Example 2. Find the volume of the body bounded by the surfaces

$$
x^{2}+y^{2}+2 x=0, \quad z=\frac{25}{4}-y^{2}, \quad z=0
$$
Solution.

1. By formula (1) with $f_{2}=25 / 4-y^{2}$ and $f_{1}=0$, the desired volume is

$$
V=\iint_{D}\left(\frac{25}{4}-y^{2}-0\right) d x d y
$$

where $D$ is the projection of the body onto the $X O Y$ plane.

2. To find $D$, we define the body using inequalities and eliminate $z$ from them. In this case, the body is defined by the inequalities

$$
\left\{\begin{array}{l}
(x+1)^{2}+y^{2} \leq 1 \\
0 \leq z \leq 25 / 4-y^{2}
\end{array}\right.
$$

From the first inequality, it is clear that $|y| \leq 1$ and, consequently, the second inequality is automatically satisfied (geometrically, this means that the projection of the surface $z=25 / 4-y^{2}$ onto the $XOY$ plane encompasses the circle $\left.(x+1)^{2}+y^{2} \leq 1\right)$. Therefore,

$$
D=\left\{(x, y):(x+1)^{2}+y^{2} \leq 1\right\}
$$

3. Since the region $D$ is bounded by circles and lines passing through the origin, it is easier to solve the problem in polar coordinates

$$
\left\{\begin{array}{l}
x=\varrho \cos \varphi \\
y=\varrho \sin \varphi
\end{array}\right.
$$

In this case, $(\varrho, \varphi) \in D^{\prime}$, and the desired volume is given by the formula

$$
V=\iint_{D}\left(\frac{25}{4}-y^{2}\right) d x d y=\iint_{D^{\prime}}\left(\frac{25}{4}-\varrho^{2} \sin ^{2} \varphi\right) \varrho d \varphi d \varrho
$$

4. To find the region $D^{\prime}$, we replace $x$ with $\varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the inequality defining the region $D$:

$$
(\varrho \cos \varphi+1)^{2}+\varrho^{2} \sin ^{2} \varphi \leq 1
$$

We obtain

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-304.jpg?height=133&width=607&top_left_y=1444&top_left_x=321)

Notice that from the inequalities $0 \leq \varrho \leq-2 \cos \varphi$ it follows that $\pi / 2 \leq \varphi \leq 3 \pi / 2$. 5. We transition from the double integral to the repeated integral:

$$
V=\iint_{D^{\prime}}\left(\frac{25}{4}-\varrho^{2} \sin ^{2} \varphi\right) \varrho d \varrho d \varphi=\int_{\pi / 2}^{3 \pi / 2} d \varphi \int_{0}^{-2 \cos \varphi}\left(\frac{25}{4}-\varrho^{2} \sin ^{2} \varphi\right) \varrho d \varrho
$$

Integrating sequentially, we get

$$
\begin{gathered}
V=\int_{\pi / 2}^{3 \pi / 2} d \varphi \int_{0}^{-2 \cos \varphi}\left(\frac{25}{4}-\varrho^{2} \sin ^{2} \varphi\right) \varrho d \varrho= \\
=\left.\int_{\pi / 2}^{3 \pi / 2}\left(\frac{25 \varrho^{2}}{8}-\frac{\varrho^{4}}{4} \sin ^{2} \varphi\right)\right|_{0} ^{-2 \cos \varphi} d \varphi= \\
=\int_{\pi / 2}^{3 \pi / 2}\left(\frac{25}{2} \cdot \cos ^{2} \varphi-4 \cos ^{4} \varphi \sin ^{2} \varphi\right) d \varphi=6 \pi
\end{gathered}
$$

Answer. $\quad V=6 \pi$ units of volume.

Conditions of the Problems. Find the volumes of the bodies bounded by the given surfaces.

1. $\quad x=\sqrt{y}, \quad x=2 \sqrt{y}, \quad z=1-y, \quad z=0$.
2. $y=\sqrt{x}, \quad y=2 \sqrt{x}, \quad z=6-x, \quad z=0$.
3. $y=\sqrt{x}, \quad y=0, \quad x=1, \quad z=x+y+1, \quad z=0$.
4. $y=x^{2}, \quad y=1, \quad z=x^{2}+y^{2}, \quad z=0$.
5. $y=6-3 x / 2, \quad y=6-3 x, y=0, \quad z=6-x-y, z=0$.
6. $\quad x^{2}+y^{2}=4, \quad z=x y, \quad z=0 \quad(x \geq 0, y \geq 0)$.
7. $y=\sqrt{x / 2}, \quad y=0 \quad z=4-x-2 y, \quad z=0$.
8. $x^{2}+y^{2}-4 y=0, \quad z=4-x^{2}, \quad z=0$.
9. $x^{2}+y^{2}-2 x=0, \quad z=x^{2}+y^{2}, \quad z=0$.
10. $x^{2}+y^{2}-2 x=0, \quad z=2 x, \quad z=4 x$.

Answers. 1. $V=4 / 15$. 2. $V=48 \sqrt{6} / 5$. 3. $V=79 / 60$. 4. $V=40 / 3$. 5. $V=12$. 6. $V=4$. 7. $V=17 / 5$. 8. $V=12 \pi$. 9. $V=3 \pi / 2$. 10. $V=2 \pi$.

## 12.6. Calculation of Areas in Cartesian Coordinates

Problem Statement. Find the area of the region $D$ bounded by the lines $f_{1}(x, y)=0, f_{2}(x, y)=0$ (and, possibly, by the lines $x=a$ and $x=b$ or $y=c$ and $y=d$).

## PLAN OF SOLUTION.

From the definition of the double integral, it follows that the desired area $S$ is numerically equal to

$$
\iint_{D} 1 \cdot d x d y
$$

1. Define the region $D$ using inequalities. For this, determine which of the inequalities

$$
f_{1}(x, y) \leq 0, \quad f_{1}(x, y) \geq 0, \quad f_{2}(x, y) \leq 0 \text { or } f_{2}(x, y) \geq 0
$$

are satisfied by the coordinates of the points in the region $D$.

Suppose, for example, that the inequalities $f_{1}(x, y) \leq 0$ and $f_{2}(x, y) \leq 0$ are such. Then

$$
D=\left\{(x, y): \begin{array}{l}
f_{1}(x, y) \leq 0 \\
f_{2}(x, y) \leq 0
\end{array}\right\}
$$

Solve the inequalities defining $D$ with respect to $x$ and $y$. We get

$$
D=\left\{\begin{array}{cc}
a & \leq x \leq b \\
(x, y): & y_{1}(x) \leq y \leq y_{2}(x)
\end{array}\right\}
$$

or

$$
D=\left\{\begin{array}{cc}
(x, y): & c \leq y \leq d \\
& x_{1}(y) \\
\leq x \leq x_{2}(y)
\end{array}\right\}
$$

2. Transition from the double integral to the repeated integral:

$$
\iint_{D} d x d y=\int_{a}^{b} d x \int_{y_{1}(x)}^{y_{2}(x)} d y
$$

or

$$
\iint_{D} d x d y=\int_{c}^{d} d y \int_{x_{1}(y)}^{x_{2}(y)} d x
$$

3. Integrate sequentially, using the properties of the definite integral.

Write the answer, not forgetting the dimension.

NOTE. If necessary, divide the region into parts and use the additivity property of the integral.