# Fubini's Theorem

### Definition

This theorem implies that a double integral over a rectangle region can be always calculated from two iterated integrals. If $z = f(x, y)$ is integrable over $D = \{ (x, y) \vert a \leq x \leq b, c \leq y \leq d \}$, the double integral over $D$ is \begin{aligned} \iint_D f(x, y)dA = \int_a^b \int_c^d f(x, y)dydx = \int_c^d \int_a^b f(x, y)dxdy \end{aligned}

If some variables are included in the interval of integration, the strong Fubini’s theorem is a workaround, which switches the order of integration.

### Definite Integral to Double Integral

The definite integral is about the limit of the area surrounded by a closed curve, while the double integral is about the limit of the volume surrounded by the closed curve.

#### Case 1. Area Surrounded by One Curve and Two Lines

As shown in the right figure, the area $S$ surrounded by the curve $f(x)$ and two lines $x = x_1$ and $x = x_2$ is \begin{aligned} S = \int_{x_1}^{x_2} f(x) dx = \int_{x_1}^{x_2} \int_0^{f(x)} dy dx \end{aligned}

#### Case 2. Area Surrounded by Two Curves and Two Lines

As shown in the right figure, the area $S$ surrounded by the curves $f(x)$ and $g(x)$, and two lines $x = x_1$ and $x = x_2$ is \begin{aligned} S = \int_{x_1}^{x_2} f(x) - g(x) dx = \int_{x_1}^{x_2} \int_{g(x)}^{f(x)} dy dx \end{aligned}

Note that the order of the interval $[g(x), f(x)]$ can be changed depending on the graphs of $f(x)$ and $g(x)$.

### Choosing Proper Axis

This is based on the fact that the area is the same whether with respect to the $x$-axis or $y$-axis.

#### Case 1. Area Below a Line

With respect to each axis, the area surrounded by the line $y=x$, $x = 1$, and $x$-axis can be obtained as follows. \begin{aligned} \begin{cases} \; S_x = \displaystyle \int_{0}^{1} x dx = \displaystyle \int_{0}^{1} \int_0^{x} dy dx & (x \text{-axis}) \\\\ \; S_y = \displaystyle \int_{0}^{1} (1 - y) dy = \displaystyle \int_{0}^{1} \int_y^{1} dx dy & (y \text{-axis}) \end{cases} \end{aligned}

#### Case 2. Area Below a Parabola

With respect to each axis, the area surrounded by the line $y=x^2$, $x = 1$, and $x$-axis can be obtained as follows. \begin{aligned} \begin{cases} \; S_x = \displaystyle \int_{0}^{1} x^2 dx = \displaystyle \int_{0}^{1} \int_0^{x^2} dy dx & (x \text{-axis}) \\\\ \; S_y = \displaystyle \int_{0}^{1} (1 - \sqrt{y}) dy = \displaystyle \int_{0}^{1} \int_{\sqrt{y}}^{1} dx dy & (y \text{-axis}) \end{cases} \end{aligned}

### Inequalities Representing the Area

To represent the area of a double integral as inequalities, use the relation with the close integral variable first. \begin{aligned} \displaystyle \int_{0}^{1} \int_0^{x^2} dy dx \quad (x \text{-axis}) &\implies D = \{ (x, y) \; \vert \; 0 \leq x \leq 1, \; 0 \leq y \leq x^2 \} \\\\ \displaystyle \int_{0}^{2} \int_{y^2}^{1} dx dy \quad (y \text{-axis}) &\implies D = \{ (y, x) \; \vert \; 0 \leq y \leq 2, \; y^2 \leq x \leq 1 \} \end{aligned}

### How to Switch the Order of Integration

When a double integral is impossible, changing the order of integration is a workaround.

#### Example 1

\begin{aligned} \displaystyle \int_{0}^{1} \int_0^{x^2} dy dx \quad (x \text{-axis}) &\implies D_x = \{ (x, y) \; \vert \; 0 \leq x \leq 1, \; 0 \leq y \leq x^2 \} \end{aligned}

After getting the inequalities, switch the base axis from the $x$-axis to the $y$-axis.

Then, update $D_x$ to $D_y$, and reorder the integration. \begin{aligned} D_y = \{ (y, x) \; \vert \; 0 \leq y \leq 1, \; \sqrt{y} \leq x \leq 1 \} &\implies \displaystyle \int_{0}^{1} \int_{\sqrt{y}}^{1} dx dy \quad (y \text{-axis}) \end{aligned}

#### Example 2

\begin{aligned} \displaystyle \int_{0}^{1} \int_{\frac{y}{2}}^{\frac{1}{2}} dx dy \quad (y \text{-axis}) &\implies D_y = \{ (y, x) \; \vert \; 0 \leq y \leq 1, \; \cfrac{y}{2} \leq x \leq \cfrac{1}{2} \} \end{aligned}

After getting the inequalities, switch the base axis from the $y$-axis to the $x$-axis.

Then, update $D_y$ to $D_x$, and reorder the integration. \begin{aligned} D_x = \{ (x, y) \; \vert \; 0 \leq x \leq \cfrac{1}{2}, \; 0 \leq y \leq 2x \} &\implies \displaystyle \int_{0}^{\frac{1}{2}} \int_{0}^{2x} dy dx \quad (x \text{-axis}) \end{aligned}

#### Example 3

\begin{aligned} \displaystyle \int_{0}^{1} \int_{\sqrt{x}}^{1} dy dx \quad (x \text{-axis}) &\implies D_x = \{ (x, y) \; \vert \; 0 \leq x \leq 1, \; \sqrt{x} \leq y \leq 1 \} \end{aligned}

After getting the inequalities, switch the base axis from the $x$-axis to the $y$-axis.

Then, update $D_x$ to $D_y$, and reorder the integration. \begin{aligned} D_y = \{ (y, x) \; \vert \; 0 \leq y \leq 1, \; 0 \leq x \leq y^2 \} &\implies \displaystyle \int_{0}^{1} \int_{0}^{y^2} dx dy \quad (y \text{-axis}) \end{aligned}

### Reordering Techniques

#### Case 1. Only The Lower Bound of the Outer Integral is $0$ (Counterclockwise Rotation)

In this case, the form of the double integral looks like $\displaystyle \int_0^a \int_{f(x)}^{f(a)} f(x,y) dy dx$. For example, the following reordering is allowed. \begin{aligned} \int_{0}^{1} \int_{\sqrt{y}}^{\color{red}{1}} dx dy = \int_{\color{blue}{0}}^{\color{red}{1}} \int_{\color{blue}{0}}^{x^2} dy dx \end{aligned}

The technique is as follows.

• The outer integral
• The lower bound is $0$.
• The upper bound is the same as the upper bound of the inner integral of the original double integral.
• The inner integral
• The lower bound is $0$.
• To determine the upper bound, set the lower bound of the inner integral of the original double integral to another variable, say, $x$. In this example, $\sqrt{y} = x$. Then, the upper bound is $y$, which is $x^2$.

Similar other examples are as follows.

• $\displaystyle \int_{0}^{1} \int_{y}^{1} dx dy = \int_{0}^{1} \int_{0}^{x} dy dx$,
• $\displaystyle \int_{0}^{\pi} \int_{y}^{\pi} dx dy = \int_{0}^{\pi} \int_{0}^{x} dy dx$,
• $\displaystyle \int_{0}^{\sqrt{\pi}} \int_{y}^{\sqrt{\pi}} dx dy = \int_{0}^{\sqrt{\pi}} \int_{0}^{x} dy dx$,
• $\displaystyle \int_{0}^{1} \int_{3x}^{3} dy dx = \int_{0}^{3} \int_{0}^{\frac{y}{3}} dx dy$,
• $\displaystyle \int_{0}^{2} \int_{\frac{y}{2}}^{1} dx dy = \int_{0}^{1} \int_{0}^{2x} dy dx$,
• $\displaystyle \int_{0}^{4} \int_{\frac{x}{2}}^{2} dy dx = \int_{0}^{2} \int_{0}^{2y} dx dy$,
• $\displaystyle \int_{0}^{2} \int_{y^2}^{4} dx dy = \int_{0}^{4} \int_{0}^{\sqrt{x}} dy dx$,
• $\displaystyle \int_{0}^{1} \int_{y^\frac{1}{3}}^{1} dx dy = \int_{0}^{1} \int_{0}^{x^3} dy dx$,
• $\displaystyle \int_{0}^{1} \int_{\sin^{-1}y}^{\frac{\pi}{2}} dx dy = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sin x} dy dx$.

#### Case 2. The Upper Bound of the Inner Integral is a Variable

For example, the following reordering is allowed. \begin{aligned} \int_{0}^{\color{red}{1}} \int_{x^2}^{\color{red}{x}} dy dx = \int_{\color{blue}{0}}^{\color{red}{1}} \int_{y}^{\sqrt{y}} dx dy \end{aligned}

The technique is as follows.

• The outer integral
• The lower bound is $0$.
• Let $a$ be the upper bound of the outer integral of the original double integral. Also, let a function $f$ be the upper bound of the inner integral of the original double integral since it is not a constant. Then, the upper bound is $f(a)$.
• The inner integral
• To determine the lower bound, set the upper bound of the inner integral of the original double integral to another variable, say, $y$. In this example, $x = y$. Then, the lower bound is $x$, which is $y$.
• To determine the upper bound, set the lower bound of the inner integral of the original double integral to another variable, say, $y$. In this example, $x^2 = y$. Then, the upper bound is $x$, which is $\sqrt{y}$.

Similar other examples are as follows.

• $\displaystyle \int_{0}^{2} \int_{x^2}^{2x} dy dx = \int_{0}^{4} \int_{\frac{y}{2}}^{\sqrt{y}} dx dy$,
• $\displaystyle \int_{0}^{8} \int_{\frac{y}{4}}^{\sqrt[3]{y}} dx dy = \int_{0}^{2} \int_{x^3}^{4x} dy dx$.

#### Case 3. The Upper Bound of the Inner Integral is $\cos$ (Not Passing Through the Origin)

For example, the following reordering is allowed. \begin{aligned} \int_{\color{red}{0}}^{\frac{\pi}{2}} \int_{0}^{\color{red}{a \cos y}} dx dy = \int_{\color{blue}{0}}^{\color{red}{a}} \int_{\color{blue}{0}}^{\cos^{-1}\frac{x}{a}} dy dx \end{aligned}

The technique is as follows.

• The outer integral
• The lower bound is $0$.
• Let $a$ be the lower bound of the outer integral of the original double integral. Also, let a function $f$ be the upper bound of the inner integral of the original double integral since it is not a constant. Then, the upper bound is $f(a)$.
• The inner integral
• The lower bound is $0$.
• To determine the upper bound, set the upper bound of the inner integral of the original double integral to another variable, say, $x$. In this example, $a \cos y = x$. Then, the upper bound is $y$, which is $\cos^{-1}(x/a)$.

Similar other examples are as follows.

• $\displaystyle \int_{0}^{1} \int_{0}^{\cos^{-1} y} dx dy = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos x} dy dx$,
• $\displaystyle \int_{0}^{2} \int_{0}^{4 - x^2} dy dx = \int_{0}^{4} \int_{0}^{\sqrt{4 - y}} dx dy$.

### Application

#### Example 1. Calculate

\begin{aligned} \int_{0}^{1} \int_{x}^{1} (1 + y^2)^{\frac{5}{2}} dy dx \end{aligned}

After getting the inequalities, switch the base axis from the $x$-axis to the $y$-axis. Then, update $D_x$ to $D_y$, and reorder the integration. \begin{aligned} &\int_{0}^{1} \int_{x}^{1} (1 + y^2)^{\frac{5}{2}} dy dx \\ \implies &D_x = \{ (x, y) \; \vert \; 0 \leq x \leq 1, \; x \leq y \leq 1 \} \\ \implies &D_y = \{ (y, x) \; \vert \; 0 \leq y \leq 1, \; 0 \leq x \leq y \} \\ \implies &\int_{0}^{1} \int_{0}^{y} (1 + y^2)^{\frac{5}{2}} dx dy \end{aligned}

Now, this integral can be calculated as follows. \begin{aligned} \int_{0}^{1} \int_{0}^{y} (1 + y^2)^{\frac{5}{2}} dx dy &= \int_{0}^{1} y (1 + y^2)^{\frac{5}{2}} dx dy = \left[ \cfrac{1}{7} (1 + y^2)^{\frac{7}{2}} \right]_0^1 = \cfrac{8\sqrt{2} - 1}{7} \end{aligned}

#### Example 2. Calculate

\begin{aligned} \int_{0}^{\pi} \int_{y}^{\pi} \cfrac{\cos x}{x} dx dy \end{aligned}

From the formula, \begin{aligned} \int_{0}^{\pi} \int_{y}^{\pi} \cfrac{\cos x}{x} dx dy = \int_{0}^{\pi} \int_{0}^{x} \cfrac{\cos x}{x} dy dx = \int_{0}^{\pi} \cfrac{\cos x}{x} x dx = \left[ \sin x \right]_0^{\pi} = 0 \end{aligned}

#### Example 3. Calculate

\begin{aligned} \int_{0}^{1} \int_{0}^{x} x e^{x^2 - y^2} dy dx + \int_{1}^{2} \int_{x - 1}^{1} x e^{x^2 - y^2} dy dx \end{aligned}

After drawing each inequality from the two integrals, it can be noticed that two regions can be merged. \begin{aligned} \int_{0}^{1} \int_{y}^{y + 1} x e^{x^2 - y^2} dx dy \end{aligned}

Now, this integral can be directly calculated as follows. \begin{aligned} \int_{0}^{1} \int_{y}^{y + 1} x e^{x^2 - y^2} dx dy = \int_{0}^{1} \cfrac{e^{2y+1} - 1}{2} dy = \cfrac{e^3 - e - 2}{4} \end{aligned}

#### Example 4. Calculate

\begin{aligned} \int_{0}^{\infty} \cfrac{\tan^{-1} \pi x - \tan^{-1} x }{x} dx \end{aligned}

Using the following known integral result, this can be changed to a double integral. \begin{aligned} &\int \cfrac{1}{1 + x^2} = \tan^{-1} x \\\\ \implies &\int_{0}^{\infty} \cfrac{\tan^{-1} \pi x - \tan^{-1} x }{x} dx = \int_{0}^{\infty} \cfrac{\left[ \tan^{-1} xy \right]_1^{\pi}}{x} dx = \int_{0}^{\infty} \int_{1}^{\pi} \cfrac{1}{1 + x^2 y^2} dy dx \end{aligned}

Then, from the definition of Fubini’s theorem, this can be calculated as follows. \begin{aligned} \int_{0}^{\infty} \int_{1}^{\pi} \cfrac{1}{1 + x^2 y^2} dy dx &= \int_{1}^{\pi} \int_{0}^{\infty} \cfrac{1}{1 + x^2 y^2} dx dy = \int_{1}^{\pi} \left[ \cfrac{\tan^{-1} xy}{y} \right]_0^{\infty} dy \\ &= \cfrac{\pi}{2} \int_{1}^{\pi} \cfrac{1}{y} \; dy = \cfrac{\pi}{2} \left[ \ln y \right]_1^{\pi} = \cfrac{\pi}{2} \ln \pi \end{aligned}