Example 4.3.1

1.

Polar coordinate (in 2-dimension).

2.

Cylindrical coordinate.

We have

3.

Spherical coordinate.

We have

Example 4.3.2 $D=\{(x_1,x_2)|1\le x_1^2-x_2^2\le 3,1\le x_1{x}_2\le 2,x_1,x_2\ge 0\}$, seek $I={\int }_D(x_1^2+x_2^2)\text{d}{x}_1\text{d}{x}_2$. Let

which is diffeomorphism of $C^1$, then $\tilde{D}=\{(y_1,y_2)|1\le y_1\le 3,2\le y_2\le 4\}$, so $$\begin{array}{rl}I& ={\int }_{\tilde{D}}\sqrt{y_1^2+y_2^2}|det\frac{\partial (x_1,x_2)}{\partial (y_1,y_2)}|\text{d}{y}_1\text{d}{y}_2={\int }_{\tilde{D}}\sqrt{y_1^2+y_2^2}{|det\frac{\partial (y_1,y_2)}{\partial (x_1,x_2)}|}^{-1}\text{d}{y}_1\text{d}{y}_2\\ & ={\int }_{\tilde{D}}\sqrt{y_1^2+y_2^2}{|det\left(\begin{array}{cc}2x_1& -2x_2\\ 2x_2& 2x_1\end{array}\right)|}^{-1}\text{d}{y}_1\text{d}{y}_2=\frac{1}{4}{\int }_{\tilde{D}}\text{d}{y}_1\text{d}{y}_2=1\end{array}$$

Example 4.3.3 $D=\{(x,y)|x^2+y^2\le x+y\}$, seek ${\int }_D\frac{x+y}{x^2+y^2}\text{d}x\text{d}y$.

Take a polar transformation $(x,y)\to (r,\theta )$, we have $\tilde{D}=\{(r,\theta )|0\le r\le \sin \theta +\cos \theta \}$ where $\theta \in [-\frac{\pi }{4},\frac{3\pi }{4}]$, then $$\begin{array}{rl}{\int }_D\frac{x+y}{x^2+y^2}\text{d}x\text{d}y& ={\int }_{-\frac{\pi }{4}}^{\frac{3\pi }{4}}\left[{\int }_0^{\sin \theta +\cos \theta }\frac{r(\sin \theta +\cos \theta )}{r^2}r\text{d}r\right]\text{d}\theta \\ & ={\int }_{-\frac{\pi }{4}}^{\frac{3\pi }{4}}(\cos \theta +\sin \theta {)}^2\text{d}\theta =\pi \end{array}$$

It is not perfectly correct, since $\frac{x+y}{x^2+y^2}$ is unbounded near $(x,y)=(0,0)$. It’s a multiple improper integral.

Example 4.3.4 $D=\{(x_1,...,x_m)|x_1+\cdots +x_m\le a\wedge x_i\ge 0,\mathrm{\forall }i\}$, seek

Take a variable transformation

then we have $$\begin{array}{rl}I& ={\int }_{0\le y_1\le \cdots \le y_m\le a}f(y_m)|det\frac{\partial (x_1,...,x_m)}{\partial (y_1,...,y_m)}|\text{d}{y}_1\cdots \text{d}{y}_m={\int }_{0\le y_1\le \cdots \le y_m\le a}f(y_m)\text{d}{y}_1\cdots \text{d}{y}_m\\ & ={\int }_0^{a}f(y_m)[{\int }_{0\le y_1\le \cdots \le y_{m-1}\le y_m}\text{d}{y}_1\cdots \text{d}{y}_{m-1}]\text{d}{y}_m\end{array}$$

How to calculate the integral in $[\cdot ]$? Firstly,

Then, consider the order of $y_1,...,y_{m-1}$. For any arrangement $\{{\sigma }_1,...,{\sigma }_{m-1}\}$ of $\{1,...,m-1\}$,

The determinant of a permutation matrix is either $1$ or $-1$, so

Notice that $$\begin{array}{rl}{\int }_{y_1,...,y_{m-1}\in [0,y_m]}\text{d}{y}_1\cdots \text{d}{y}_{m-1}& =\sum _{\{{\sigma }_1,...,{\sigma }_{m-1}\}=\{1,...,m-1\}}{\int }_{0\le y_{{\sigma }_1}\le \cdots \le y_{{\sigma }_{m-1}}\le y_m}\text{d}{y}_{{\sigma }_1}\cdots \text{d}{y}_{{\sigma }_{m-1}}\\ & =(m-1)!{\int }_{0\le y_1\le \cdots \le y_{m-1}\le y_m}\text{d}{y}_1\cdots \text{d}{y}_{m-1}\end{array}$$

Therefore

Terminally

Example 4.3.5 Assuming $\mu \in {\mathbb{R}}^m$, $\mathrm{\Sigma }$ is a symmetric positive-definite matrix of $m^{\text{th}}$-order, prove that

According to spectral theorem, $\mathrm{\Sigma }$ is symmetric and positive-definite, so there exists an orthogonal matrix $Q$ such that

where ${\sigma }_k>0$. Then

where $A$ is invertible. Let $y=A^{-1}(x-\mu )$, then $x=Ay+\mu$ where $x$ and $y$ are diffeomorphism, hence

Notice that

therefore $$\begin{array}{rl}I& ={\int }_{{\mathbb{R}}^m}\frac{1}{\sqrt{(2\pi {)}^m(detAdet{A}^T)}}\exp (-\frac{1}{2}{y}^{T}y)|detA|\text{d}{y}_1\cdots \text{d}{y}_m\\ & ={\int }_{{\mathbb{R}}^m}\frac{1}{\sqrt{(2\pi {)}^m}}\exp [-\frac{1}{2}(y_1^2+\cdots +y_n^2)]\text{d}{y}_1\cdots \text{d}{y}_m\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{\text{e}}^{-\frac{y_1^2}{2}}\text{d}{y}_1\cdots {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{\text{e}}^{-\frac{y_n^2}{2}}\text{d}{y}_n\\ & ={\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{\text{e}}^{-\frac{t^2}{2}}\text{d}t\right)}^m\end{array}$$

How to calculate the last integral? Consider $$\begin{array}{rl}{\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{\text{e}}^{-\frac{t^2}{2}}\text{d}t\right)}^2& ={\int }_{{\mathbb{R}}^2}\frac{1}{2\pi }{\text{e}}^{-\frac{1}{2}(y_1^2+y_2^2)}\text{d}{y}_1\text{d}{y}_2\stackrel{\text{Polar coordinate}}{\to }\\ & ={\int }_0^{+\mathrm{\infty }}{\int }_0^{2\pi }\frac{1}{2\pi }{\text{e}}^{\frac{1}{2}{r}^2}r\text{d}r\text{d}\theta =1\end{array}$$

Terminally $I=1$. $◻$

Example 4.3.6 Assuming $\mathrm{\Omega }=\{x=(x_1,...,x_m)\in {\mathbb{R}}^m|x_m\ge 0,x_1^2+\cdots +x_m^2\le R^2\}$, its mass is evenly distributed, i.e. $\rho (x)=\rho =\text{const}$. Seek the center of mass of $\mathrm{\Omega }$.

According to the definition, we have

Firstly, $$\begin{array}{rl}{V}_m(R)& ={\int }_{x_1^2+\cdots +x_m^2\le {\mathbb{R}}^2}={\int }_{-R}^R\text{d}{x}_m{\int }_{x_1^2+\cdots +x_{m-1}^2\le R^2-x_m^2}\text{d}{x}_1\cdots \text{d}{x}_{m-1}\\ & ={\int }_{-R}^R{V}_{m-1}\left(\sqrt{R^2-x_m^2}\right)\text{d}{x}_m\end{array}$$

$V_1(R)=2R$. Assuming $V_m(R)=A_m{\text{R}}^m$ where $A_m$ has nothing to do with $R$ (proven by mathematical induction), we have

Therefore, $$\begin{array}{rl}{A}_m& =A_{m-1}{\int }_{-1}^1(1-t^2{)}^{\frac{m-1}{2}}\text{d}t\stackrel{x=t^2}{\to }{A}_{m-1}{\int }_0^1(1-x{)}^{\frac{m-1}{2}}{x}^{-\frac{1}{2}}\text{d}x\\ & =A_{m-1}B(\frac{1}{2},\frac{m+1}{2})=A_{m-1}\frac{\mathrm{\Gamma }(\frac{1}{2})\mathrm{\Gamma }(\frac{m+1}{2})}{\mathrm{\Gamma }(\frac{m+2}{2})}\\ A_m& =\frac{A_m}{A_{m-1}}\frac{A_{m-1}}{A_{m-2}}\cdots \frac{A_2}{A_1}{A}_1\\ & =\frac{\mathrm{\Gamma }(\frac{1}{2})\mathrm{\Gamma }(\frac{m+1}{2})}{\mathrm{\Gamma }(\frac{m+2}{2})}\frac{\mathrm{\Gamma }(\frac{1}{2})\mathrm{\Gamma }(\frac{m}{2})}{\mathrm{\Gamma }(\frac{m+1}{2})}\cdots \frac{\mathrm{\Gamma }(\frac{1}{2})\mathrm{\Gamma }(\frac{3}{2})}{\mathrm{\Gamma }(\frac{4}{2})}2=\frac{{\mathrm{\Gamma }}^m(\frac{1}{2})}{\mathrm{\Gamma }(\frac{m+2}{2})}\end{array}$$

According to symmetry,

As for $x_m$, we have $$\begin{array}{rl}{\int }_{\mathrm{\Omega }}{x}_m\text{d}{x}_1\cdots \text{d}{x}_m& ={\int }_0^R({\int }_{x_1^2+\cdots +x_{m-1}^2\le R^2-x_m^2}\text{d}{x}_1\cdots \text{d}{x}_{m-1})x_m\text{d}{x}_m\\ & ={\int }_0^R\frac{{\mathrm{\Gamma }}^{m-1}(\frac{1}{2})}{\mathrm{\Gamma }(\frac{m+1}{2})}(R^2-t^2{)}^{\frac{m-1}{2}}t\text{d}t\stackrel{t^2=R^{2}s}{\to }\\ & =\frac{{\mathrm{\Gamma }}^{m-1}(\frac{1}{2})}{2\mathrm{\Gamma }(\frac{m+1}{2})}{R}^{m+1}{\int }_0^1(1-s{)}^{\frac{m-1}{2}\text{d}s}\\ & =\frac{{\mathrm{\Gamma }}^{m-1}(\frac{1}{2})}{(m+1)\mathrm{\Gamma }(\frac{m+1}{2})}{R}^{m+1}{[-(1-s{)}^{\frac{m+1}{2}}]}_0^1=\frac{{\mathrm{\Gamma }}^{m-1}(\frac{1}{2})}{(m+1)\mathrm{\Gamma }(\frac{m+1}{2})}{R}^{m+1}\end{array}$$

Terminally,

Example 4.3.7 Select $n$ numbers at random in $[0,1]$ (independent and evenly distributed), seek the mean value (expectation) of the minimum value.

The probability density is $f(x)=1_{0\le x\le 1}$, the joint density of $n$ random numbers is

, then the expectation of the minimum value is $$\begin{array}{rl}{\bar{x}}_{min}& ={\int }_{{\mathbb{R}}^n}\underset{1\le i\le n}{min}{x}_i\cdot f(x_1)\cdots f(x_n)\text{d}{x}_1\cdots \text{d}{x}_n={\int }_{[0,1{]}^n}\underset{1\le i\le n}{min}{x}_i\cdot \text{d}{x}_1\cdots \text{d}{x}_n\\ & =\sum _{\{{\sigma }_1,...,{\sigma }_m\}=\{1,...,m\}}{\int }_{0\le x_{{\sigma }_1}\le \cdots \le x_{{\sigma }_m}\le 1}{x}_{{\sigma }_1}|det\frac{\partial (x_1,...,x_n)}{\partial (x_{{\sigma }_1},...,x_{{\sigma }_n})}|\text{d}{x}_{{\sigma }_1}\cdots \text{d}{x}_{{\sigma }_n}\\ & =\sum _{\{{\sigma }_1,...,{\sigma }_m\}=\{1,...,m\}}{\int }_0^1({\int }_{x_{{\sigma }_1}\le \cdots \le x_{{\sigma }_m}\le 1}\text{d}{x}_{{\sigma }_2}\cdots \text{d}{x}_{{\sigma }_n})x_{{\sigma }_1}\text{d}{x}_{{\sigma }_1}\\ & =\sum _{\{{\sigma }_1,...,{\sigma }_m\}=\{1,...,m\}}{\int }_0^1\frac{1}{(n-1)!}(1-x_{{\sigma }_1}{)}^{n-1}{x}_{{\sigma }_1}\text{d}{\sigma }_1\\ & =n!\cdot \frac{1}{(n-1)!}B(2,n)=\frac{1}{n+1}\end{array}$$

Example 4.3.8 Assuming the radius and density of a 3-dimension ball are respectively $R,\rho =\text{const}$, seek its gravitational force applied to a mass point outside the ball.

Construct the coordinate system where the center of the ball is put at origin and the mass point $m$ is put at $(0,0,a)$. For a point $\mathbf{x}$ in the ball, we have

Therefore, $$\begin{array}{rl}\mathbf{F}& ={\int }_{\mathrm{\Omega }}\text{d}\mathbf{F}(\mathbf{x})=Gm\rho {\int }_{x_1^2+x_2^2+x_3^2\le R^2}\frac{(x_1,x_2,x_3-a{)}^T\text{d}{x}_1\text{d}{x}_2{\text{d}}_3}{{(x_1^2+x_2^2+(x_3-a{)}^2)}^{\frac{3}{2}}}\\ & ={\mathbf{e}}_{3}Gm\rho {\int }_{-R}^R(x_3-a)\left[{\int }_{x_1^2+x_2^2\le R^2-x_3^2}\frac{\text{d}{x}_1\text{d}{x}_2}{{(x_1^2+x_2^2+(x_3-a{)}^2)}^{\frac{3}{2}}}\right]\text{d}{x}_3\\ & ={\mathbf{e}}_{3}Gm\rho {\int }_{-R}^R(t-a)\left[{\int }_0^{\sqrt{R^2-t^2}}{\int }_0^{2\pi }\frac{r\text{d}r\text{d}\theta }{{(r^2+(t-a{)}^2)}^{\frac{3}{2}}}\right]\text{d}t\\ & =-\frac{Gm\rho }{a^2}\frac{4\pi }{3}{R}^3{\mathbf{e}}_3\end{array}$$

Example 4.3.9 Kepler II law for centripetal (centrifugal) force field. The motion of a mass point on a plain $\mathbf{u}:[a,b]\to {\mathbb{R}}^2$ where $\mathbf{u}(t)=(x_1(t),x_2(t))$. Assuming $\mathrm{\Omega }$ is covered by a line connecting a planet to the sun over certain period of time $[a,b]$, let $(x_1,x_2)=s\mathbf{u}(t)$, we have

then the area of $\mathrm{\Omega }$ is

Kepler II:

By taking the derivative, we have

meaning ${\mathbf{u}}^{″}(t),\mathbf{u}(t)$ are linear dependent, i.e. the force field is centripetal (centrifugal).