Example 2. Calculate the area of the figure bounded by the closed curve $\left(x^{2}+y^{2}\right)^{2}=4 a y^{3}, a>0$.
Solution. We transition to polar coordinates:

$$
\left\{\begin{array}{l}
x=r \cos \varphi \\
y=r \sin \varphi
\end{array}\right.
$$

Then the equation of the given curve takes the form

$$
\left(r^{2}\left(\cos ^{2} \varphi+\sin ^{2} \varphi\right)\right)^{2}=4 a \cdot r^{3} \sin ^{3} \varphi, \quad \text { or } \quad r=4 a \sin ^{3} \varphi
$$

Since the original equation is meaningful for $y \geqslant 0$, the equation in polar coordinates is defined for $\varphi \in[0, \pi]$.

The area of the corresponding figure (Fig. 3.13) is computed by transitioning to polar coordinates in the integral:

$$
\begin{aligned}
& S=\iint_{D} d x d y=\iint_{D} r d r d \varphi=\int_{0}^{\pi} d \varphi \int_{0}^{4 a \sin ^{3} \varphi} r d r=\left.\frac{1}{2} \int_{0}^{\pi} d \varphi r^{2}\right|_{0} ^{4 a \sin ^{3} \varphi}= \\
& =8 a^{2} \int_{0}^{\pi} \sin ^{6} \varphi d \varphi=8 a^{2} \int_{0}^{\pi}\left(\frac{1-\cos 2 \varphi}{2}\right)^{3} d \varphi= \\
& =a^{2} \int_{0}^{\pi}\left(1-3 \cos 2 \varphi+3 \cdot \frac{1+\cos 4 \varphi}{2}-\frac{1+\cos 4 \varphi}{2} \cdot \cos 2 \varphi\right) d \varphi=\frac{5}{2} \pi a^{2} \\
& \text { Answer. } S=\frac{5}{2} \pi a^{2} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-055.jpg?height=271&width=346&top_left_y=253&top_left_x=1456)

Fig. 3.13

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-055.jpg?height=349&width=395&top_left_y=178&top_left_x=1984)

Fig. 3.14