2.4. Compute the line integrals of the second kind $\int_{L} P d x+Q d y$, if:
a) $P=y(x-y) ; Q=x ; L: y=2 x, x \in[0,1]$

b) $P=\left(x^{2}-y\right) ; Q=0 ; L-$ perimeter of the rectangle formed by the lines $x=0, y=0, x=1, y=2$, traversed clockwise;

c) $P=2 x(y-1) ; Q=x^{2} ; L-$ contour of the figure formed by the lines $y=x^{2}, y=9$ (when traversing the contour, the figure remains on the left).
Solution.

a) By substituting $y=2 x, d y=2 d x$, the curvilinear integral over the given segment is reduced to a definite integral:

$$
\begin{aligned}
& \int_{L} P d x+Q d y=\int_{0}^{1}(2 x(x-2 x) d x+x 2 d x)= \\
& =\int_{0}^{1}\left(2 x-2 x^{2}\right) d x=\left.\left(x^{2}-\frac{2}{3} x^{3}\right)\right|_{0} ^{1}=\frac{1}{3}
\end{aligned}
$$

b) The contour is shown in Fig. 2.6. On segments $O A$ and $B C$, $d x=0$, so only segments $A B$ and $C O$ remain.

$$
\begin{aligned}
& \int_{O A B C}\left(x^{2}-y\right) d x=\int_{A B}\left(x^{2}-y\right) d x+\int_{C O}\left(x^{2}-y\right) d x=\int_{0}^{1}\left(x^{2}-2\right) d x+ \\
&+\int_{1}^{0} x^{2} d x=\left.\left(\frac{x^{3}}{3}-2 x\right)\right|_{0} ^{1}+\left.\frac{x^{3}}{3}\right|_{1} ^{0}=\frac{1}{3}-2-\frac{1}{3}=-2 .
\end{aligned}
$$

c) The contour bounding the figure is shown in Fig. 2.7. On the curve $B O A$, $y=x^{2}, d y=2 x d x, x \in[-3,3]$.

On the segment $A B$, $y=9, d y=0, x \in(-3,3)$.

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-124.jpg?height=317&width=384&top_left_y=1395&top_left_x=246)

Fig. 2.6

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-124.jpg?height=293&width=343&top_left_y=1411&top_left_x=747)

Fig. 2.7

Using the additivity of the curvilinear integral, we get

$$
\begin{gathered}
\int_{L} P d x+Q d y=\int_{B O A B}\left(2 x(y-1) d x+x^{2} d y\right)= \\
=\int_{B O A}\left(2 x(y-1) d x+x^{2} d y\right)+\int_{A B}\left(2 x(y-1) d x+x^{2} d y\right)= \\
=\int_{3}^{3}\left(2 x\left(x^{2}-1\right) d x+2 x^{3} d x\right)+\int_{-3}^{3} 2 x(9-1) d x= \\
=\int_{-3}^{3}\left(4 x^{3}-2 x\right) d x+16 \int_{-3}^{3} x d x=\left.\left(x^{4}-x^{2}\right)\right|_{-3} ^{3}+\left.8 x^{2}\right|_{-3} ^{3}=0
\end{gathered}
$$