Double integral into the polar form...

[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?

thanks in advance.

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

[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).

[nemesis24]

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

[Kummer]

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

[HallsofIvy]

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