1.7. Convert to polar coordinates and evaluate the double integrals.

a) $\iint_{\Omega} x y^{2} d x d y$, where the region $\Omega$ is bounded by the circles $x^{2}+(y-1)^{2}=1$ and $x^{2}+y^{2}=4 y$

b) $\iint_{\Omega} e^{-x^{2}-y^{2}} d x d y$, where $\Omega-$ is the circle $x^{2}+y^{2} \leqslant R^{2}$.
Solution.

a) First, let's depict the region $\Omega$ in the Cartesian coordinate system (Fig. 1.32).

The equations of the boundary circles in polar coordinates are obtained after the substitution $x=\rho \cos \varphi, y=\rho \sin \varphi$:

$$
\begin{aligned}
x^{2}+(y-1)^{2}=1 \Rightarrow x^{2}+y^{2} & =2 y \Rightarrow \rho=2 \sin \varphi, 0 \leqslant \varphi \leqslant \pi \\
x^{2}+y^{2} & =4 y \Rightarrow \rho=4 \sin \varphi, 0 \leqslant \varphi \leqslant \pi
\end{aligned}
$$

The region $\Omega$ in coordinates $x, y$ is transformed into the region $D$ in coordinates $\varphi, \rho: D=\{(\varphi, \rho): 0 \leqslant \varphi \leqslant \pi ; 2 \sin \varphi \leqslant \rho \leqslant 4 \sin \varphi\}$. We will perform the transformation to coordinates $\varphi, \rho$ and compute the original double integral:

$$
\begin{gathered}
\iint_{\Omega} x y^{2} d x d y=\iint_{D} \rho^{3} \cos \varphi \sin ^{2} \varphi \rho d \rho d \varphi= \\
=\int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi d \varphi \int_{2 \sin \varphi}^{4 \sin \varphi} \rho^{4} d \rho=\int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi\left(\left.\frac{\rho^{5}}{5}\right|_{2 \sin \varphi} ^{4 \sin \varphi}\right) d \varphi= \\
=\frac{992}{5} \int_{0}^{\pi} \sin ^{7} \varphi \cos \varphi d \varphi=\frac{992}{5} \int_{0}^{\pi} \sin ^{7} \varphi d(\sin \varphi)= \\
=\left.\frac{992}{5} \frac{\sin ^{8} \varphi}{8}\right|_{0} ^{\pi}=0
\end{gathered}
$$

b) Transform the integral to polar coordinates $x=\rho \cos \varphi, y=\rho \sin \varphi:$

$$
\begin{gathered}
\iint_{\Omega} e^{-x^{2}-y^{2}} d x d y=\iint_{D} e^{-\rho^{2}} \rho d \rho d \varphi= \\
=\int_{0}^{2 \pi} d \varphi \int_{0}^{R} e^{-\rho^{2}} \rho d \rho=\left.\int_{0}^{2 \pi}\left(-\frac{e^{-\rho^{2}}}{2}\right)\right|_{0} ^{R} d \varphi=\pi\left(1-e^{-R^{2}}\right)
\end{gathered}
$$