Question

Convert the integral below to polar coordinates and evaluate the integral.

Answer 1

I suppose the integral is

$\displaystyle\int_0^5\int_y^{\sqrt{25-y^2}} xy\,\mathrm dx\,\mathrm dy$

The integration region corresponds to a sector of a cirlce with radius 5 subtended by a central angle of π/4 rad. We can capture this region in polar coordinates by the set

$R=\left\{(r,\theta)\mid0\le r\le5,0\le\theta\le\dfrac\pi4\right\}$

Then $x=r\cos\theta$, $y=r\sin\theta$, and $\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$. So the integral becomes

$\displaystyle\iint_Rxy\,\mathrm dx\,\mathrm dy=\int_0^{\pi/4}\int_0^5r^3\sin\theta\cos\theta\,\mathrm dr\,\mathrm d\theta=\frac{625}{16}$