Example 9. Calculate the volume of the solid of revolution around the $O x$ axis of the figure bounded by the lines $2 y=x^{2}$ and $2 x+2 y=3$.
Solution. The desired volume $V$ is the difference $V_{1}-V_{2}$ of the volumes of two bodies. The first is obtained by rotating the segment $A B$ around the $O x$ axis, the second by rotating the arc of the parabola $A O B$ (Fig. 2.9). The limits of integration are the projections of points $A$ and $B$ onto the $O x$ axis. We find them:

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-039.jpg?height=346&width=499&top_left_y=490&top_left_x=77)

Fig 2.9

$$
\begin{aligned}
& \left\{\begin{array}{l}
2 y=x^{2} \\
2 y=3-2 x
\end{array} \Rightarrow\right. \\
& \Rightarrow \begin{array}{l}
x^{2}=3-2 x, x^{2}+2 x-3=0 \\
x_{1}=-3, x_{2}=1
\end{array}
\end{aligned}
$$

Thus,

$$
V_{1}=\pi \int_{-3}^{1}\left(\frac{3-2 x}{2}\right)^{2} d x=
$$

$$
\begin{gathered}
=\frac{\pi}{4} \int_{-3}^{1}\left(9-12 x+4 x^{2}\right) d x=\left.\frac{\pi}{4}\left(9 x-6 x^{2}+\frac{4 x^{3}}{3}\right)\right|_{-3} ^{1}=\frac{91}{3} \pi \\
V_{2}=\pi \int_{-3}^{1}\left(\frac{x^{2}}{2}\right)^{2} d x=\left.\frac{\pi}{4} \cdot \frac{x^{5}}{5}\right|_{-3} ^{1}=\frac{61}{5} \pi
\end{gathered}
$$

Finally: $V=V_{1}-V_{2}=\frac{272}{15} \pi$.

Answer. $V=\frac{272}{15} \pi$.