Convert integral $\int_{0}^{1}\int_{y}^{\sqrt{4-y^2}} x^2+y^2 dx dy$ into polar form

Question

Convert the following double integral to an equivalent polar form but do NOT evaluate:

$$\int_{0}^{1}\int_{y}^{\sqrt{4-y^2}} x^2+y^2 \; dx \; dy.$$

Do we let $x=r\cos{(\theta)}$ and $y=r\sin{(\theta)}$ so $x^2+y^2=r^2$ and substitute in? Thanks for your help.

Answer 1

Let $x=r\cos\theta$, $y=r\sin\theta$ and $dxdy=rdr d\theta$. Note the integration region is a truncated circle sector as shown in the graph. Thus, $$\int_{0}^{1}\int_{y}^{\sqrt{4-y^2}}( x^2+y^2 )\; dx \; dy =\int_{0}^{\frac\pi6}\int_{0}^2 r^3drd\theta +\int_{\frac\pi6}^{\frac\pi4}\int_{0}^{1/\sin\theta} r^3drd\theta$$