Example. Change the order of integration

$$
I=\int_{0}^{1} d y \int_{0}^{\sqrt{y}} f(x, y) d x+\int_{1}^{\sqrt{2}} d y \int_{0}^{\sqrt{2-y^{2}}} f(x, y) d x
$$
Solution.

1. The region of integration consists of two regions $D_{1}$ and $D_{2}$. Define them using inequalities

$$
\begin{gathered}
D_{1}=\left\{\begin{array}{cc}
(x, y): \begin{array}{c}
0 \leq y \leq 1 \\
0 \leq x \leq \sqrt{y}
\end{array}
\end{array}\right\} \\
D_{2}=\left\{\begin{array}{cc}
(x, y): & 1 \leq y \leq \sqrt{2} \\
0 \leq x \leq \sqrt{2-y^{2}}
\end{array}\right\}
\end{gathered}
$$

2. Solve the systems of inequalities defining regions $D_{1}$ and $D_{2}$ with respect to $y$ and obtain

$$
x^{2} \leq y \leq 1, \quad 1 \leq y \leq \sqrt{2-x^{2}}
$$

3. Determine the boundaries of variation for $x$ by solving the inequalities

$$
x^{2} \leq 1, \quad 1 \leq \sqrt{2-x^{2}}
$$

Considering that $x \geq 0$, in both cases we get $0 \leq x \leq 1$.

4. The regions $D_{1}$ and $D_{2}$ can be represented as

$$
\begin{gathered}
D_{1}=\left\{(x, y): \begin{array}{c}
0 \leq x \leq 1 \\
x^{2} \leq y \leq 1
\end{array}\right\} \\
D_{2}=\left\{(x, y): \begin{array}{c}
0 \leq x \leq 1 \\
1 \leq y \leq \sqrt{2-x^{2}}
\end{array}\right\}
\end{gathered}
$$

5. Write the integrals $I$ with the changed order of integration:

$$
I=\int_{0}^{1} d x \int_{x^{2}}^{1} f(x, y) d y+\int_{0}^{1} d x \int_{1}^{\sqrt{2-x^{2}}} f(x, y) d y
$$

6. Using the linearity and additivity of integrals, we get

$$
\begin{aligned}
I= & \int_{0}^{1} d x\left\{\int_{x^{2}}^{1} f(x, y) d y+\int_{1}^{\sqrt{2-x^{2}}} f(x, y) d y\right\}=\int_{0}^{1} d x \int_{x^{2}}^{\sqrt{2-x^{2}}} f(x, y) d y \\
& \text { Answer. } I=\int_{0}^{1} d x \int_{x^{2}}^{\sqrt{2-x^{2}}} f(x, y) d y
\end{aligned}
$$

Conditions of the Problem. Change the order of integration.

1. $\int_{0}^{1} d x \int_{1}^{2^{x}} f(x, y) d y+\int_{1}^{2} d x \int_{1}^{2 / x} f(x, y) d y$.
2. $\int_{1 / 4}^{1} d y \int_{1 / y}^{4} f(x, y) d x+\int_{1}^{2} d y \int_{y^{2}}^{4} f(x, y) d y$.
3. $\int_{-6}^{-3} d y \int_{0}^{\sqrt{36-y^{2}}} f(x, y) d x+\int_{-3}^{0} d y \int_{0}^{\sqrt{-y^{2}-12 y}} f(x, y) d y$.
4. $\int_{0}^{16} d y \int_{-y / 4}^{0} f(x, y) d x+\int_{16}^{32} d y \int_{-\sqrt{32-y}}^{0} f(x, y) d x$.
5. $\int_{0}^{\sqrt{6}} d x \int_{0}^{\sqrt[4]{6 x^{2}}} f(x, y) d y+\int_{\sqrt{6}}^{2 \sqrt{3}} d x \int_{0}^{\sqrt{12-x^{2}}} f(x, y) d y$.
6. $\int_{0}^{1} d y \int_{-y^{2}}^{0} f(x, y) d x+\int_{1}^{2} d y \int_{-\sqrt{2 y-y^{2}}}^{0} f(x, y) d x$.
7. $\int_{0}^{1} d x \int_{0}^{\sqrt{x}} f(x, y) d y+\int_{1}^{2} d x \int_{0}^{2-x} f(x, y) d y$.
8. $\int_{-1}^{0} d y \int_{0}^{\sqrt{1-y^{2}}} f(x, y) d x+\int_{0}^{1} d y \int_{0}^{\sqrt{1-y}} f(x, y) d x$.
9. $\int_{0}^{1} d y \int_{1}^{e^{y}} f(x, y) d x+\int_{1}^{e} d y \int_{1}^{e / y} f(x, y) d x$.
10. $\int_{-2}^{-1} d x \int_{-2-x}^{0} f(x, y) d y+\int_{-1}^{0} d x \int_{x^{3}}^{0} f(x, y) d y$.

Answers.

1. $\int_{1}^{2} d y \int_{\log _{2} y}^{2 / y} f(x, y) d x$.
2. $\int_{1}^{4} d x \int_{1 / x}^{\sqrt{x}} f(x, y) d y$.
3. $\int_{0}^{3 \sqrt{3}} d x \int_{-\sqrt{36-x^{2}}}^{-6+\sqrt{36-x^{2}}} f(x, y) d y$.
4. $\int_{-4}^{0} d x \int_{-4 x}^{32-x^{2}} f(x, y) d y$.
5. $\int_{0}^{\sqrt{6}} d y \int_{y^{2} / \sqrt{6}}^{\sqrt{12-y^{2}}} f(x, y) d x$.
6. $\int_{-1}^{0} d x \int_{\sqrt{-x}}^{1+\sqrt{1-x^{2}}} f(x, y) d y$.
7. $\int_{0}^{1} d y \int_{y^{2}}^{2-y} f(x, y) d x$.
8. $\int_{0}^{1} d x \int_{-\sqrt{1-x^{2}}}^{1-x^{2}} f(x, y) d y$.
9. $\int_{1}^{e} d x \int_{\ln x}^{e / x} f(x, y) d y$.
10. $\int_{-1}^{0} d y \int_{-2-y}^{\sqrt[3]{y}} f(x, y) d x$.

## 12.2. Double Integral

## in Cartesian Coordinates

Problem Statement. Compute the double integral

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

where the region $D$ is 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.

1. Define the region $D$ using inequalities. Determine which of the inequalities

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

For example, let the inequalities be $f_{1}(x, y) \leq 0$ and $f_{2}(x, y) \leq 0$. 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

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-291.jpg?height=109&width=594&top_left_y=572&top_left_x=322)

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

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

or

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

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

Write the answer.

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