Example. Evaluate the double integral

$$
\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y
$$

where the region $D$ is bounded by the lines $x=1, y=x^{3}$ and $y=-\sqrt{x}$.
Solution.

1. Let's define the region $D$ by inequalities. It is obvious that $-\sqrt{x} \leq x^{3}$. Therefore, $-\sqrt{x} \leq y \leq x^{3}$. Since $x$ appears under the square root, $x \geq 0$. For $x$, the possible inequalities are $0 \leq x \leq 1$ or $1 \leq x$. In the second case, the region is unbounded, which is unacceptable.

Thus,

$$
D=\left\{\begin{array}{cc}
(x, y): & 0 \leq x \leq 1 \\
-\sqrt{x} \leq y \leq x^{3}
\end{array}\right\}
$$

2. We transition from the double integral to the iterated integral:

$$
\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y=\int_{0}^{1} d x \int_{-\sqrt{x}}^{x^{3}}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d y
$$

3. Using the properties of the definite integral, we sequentially integrate first with respect to $y$ (considering $x$ as a constant), and then with respect to $x$:

$$
\begin{aligned}
& \int_{0}^{1} d x \int_{-\sqrt{x}}^{x^{3}}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d y= \\
& =\int_{0}^{1}\left[54 x^{2} \int_{-\sqrt{x}}^{x^{3}} y^{2} d y+150 x^{4} \int_{-\sqrt{x}}^{x^{3}} y^{4} d y\right] d x= \\
& =\int_{0}^{1}\left[\left.54 x^{2} \frac{y^{3}}{3}\right|_{-\sqrt{x}} ^{x^{3}}+\left.150 x^{4} \frac{y^{5}}{5}\right|_{-\sqrt{x}} ^{x^{3}}\right] d x= \\
& =\int_{0}^{1}\left[18 x^{11}+18 x^{7 / 2}+30 x^{19}+30 x^{13 / 2}\right] d x=11
\end{aligned}
$$

Answer. $\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y=11$.

Conditions of the Problems. Calculate the double integrals over the regions $D$ bounded by the given lines.

1. $\iint_{D}(2 x-y) d x d y, \quad y=x^{2}, y=\sqrt{x}$.
2. $\iint_{D}(x-y) d x d y, \quad y=2-x^{2}, y=2 x-1$.
3. $\quad \iint_{D}(y \ln x) d x d y, \quad y=\frac{1}{x}, y=\sqrt{x}, x=2$.
4. $\quad \iint_{D}(\cos 2 x+\sin y) d x d y, \quad y=\frac{\pi}{4}-x, y=0, x=0$.
5. $\quad \iint_{D} \sin (x+y) d x d y, \quad y=x, y=\frac{\pi}{2}, x=0$.
6. $\iint_{D} \frac{x^{2}}{y^{2}} d x d y, \quad y=\frac{1}{x}, y=x, x=2$.
7. $\iint_{D}\left(x^{2}+y\right) d x d y, \quad y=x^{2}, y=\sqrt{x}$.
8. $\quad \iint_{D}(2 x-y) d x d y, \quad y=x^{2}, y=x, x=1, x=2$.
9. $\quad \int_{D} \sqrt{1-x^{2}-y^{2}} d x d y, \quad y=\sqrt{1-x^{2}}, \quad y=0, x=0$.
10. $\iint_{D} x^{2} y^{2} \sqrt{1-x^{3}-y^{3}} d x d y, \quad y=\sqrt[3]{1-x^{3}}, \quad y=0, \quad x=0$.

Answers. 1. $I=1 / 10 . \quad 2 . \quad I=64 / 15 . \quad$ 3. $I=5(2 \ln 2-1) / 8$.

4. $I=(\pi+1-2 \sqrt{2}) / 4 . \quad$ 5. $I=1 . \quad 6 . \quad I=9 / 4 . \quad$ 7. $I=33 / 140$.
5. $I=9 / 10$. $\quad 9 . \quad I=\pi / 6$. $\quad 10 . \quad I=4 / 135$.

## 12.3. Double Integral

## in Polar Coordinates

Problem Statement. Calculate the double integral

$$
\iint_{D} f(x, y) d x d y
$$

where the region $D$ is bounded by two circles

$$
y^{2}+a_{1} y+b_{1} x+x^{2}=0, \quad y^{2}+a_{2} y+b_{2} x+x^{2}=0
$$

$$
\left(a_{1}=0, a_{2}=0, \quad b_{1} b_{2}>0 \quad \text { or } \quad b_{1}=0, b_{2}=0, \quad a_{1} a_{2}>0\right)
$$

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.

1. Define the region $D$ by inequalities in the Cartesian coordinate system.

Notice that the circles $y^{2}+a_{1} y+b_{1} x+x^{2}=0$ and $y^{2}+a_{2} y+b_{2} x+x^{2}=0$ pass through the origin and their centers are located on the $O X$ axis (when $a_{1}=0, a_{2}=0$) or on the $O Y$ axis (when $b_{1}=0, b_{2}=0$) on the same side of the origin (since $b_{1} b_{2}>0$ or $a_{1} a_{2}>0$). Therefore, the circle with the smaller radius is located inside the other. Suppose, for example, this is the circle $y^{2}+a_{1} y+b_{1} x+x^{2}=0$. The region $D$ is between the circles, so the coordinates of the points in the region $D$ satisfy the inequalities

$$
y^{2}+a_{1} y+b_{1} x+x^{2} \geq 0, \quad y^{2}+a_{2} y+b_{2} x+x^{2} \leq 0
$$

The lines $m_{1} y+k_{1} x=0$ and $m_{2} y+k_{2} x=0$ pass through the origin. The region $D$ is located between them. Considering in which half-plane the circles and, consequently, the region $D$ are located, we determine which of the following pairs of inequalities the coordinates of the points in the region $D$ satisfy:

$$
\begin{array}{ll}
m_{1} y+k_{1} x \geq 0, & m_{2} y+k_{2} x \geq 0 \\
m_{1} y+k_{1} x \leq 0, & m_{2} y+k_{2} x \geq 0 \\
m_{1} y+k_{1} x \geq 0, & m_{2} y+k_{2} x \leq 0 \\
m_{1} y+k_{1} x \leq 0, & m_{2} y+k_{2} x \leq 0
\end{array}
$$

2. Since the region $D$ is bounded by circles and lines passing through the origin, it is easier to solve the given 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 integral is defined by the formula

$$
\iint_{D} f(x, y) d x d y=\iint_{D^{\prime}} f(\varrho \cos \varphi, \varrho \sin \varphi) \varrho d \varrho d \varphi
$$

3. To find the region $D^{\prime}$, we replace $x$ with $\varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the inequalities defining the region $D$. Then we solve the obtained inequalities for $\varrho$ and $\varphi$. Thus, we get

$$
D^{\prime}=\left\{\begin{array}{rlrl} 
& =\varphi_{1} & \leq \varphi \leq \varphi_{2} \\
\varrho_{1}(\varphi) & \leq \varrho \leq \varrho_{2}(\varphi)
\end{array}\right\}
$$

4. We transition from the double integral to the iterated integral:

$$
S=\int_{\varphi_{1}}^{\varphi_{2}} d \varphi \int_{\varrho_{1}(\varphi)}^{\varrho_{2}(\varphi)} f(\varrho \cos \varphi, \varrho \