Double integral into the polar form

1. Nov 10, 2007

nemesis24

hello i have this problem about polar form, i am aware that when you have a problem like $$\int\int$$ x^2 + y^2 dxdy you use r^2 = x^2 + y^2 but i what would you do if you had a problem like $$\int\int$$ xy dxdy?

edit: i know the limits if you need them plz let me know but i was more interested in the concept behind it

2. Nov 10, 2007

Kummer

If you have,
$$\iint_R xy \ dA$$ then since $$x=r\cos \phi$$ and $$y = r\sin \phi$$ it means, $$xy = r^2 \sin \phi \cos \phi = \frac{1}{2} r^2 \sin (2\phi)$$.

3. Nov 10, 2007

nemesis24

so you would just integrate 1/2r^2sin(2(teta)

4. Nov 10, 2007

Kummer

No you also have to remember the factor of $$r$$ whichs appears in the Jacobian.

5. Nov 11, 2007

HallsofIvy

Staff Emeritus
That is, the "differential of area" in polar coordinates is $r dr d\theta$.