Example 2.54. Calculate the area of one petal of the curve $\rho=\sin ^{2} \varphi$.
Solution. When $\varphi$ changes from 0 to $\pi$, one petal of the given curve is obtained (the shaded area in Fig. 2.8)

$$
\begin{aligned}
S= & 0.5 \int_{0}^{\pi}(\sin \varphi)^{2} d \varphi=0.125 \int_{0}^{\pi}(1-\cos 2 \varphi)^{2} d \varphi= \\
& =0.125 \int_{0}^{\pi}\left(1-2 \cos 2 \varphi+\cos ^{2} 2 \varphi\right) d \varphi= \\
= & 0.125 \int_{0}^{\pi}(1-2 \cos 2 \varphi+0.5(1+\cos 4 \varphi) d \varphi= \\
= & 0.125\left(1.5 \varphi-\sin 2 \varphi+0.125 \sin 4 \varphi\right)\bigg|_{0} ^{\pi}=\frac{3 \pi}{16}
\end{aligned}
$$