Fubini's Theorem

• Definition
• Definite Integral to Double Integral
• Choosing Proper Axis
• Inequalities Representing the Area
• How to Switch the Order of Integration
• Reordering Techniques
• Application

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.

• First, find the inequalities representing the area.
• Graph these inequalities.
• By switching the base axis, change the inequalities.
• Reorder the integration.

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

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

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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

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

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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

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

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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

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

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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. $$\begin{aligned} \int_{0}^{\infty} \int_{1}^{\pi} \cfrac{1}{1 + x^2 y^2} dy dx &= \int_{1}^{\pi} \int_{0}^{\infty} \cfrac{1}{1 + x^2 y^2} dx dy = \int_{1}^{\pi} \left[ \cfrac{\tan^{-1} xy}{y} \right]_0^{\infty} dy \\ &= \cfrac{\pi}{2} \int_{1}^{\pi} \cfrac{1}{y} \; dy = \cfrac{\pi}{2} \left[ \ln y \right]_1^{\pi} = \cfrac{\pi}{2} \ln \pi \end{aligned}$$