Question

Change the Cartesian integral \[\int\limits_{0}^{2}\int\limits_{y}^{\sqrt{4-y^2}}xdxdy\]into an equivalent polar integral. Then evaluate the polar integral.
I know (or think rather) the integrand should be \[r^2\sin( \Theta )drd \Theta \] What I'm having trouble with is converting the limits of integration. Any help would be appreciated.

Answer 1

drawing the region of integration always helps:

Answer 2

So we talk about a circle with radius 2 and you want to integrate that until y=x which is at exactly 45 degree, or pi/4

\[0 \leq r \leq 2\]

and \[0 \leq \theta \leq \frac{ \pi }{ 4}\]

Answer 3

I disagree with your integrand by the way, remember that:

\[x=r \cos \theta\] that will turn your integrand into

\[r^2\cos \theta dr d \theta \]

Answer 4

Thanks for that. And yeah I actually meant cos. Didn't go back and reread what I wrote. Thanks for pointing that out.

Answer 5

you're very welcome.

Answer 6

note your limits
\[ \int\limits_{0}^{2}\int\limits_{y}^{\sqrt{4-y^2}}xdxdy = \int_0^{\sqrt2}\int_y^{\sqrt{4-y^2}}x dx dy + \int_{\sqrt2}^2\int_{\sqrt{4-y^2}}^yx dx dy\]

Answer 7

the first term goes in above from 0 to pi/4

Answer 8

better make change of variables than to change it into polar coordinates