Example 3. Calculate the volume of the body bounded by the surfaces $a z=a^{2}-x^{2}-y^{2}, z=0, x^{2}+y^{2}+a x=0(a>0)$.
Solution. The given body is a part of a paraboloid of revolution $a z=a^{2}-x^{2}-y^{2}$, enclosed within the cylinder $x^{2}+y^{2}+a x=0(a>0)$.

The integration domain $D$ is the circle $x^{2}+y^{2}+a x \leqslant 0$, or $\left(x+\frac{a}{2}\right)^{2}+y^{2} \leqslant \frac{a^{2}}{4}$, located in the II and III quadrants, i.e., the angular polar coordinate $\varphi$ varies within the limits: $\frac{\pi}{2} \leqslant \varphi \leqslant \frac{3 \pi}{2}$ (Fig. 3.14).

The volume of the body is calculated by the formula

$$
V=\iint_{D} \frac{1}{a}\left(a^{2}-x^{2}-y^{2}\right) d x d y=\frac{1}{a} \iint_{G}\left(a^{2}-r^{2}\right) r d r d \varphi
$$

To determine the limits of integration in polar coordinates, we write the region $G$ as inequalities. This is the circle bounded by the circumference $x^{2}+y^{2}+a x=0$, i.e., $r^{2}+r a \cos \varphi=0$, or $r=-a \cos \varphi$ $(\cos \varphi \leqslant 0)$. Therefore,

$$
G: 0 \leqslant r \leqslant-a \cos \varphi, \quad \frac{\pi}{2} \leqslant \varphi \leqslant \frac{3 \pi}{2}
$$

$$
\begin{aligned}
& \begin{aligned}
& \text { Therefore } \\
& V \frac{1}{a} \int_{\pi / 2}^{3 \pi / 2} d \varphi \int_{0}^{-a \cos \varphi} r\left(a^{2}-r^{2}\right) d r=\frac{1}{a} \int_{\pi / 2}^{3 \pi / 2} d \varphi\left[-\left.\frac{\left(a^{2}-r^{2}\right)^{2}}{4}\right|_{0} ^{-a \cos \varphi}\right]= \\
&=\frac{1}{4 a} \int_{\pi / 2}^{3 \pi / 2}\left[a^{4}-a^{4} \sin ^{4} \varphi\right] d \varphi=\left.\frac{1}{4} a^{3} \varphi\right|_{\pi / 2} ^{3 \pi / 2}-\frac{a^{3}}{4} \int_{\pi / 2}^{3 \pi / 2}\left(\frac{1-\cos 2 \varphi}{2}\right)^{2} d \varphi= \\
&=\frac{\pi a^{3}}{4}-\frac{a^{3}}{16} \int_{\pi / 2}^{3 \pi / 2}\left(1-2 \cos 2 \varphi+\frac{1+\cos 4 \varphi}{2}\right) d \varphi=\frac{5}{32} \pi a^{3}
\end{aligned} \\
& \text { Answer. } V=\frac{5 \pi a^{3}}{32}
\end{aligned}
$$

## Exercises

Calculate the double integrals, switching to polar coordinates.

1. $\iint_{D} \sqrt{x^{2}+y^{2}} d x d y, D: x^{2}+y^{2} \leqslant r^{2}$.
2. $\int_{D} \int \sqrt{1-x^{2}-y^{2}} d x d y, D: x^{2}+y^{2} \leqslant r^{2}$.
3. $\iint_{D} \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}} d x d y, D: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1, a>0, b>0$.
4. $\iint_{D} \sqrt{1+x^{2}+y^{2}} d x d y, D-$ interior of the lemniscate

$$
\left(x^{2}+y^{2}\right)^{2}=x^{2}-y^{2}
$$

Calculate the areas of figures bounded by the given lines.
5. $x^{2}+y^{2}=2 x, x^{2}+y^{2}=4 x, y=x, y=0$.
6. $(x-2 y+3)^{2}+(3 x+4 y+1)^{2}=64$.
7. $\left(\frac{x^{2}}{4}+\frac{y^{2}}{9}\right)^{2}=\frac{x^{2}}{4}-\frac{y^{2}}{9}$.
8. $y^{2}=a x, y^{2}=16 a x, y^{2}=x^{3}, 16 a y^{2}=x^{3}$.
9. $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$.

Calculate the volumes of bodies bounded by the given surfaces.
10. $z=x^{2}+y^{2}, y=x^{2}, z=0, y=1$.
11. $x^{2}+y^{2}=R^{2}, x^{2}+y^{2}-z^{2}+R^{2}=0$.
12. $x^{2}+y^{2}-\frac{z^{2}}{2}=0, x^{2}+y^{2}-z^{2}+R^{2}=0$.
13. $z=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}, \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2 \frac{x}{a}, a>0, b>0$.
14. $z=0, x^{2}+y^{2}-2 R x=0, x^{2}+y^{2}-z^{2}=0$.
15. $z=2 x^{2}+y^{2}+1, x+y=1, x=0, y=0, z=0$.

## Answers

1. $\frac{\pi r^{3}}{6}$. 2. $\frac{2}{3} \pi r^{2}$. 3. $\frac{2 \pi a b}{3}$. Hint. Put $x=\operatorname{ar} \cos \varphi, y=b r \sin \varphi$; in this case the Jacobian $J=a b c$. 4. $\frac{20}{9}-\frac{\pi}{3}$. 5. $\frac{3 \pi}{4}+\frac{3}{