Example. Find the area of the region $D$ bounded by the lines

$$
x^{2}+y^{2}=12, \quad x \sqrt{6}=y^{2} \quad(x \geq 0)
$$
Solution.

1. Let's define the region $D$ by inequalities. The region cannot be outside the circle, as it would then be unbounded. The region cannot be to the left of the parabola, as in this case, its points could have negative abscissas, which is excluded by the condition $x \geq 0$. Therefore,

$$
D=\left\{\begin{array}{ll}
(x, y): & \begin{array}{l}
x^{2}+y^{2} \leq 12 \\
x \geq y^{2} / \sqrt{6}
\end{array}
\end{array}\right\}
$$

Solving the inequalities defining $D$ with respect to $x$ and $y$, we get

$$
\frac{y^{2}}{\sqrt{6}} \leq x \leq \sqrt{12-y^{2}}
$$

Thus, $y^{2} / \sqrt{6} \leq \sqrt{12-y^{2}}$. From this, $-\sqrt{6} \leq y \leq \sqrt{6}$. Therefore,

$$
D=\left\{\begin{array}{ll}
(x, y): \begin{array}{l}
-\sqrt{6} \leq y \leq \sqrt{6} \\
y^{2} / \sqrt{6} \leq x \leq \sqrt{12-y^{2}}
\end{array}
\end{array}\right\}
$$

2. We calculate the area of the region $D$ using formula (1). Transitioning from a double integral to a repeated integral, we get

$$
S=\iint_{D} d x d y=\int_{-\sqrt{6}}^{\sqrt{6}} d y \int_{y^{2} / \sqrt{6}}^{\sqrt{12-y^{2}}} d x
$$

3. Using the properties of the definite integral, we integrate sequentially:

$$
S=\int_{-\sqrt{6}}^{\sqrt{6}} d y \int_{y^{2} / \sqrt{6}}^{\sqrt{12-y^{2}}} d x=\int_{-\sqrt{6}}^{\sqrt{6}}\left(\sqrt{12-y^{2}}-\frac{y^{2}}{\sqrt{6}}\right) d y=3 \pi+2
$$

Answer. $S=(3 \pi+2)$ (units of length) $)^{2}$.

Conditions of the problems. Find the areas of figures bounded by the given lines.

1. $y=2 / x, y=4 e^{x}, y=2, y=4$.
2. $y=1 / x, y=2 e^{x}, y=1, y=2$.
3. $y=2 / x, y=2 \sqrt{x}, x=4$.
4. $x^{2}+y^{2}=2, y=-x^{2}(y \leq 0)$.
5. $y=\sqrt{x}, y=0, x=4$.
6. $x=2-y^{2}, x=-y$.
7. $y=\sin x, y=\cos x, x=0, x=\pi / 4$.
8. $y=2 x^{2}-1, y=x$.
9. $x=\sqrt{4-y^{2}}, x=y^{2} / 3$.
10. $y=\ln x, y=e / x, x=1$.

Answers. 1. $S=2$. 2. $S=1$. 3. $S=28 / 3-\ln 16$. 4. $S=\pi / 2+1 / 3$. 5. $S=16 / 3$. 6. $S=7 / 6$. 7. $S=\sqrt{2}-1$. 8. $S=27 / 24$. 9. $S=2 \pi / 3-\sqrt{3} / 6$. 10. $S=e-1$.

## 12.7. Calculation of Areas in Polar Coordinates

Problem Statement. Find the area of the region $D$ bounded by two circles

$$
\begin{gathered}
y^{2}+a_{1} y+b_{1} x+x^{2}=0, \quad y^{2}+a_{2} y+b_{2} x+x^{2}=0, \\
\left(a_{1}=0, a_{2}=0, \quad b_{1} b_{2}>0 \quad \text { or } \quad b_{1}=0, b_{2}=0, \quad a_{1} a_{2}>0\right)
\end{gathered}
$$

and two lines

$$
m_{1} y+k_{1} x=0, \quad\left(m_{1}^{2}+k_{1}^{2} \neq 0\right), \quad m_{2} y+k_{2} x=0, \quad\left(m_{2}^{2}+k_{2}^{2} \neq 0\right)
$$

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

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

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

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

and writing the equations of the boundaries in polar coordinates.

In this case, the region $D$ will transition to the region $D^{\prime}$, and the desired area will be equal to

$$
S=\iint_{D^{\prime}} \rho d \rho d \varphi
$$

2. Let's define the region $D^{\prime}$ in polar coordinates by inequalities:

$$
D^{\prime}=\left\{\begin{array}{cc}
(\rho, \varphi): & \varphi_{1} \leq \varphi \leq \varphi_{2} \\
\rho_{1}(\varphi) \leq \rho \leq \rho_{2}(\varphi)
\end{array}\right\}
$$

3. Transition from the double integral to a repeated integral:

$$
S=\int_{\varphi_{1}}^{\varphi_{2}} d \varphi \int_{\rho_{1}(\varphi)}^{\rho_{2}(\varphi)} \rho d \rho
$$

and compute it using the properties of the definite integral.

Write the answer, not forgetting the units of measurement.