2.11. Compute the line integral of the second kind

\[
\int_{L} x^{2} y d x - y^{2} x d y
\]

where \( L: \left(x^{2}+y^{2}\right)^{2}=a^{2}\left(x^{2}-y^{2}\right)^{2}, y \geqslant 0 \) from \( A(0,0) \) to \( B(a, 0) \).
Solution. To get an idea of the curve $L$, called a lemniscate, it is easiest to switch to polar coordinates $x=r \cos \varphi, y=$ $=r \sin \varphi$ (Fig. 2.12).

$$
L: r^{2}=a^{2} \cos 2 \varphi, \varphi \in\left[0, \frac{\pi}{4}\right]
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-134.jpg?height=304&width=422&top_left_y=124&top_left_x=717)

Fig. 2.12

By connecting points $A$ and $B$ with a segment of the $O X$ axis, we obtain a closed contour $C$ that bounds the region $\Omega$ (see Fig. 2.7). The vector field $\mathcal{A}=(P, Q)=\left(x^{2} y,-y^{2} x\right)$ is continuously differentiable in $\bar{\Omega}$, so we can apply Green's formula to the contour integral over $C$. We get

$$
\oint_{C} x^{2} y d x-y^{2} x d y=\iint_{\Omega}\left(-x^{2}-y^{2}\right) d x d y
$$

Using the property of line integrals, we can write:

$$
\oint_{c} x^{2} y d x-y^{2} x d y=\int_{A B} x^{2} y d x-y^{z^{2}} x d y-\int_{L} x^{2} y d x-y^{2} x d y
$$

Note that the integral over the segment $A B$ is zero ( $y=0, d y=0$ on the $O X$ axis), so the desired integral is

$$
\int_{L} x^{2} y d x-y^{2} x d y=\iint_{\Omega}\left(x^{2}+y^{2}\right) d x d y
$$

To compute the double integral, we switch to polar coordinates:

$$
\begin{gathered}
\iint_{\Omega}\left(x^{2}+y^{2}\right) d x d y=\iint_{D} r^{3} d r d \varphi=\int_{0}^{\frac{\pi}{4}} d \varphi \int_{0}^{a \sqrt{\cos 2 \varphi}} r^{3} d r=\frac{a^{4}}{4} \int_{0}^{\frac{\pi}{4}} \cos ^{2} 2 \varphi d \varphi= \\
=\frac{a^{4}}{4} \int_{0}^{\frac{\pi}{4}} \frac{1+\cos 4 \varphi}{2} d \varphi=\left.\frac{a^{4}}{8}\left(\varphi+\frac{\sin 4 \varphi}{4}\right)\right|_{0} ^{\frac{\pi}{4}}=\frac{\pi a^{4}}{32}
\end{gathered}
$$

Problems on the computation of surface integrals of the first kind.