By changing to polar coordinates, evaluate the double integral∬(x^(2)+y(2))^(5/2)dxdy
where D is the disk x^(2)+y^(2)≤16.

Question

By changing to polar coordinates, evaluate the double integral∬(x^(2)+y(2))^(5/2)dxdy
where D is the disk x^(2)+y^(2)≤16.

zepdrix · Answer

|dw:1384825626136:dw|Mmmm ok so this is the region we're integrating over, yes?

zepdrix · Answer

So in polar our function will be $\Large f(r,\;\theta)$
Let's use the picture to try and change the limits of integration.
So what will our boundaries be for r and theta?
Lemme know if that's too confusing.

anonymous · Answer

theta is 0 to 2pi, r is 0 to 4?

zepdrix · Answer

Good good good.
Then we'll use these to change from cartesian to polar:$$\Large x=r \cos \theta, \qquad\qquad\qquad y=r \sin \theta$$

zepdrix · Answer

When we change our differentials, we need to remember that an extra factor of r shows up.$$\Large dy\;dx \qquad\to\qquad r\;dr\;d\theta$$

anonymous · Answer

so the new integral is $$\int\limits_{0}^{2\pi}\int\limits_{0}^{4}r^5rdrd \theta$$

zepdrix · Answer

ah yes! very good :)

anonymous · Answer

Thank you very much! You have been very helpful. Could you help me with another one?

zepdrix · Answer

sure

anonymous · Answer

Calculate the integral over the given region by changing to polar coordinates: 
$$f(x,y)=\left| 16xy \right|,    x^2+y^2\le 81$$

zepdrix · Answer

Hmm ok lemme think a sec

zepdrix · Answer

Hmm so we have nice easy conversion of the boundaries again, right?

anonymous · Answer

Yes, r spans from 0 to 9. I am unsure of the boundaries for theta though. Does the absolute value affect them? Or are they still 0 to 2pi?

zepdrix · Answer

|dw:1384827526770:dw|We still want to cover this entire region.
So theta will still be 0 to 2pi.