Double Integrals - Calculus

we change the integral to the polar coordinate as the following: $x=rcos\phi$ and $y=rsin\phi$

Because the domain is the semi circle, then we have $r:0\to 1$ , $\phi:-\frac{\pi}{2}\to\frac{\pi}{2}$ and $dxdy=rdrd\phi$ The integral becomes $\int\int_Dr^2(cos\phi-sin\phi)drd\phi=(\int_0^1r^2dr)\cdot(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(cos\phi-sin\phi))=\frac{r^3}{3}|_0^1(sin\phi+cos\phi)|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{2}{3}$

$$ $B$ is the region between $x=0$ and $x=\sqrt{1-y^2}$ for $-1 \le y \le 1$ . Therefore $$ $\begin{aligned} \int_{y=-1}^1 \int_{x=0}^{\sqrt{1-y^2}} (x-y)\, dxdy &= \int_{y=-1}^1 \left[\frac{1}{2}x^2-yx\right]_{x=0}^{\sqrt{1-y^2}}\, dy\\ &= \int_{y=-1}^1 \left(\frac{1}{2}(1-y^2)-y\sqrt{1-y^2}\right)\, dy\\ &= \left[\frac{1}{2}y-\frac{1}{6}y^3+\frac{1}{3}(1-y^2)^\frac{3}{2}\right]_{y\,=-1}^1\\ &=1-\frac{1}{3}+0\\ &=\frac{2}{3} \end{aligned}$ $$ Thus, $a=2$ , $b=3$ and $a+b = \boxed{5}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$