Evaluate this integral involving circle

Question

Let $D$ be the circle of radius $a$ centered at the origin. Evaluate $$I=\int\int_D(1-x^2-y^2)\:\:dxdy$$ My work:

Let $x=r\:\:\text{cos}\theta$ and $y=r\:\:\text{sin}\theta$ $\implies x^2+y^2=r^2$ or $(1-x^2-y^2)=1-r^2$ Our integral becomes $$I=\int_{r=0}^{a}\int_{\theta=0}^{2\pi}(1-r^2)\:\:rd\theta dr$$ I am stuck after this. Any help will be greatly appreciated.

Answer 1

You have made the correct subtitution, just integrate as usual $$I=\int_{0}^{2\pi}\int_{r=0}^{a}(1-r^2)\:\:rdr d\theta=\int_{0}^{2\pi}-\dfrac{a^3-3a}{3}d\theta=\frac{2\pi(3a-a^3)}{3}$$