.

$x^{3}+xy^{2}$

$r^{3}cos^{3}{\theta}+rcos{\theta} \cdot r^{2}sin^{2}{\theta}=r^{3}cos{\theta}$

$r \cdot r^{3}cos{\theta}$
• Nov 13th 2009, 06:30 AM
billym
Should the answer be the same as $\frac{\pi-\pi \frac{1}{2}^2}{4}$ or am I missing the point here?
• Nov 13th 2009, 08:17 AM
billym
Or rather, that is R. Which is basically the upper left portion of a doughnut. But what does the area of integration look like?
• Nov 14th 2009, 02:50 AM
billym
Pardon me for being the village idiot, but could somebody please answer this one question for me:

When sketching the area of integration, do I just sketch $R$ , or do I have to consider $x^3+xy^2$ ?
• Nov 14th 2009, 04:51 AM
rn443
Quote:

Originally Posted by billym
Should the answer be the same as $\frac{\pi-\pi \frac{1}{2}^2}{4}$ or am I missing the point here?

The answer shouldn't have pi in it. $\int_{\frac{1}{2}}^1 \int_{\frac{\pi}{2}}^\pi r^4 \cos (\theta ) d\theta dr = \int_{\frac{1}{2}}^1 -r^4 dr$.
• Nov 14th 2009, 06:03 AM
billym
yeah I realized that last night. Anyway, I've got the right answer now. Thanks dudes.