1.10. Find the areas of the figures bounded by the lines:
a) $\left(x^{2}+y^{2}\right)^{2}=2 y^{2}$

b) $\left(x^{2}+y^{2}\right)^{3}=x^{4}+y^{4}$.
Solution.

a) Let's switch to polar coordinates $x=\rho \cos \varphi ; y=$ $=\rho \sin \varphi$, we get the equation of the line bounding $\Omega$ in polar coordinates:

$$
\begin{gathered}
\rho^{4}=2 \rho^{2} \sin ^{2} \varphi \\
\rho^{2}=2 \sin ^{2} \varphi \\
\rho=\sqrt{2 \sin ^{2} \varphi}=\sqrt{2}|\sin \varphi|
\end{gathered}
$$

This figure $\Omega$ is a two-leafed shape. We find the area of one leaf and double it, in this case

$$
\begin{gathered}
D=\{(\varphi, \rho): \quad 0 \leqslant \varphi \leqslant \pi ; \quad 0 \leqslant \rho \leqslant \sqrt{2} \sin \varphi\} \\
S_{\Omega}=2 \iint_{D} \rho d \rho d \varphi=2 \int_{0}^{\pi} d \varphi \int_{0}^{\sqrt{2} \sin \varphi} \rho d \rho=2 \int_{0}^{\pi} \sin ^{2} \varphi d \varphi=\int_{0}^{\pi}(1-\cos 2 \varphi) d \varphi=\pi
\end{gathered}
$$

b) It is easiest, as in example a), to switch to polar coordinates $x=\rho \cos \varphi ; y=\rho \sin \varphi$, we get:

$$
\rho^{2}=\cos ^{4} \varphi+\sin ^{4} \varphi=1-2 \sin ^{2} \varphi \cos ^{2} \varphi=1-\frac{1}{2} \sin ^{2} 2 \varphi
$$

The domain of variation of the variables $\rho, \varphi$ :

$$
D=\left\{(\varphi, \rho): \quad 0 \leqslant \varphi \leqslant 2 \pi ; \quad 0 \leqslant \rho \leqslant \sqrt{1-\frac{1}{2} \sin ^{2} 2 \varphi}\right\} .
$$

We find the area of $\Omega$ in polar coordinates:

$$
\begin{aligned}
S_{\Omega 2}= & 2 \int_{0}^{2 \pi} \int_{0}^{\sqrt{1-\frac{1}{2} \sin ^{2} 2 \varphi}} \rho d \rho d \varphi=\frac{1}{2} \int_{0}^{2 \pi}\left(1-\frac{1}{2} \sin ^{2} 2 \varphi\right) d \varphi= \\
& =\pi-\frac{1}{8} \int_{0}^{2 \pi}(1-\cos 4 \varphi) d \varphi=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}
\end{aligned}
$$