1.9. Find the areas of figures bounded by lines in polar coordinates:

a) $\rho=\cos 2 \varphi$

b) $\rho=4 \sin \varphi, \rho=2 \sin \varphi$

c) $\rho=\sin 3 \varphi$.
Solution. When calculating the areas of figures bounded by lines in polar coordinates, the formula used is

$$
S_{\Omega}=\iint_{D} \rho d \rho d \varphi
$$

where $D$ is the domain of variation of the variables $\rho$ and $\varphi$.

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-054.jpg?height=234&width=436&top_left_y=222&top_left_x=191)

Fig. 1.34

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-054.jpg?height=394&width=448&top_left_y=129&top_left_x=645)

Fig. 1.35

a) On the $X O Y$ plane, the curve $\rho=\cos 2 \varphi$ (a two-leaf rose) is depicted in Fig. 1.34. Due to symmetry, we find the area of one petal and double the result, so

$$
\begin{gathered}
D=\left\{(\varphi, \rho):-\frac{\pi}{4} \leqslant \varphi \leqslant \frac{\pi}{4}, \quad 0 \leqslant \rho \leqslant \cos 2 \varphi\right\} \\
S_{\Omega}=2 \iint_{D} \rho d \rho d \varphi=2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} d \varphi \int_{0}^{\cos 2 \varphi} \rho d \rho=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\cos ^{2} 2 \varphi\right) d \varphi=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos 4 \varphi+1}{2} d \varphi= \\
=\left.\left(\frac{\sin 4 \varphi}{8}+\frac{\varphi}{2}\right)\right|_{-\frac{\pi}{4}} ^{\frac{\pi}{4}}=\frac{\pi}{4}
\end{gathered}
$$

b) The domain of variation of the variables $\rho, \varphi$ in this case is $D=\{(\varphi, \rho): 0 \leqslant \varphi \leqslant \pi ; 2 \sin \varphi \leqslant \rho \leqslant 4 \sin \varphi\}$, therefore

$$
\begin{gathered}
S_{\Omega}=\iint_{D} \rho d \rho d \varphi=\int_{0}^{\pi} d \varphi \int_{2 \sin \varphi}^{4 \sin \varphi} \rho d \rho=\frac{1}{2} \int_{0}^{\pi} 12 \sin ^{2} \varphi d \varphi=3 \int_{0}^{\pi}(1-\cos 2 \varphi) d \varphi= \\
=\left.3\left(\varphi-\frac{\sin 2 \varphi}{2}\right)\right|_{0} ^{\pi}=3 \pi .
\end{gathered}
$$

c) In the $O X Y$ plane, the figure $\Omega$ is a three-leaf rose (Fig. 1.35), and

$$
\begin{gathered}
D=\left\{(\varphi, \rho): 0 \leqslant \varphi \leqslant \frac{\pi}{3} ; 0 \leqslant \rho \leqslant \sin 3 \varphi\right\}: \\
S_{\Omega}=3 \iint_{D} \rho d \rho d \varphi=3 \int_{0}^{\pi / 3} d \varphi \int_{0}^{\sin 3 \varphi} \rho d \rho=\frac{3}{2} \int_{0}^{\pi / 3} \sin ^{2} 3 \varphi d \varphi=3 \int_{0}^{\pi / 3}(1-\cos 6 \varphi) d \varphi= \\
=\left.3\left(\varphi-\frac{\sin 6 \varphi}{6}\right)\right|_{0} ^{\pi / 3}=\pi .
\end{gathered}
$$