重积分

二重积分

性质

线性、可加性、单调性、估值不等式、绝对值不等式

与一元定积分类似.

中值定理

设 \(D\) 是连通有界闭集，\(f(x,y)\in C(D)\)，则 \(\exists (\xi ,\eta)\in D\) 使得

\[ \iint_Df(x,y)d\sigma = f(\xi ,\eta)A_D \]

计算

矩形区域

若 \(f(x,y)\) 在 \([a,b]\times [c,d]\) 可积且连续，则有

\[ \iint_{[a,b]\times [c,d]} f(x,y)\text{d}\sigma = \int_{a}^{b}\text{d}x\int_{c}^{d}f(x,y)\text{d}y=\int_{c}^{d}\text{d}y\int_{a}^{b}f(x,y)\text{d}x \]

设 \(f\in C[a,b],g\in C[c,d]\)，则有

\[ \iint_{[a,b]\times [c,d]} f(x,y)\text{d}x\text{d}y = \int_a^bf(x)\text{d}x\cdot \int_c^dg(y)\text{d}y \]

\(x\) 型区域

\[\iint_Df(x,y)\text{d}x\text{d}y = \int_{a}^{b}\text{d}x\int_{y_1(x)}^{y_2(x)}f(x,y)\text{d}y\]

\(y\) 型区域

\[\begin{aligned}\iint_Df(x,y)\text{d}x\text{d}y = \int_{c}^{d}\text{d}y\int_{x_1(y)}^{x_2(y)}f(x,y)\text{d}x\end{aligned}\]

交换二次积分次序

• 将二次积分还原微二重积分；
• 由积分限确定积分区域 \(D\)，并按另一类型表示它；
• 化二重积分为另一次序的二次积分.

简化计算

用积分区域对称性和被积函数奇偶性简化积分计算.

若 \(D\) 关于 \(y\) 轴对称，记 \(D_1\) 为 \(D\) 中 \(x\geq0\) 部分，则

\[\iint_Df(x,y)\text{d}x\text{d}y = \begin{cases} 0,& f(-x,y) = -f(x,y)\\ \displaystyle 2\iint_{D_1}f(x,y)\text{d}x\text{d}y, &f(-x,y) = f(x,y) \end{cases}\]

换元

极坐标形式

使 \(x=r\cos \theta, y=r\sin \theta\),则

\[\iint_Df(x,y)\text{d}x\text{d}y=\iint_{D'}f(r\cos\theta,r\sin\theta)r\text{d}r\text{d}\theta\]

特别地，若 \(D'=\{(r,\theta)\mid \alpha\leq\theta\leq\beta,r_1(\theta)\leq r\leq r_2(\theta)\}\)，则有

\[\iint_{D'}f(r\cos\theta,r\sin\theta)r\text{d}r\text{d}\theta = \int_{\alpha}^{\beta}\text{d}\theta\int_{r_1(\theta)}^{r_2(\theta)}f(r\cos\theta,r\sin\theta)r\text{d}r\]

注意

由于 \(r\neq 0\)，显然这里的 \(D'\) 并不包括 \(z\) 轴，即 \(\mathbf{F}: (x,y) \mapsto (r,\theta)\) 并不是一对一的。

一般形式换元

设变换 \(\mathbf{F}:\begin{cases}x=x(u,v)\\y=y(u,v)\end{cases}\) 有连续偏导数，且满足 \(\begin{aligned}\mathbf{J} = \frac{\partial(x,y)}{\partial(u,v)}\neq 0\end{aligned}\)，又 \(f(x,y)\in C(D)\)，则

\[\iint_Df(x,y)\text{d}x\text{d}y = \iint_{D'}f(x(u,v),y(u,v))|\mathbf{J}|\text{d}u\text{d}v\]

其中 \(\mathbf{F}\) 将 \(D'\) 变为 \(D\).

三重积分

计算

长方体区域

设 \(f(x,y,z)\) 在 \([a,b]\times [c,d]\times [e,h]\) 可积.

1. 若 \(\forall (x,y)\in [a,b]\times [c,d]\)，存在首次积分 \(\displaystyle \mu (x,y) = \int_e^hf(x,y,z)\text{d}z\)，则

\[\iiint_{[a,b]\times [c,d]\times [e,h]}f(x,y,z)\text{d}V = \iint_{[a,b]\times [c,d]}\text{d}x\text{d}y\int_e^hf(x,y,z)\text{d}z\]

1. 若 \(\forall z\in [e,h]\)，存在二重积分 \(\displaystyle \mu(z) = \iint_{[a,b]\times [c,d]}f(x,y,z)\text{d}x\text{d}y\)，则

\[\iiint_{[a,b]\times [c,d]\times [e,h]}f(x,y,z)\text{d}V = \int_e^h\text{d}z\iint_{[a,b]\times [c,d]}f(x,y,z)\text{d}x\text{d}y\]

柱线法（坐标面投影法）

• \(\Omega = \{(x,y,z)\mid z_1(x,y)\leq z\leq z_2(x,y),(x,y)\in D\}\)

• \(D = \{(x,y)\mid a\leq x\leq b, y_1(x)\leq y\leq y_2(x)\}\)

\[\begin{aligned} \iiint_{\Omega}f(x,y,z)\text{d}x\text{d}y\text{d}z &= \iint_D\left(\int_{z_1(x,y)}^{z_2(x,y)}f(x,y,z)\text{d}z\right)\text{d}x\text{d}y\\ &=\iint_D\text{d}x\text{d}y\int_{z_1(x,y)}^{z_2(x,y)}f(x,y,z)\text{d}z\\ &=\int_a^b\text{d}x\int_{y_1(x)}^{y_2(x)}\text{d}y\int_{z_1(x,y)}^{z_2(x,y)}f(x,y,z)\text{d}z \end{aligned}\]

截面法（坐标轴投影法）

• \(\Omega = \{(x,y,z)\mid h_1\leq z\leq h_2,(x,y)\in D_2\}\)

\[\begin{aligned}\iiint_{\Omega}f(x,y,z)\text{d}x\text{d}y\text{d}z &= \int_{h_1}^{h_2}\left(\iint_{D_z}f(x,y,z)\text{d}x\text{d}y\right)\text{d}z\\ &=\int_{h_1}^{h_2}\text{d}z\iint_{D_z}f(x,y,z)\text{d}x\text{d}y \end{aligned}\]

换元

一般形式换元

设变换 \(\mathbf{T}:\begin{cases}x=x(u,v,w)\\ y=y(u,v,w)\\ z=z(u,v,w) \end{cases}\) 有连续偏导数，且满足 \(\begin{aligned}\mathbf{J} = \frac{\partial(x,y,z)}{\partial(u,v,w)}\neq 0\end{aligned}\)，又 \(f(x,y,z)\in C(\Omega)\)，则

\[\iiint_{\Omega}f(x,y,z)\text{d}x\text{d}y\text{d}z = \iiint_{\Omega'}f(x(u,v,w),y(u,v,w),z(u,v,w))|\mathbf{J}|\text{d}u\text{d}v\text{d}w\]

其中 \(\mathbf{T}\) 将 \(\Omega'\) 变为 \(\Omega\).

柱面坐标系

\(\begin{cases} x=r\cos\theta\\ y=r\sin\theta\\ z=z \end{cases},\quad \begin{aligned}\frac{\partial (x,y,z)}{\partial (r,\theta,z)}=r\end{aligned}\)

\[\iiint_{\Omega}f(x,y,z)\text{d}V=\iiint_{\Omega'}f(r\cos\theta,r\sin\theta,z)r\text{d}r\text{d}\theta\text{d}z\]

注意

同二重积分，由于 \(r\neq 0\)，显然这里的 \(\Omega'\) 并不包括 \(z\) 轴，即 \(\mathbf{T}: (x,y,z) \mapsto (r,\theta,z)\) 并不是一对一的。

球面坐标系

\(\begin{cases} x=\rho\sin\varphi\cos\theta\\ y=\rho\sin\varphi\sin\theta\\ z=\rho\cos\varphi \end{cases},\quad \begin{aligned}\frac{\partial (x,y,z)}{\partial (\rho,\varphi,\theta)}=\rho^2\sin\varphi\end{aligned}\)

\[\iiint_{\Omega}f(x,y,z)\text{d}V = \iiint_{\Omega'}f(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\rho^2\sin\varphi\text{d}\rho\text{d}\varphi\text{d}\theta\]

注

1. \(V_{椭球} = \begin{aligned}\frac{4\pi abc}{3}\end{aligned}\)

2. (轮换对称性) \(\begin{aligned} \iiint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\leq 1} z^2\text{d}V = \frac{4\pi abc^3}{15}\quad \iiint_{\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{x^2}{a^2}\leq 1} x^2\text{d}V = \frac{4\pi a^3bc}{15} \end{aligned}\)

3. 令 \(V:x^2+y^2+z^2\leq R^2\) 则

\[\iiint_Vx^2\text{d}V = \iiint_Vy^2\text{d}V = \iiint_Vz^2\text{d}V = \frac{1}{3}\iiint_V(x^2+y^2+z^2)\text{d}V = \frac{4\pi R^5}{15}\]

\(n\) 重积分

计算

长方体区域

设 \(f\) 在 \(\Omega=[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n]\) 上连续，则

\[\begin{aligned} &\int_{\Omega}f(x_1,x_2,\dots,x_n)\text{d}x_1\text{d}x_2\cdots \text{d}x_n\\ =& \int_{a_1}^{b_1}\text{d}x_1\int_{[a_2,b_2]\times\cdots\times [a_n,b_n]}f(x_1,x_2,\dots,x_n)\text{d}x_2\cdots \text{d}x_n\\ =& \int_{[a_1,b_1]\times\cdots\times [a_{n-1},b_{n-1}]}\text{d}x_1\cdots \text{d}x_{n-1}\int_{a_n}^{b_n}f(x_1,x_2,\dots,x_n)\text{d}x_n \end{aligned}\]

\(n\) 维单纯形

\(T_n=\{(x_1,x_2,\dots,x_n)\mid x_1+x_2+\cdots +x_n\leq h, x_i\geq 0(i=1,2,\dots,n)\}\) 的体积

\[V_n = \frac{h^n}{n!}\]

\(n\) 维球体积

提示

换元为多维球面坐标系推导

3维球4维球\(n\) 维球

\(T_3 = \{(x,y,z)\mid x^2+y^2+z^2\leq R^2\}\)

\(\begin{cases}x = \rho\sin\varphi\cos\theta\newline y = \rho\sin\varphi\sin\theta\newline z = \rho\cos\varphi \end{cases} \quad \textbf{J} = \begin{aligned}\frac{\partial (x,y,z)}{\partial (\rho,\varphi,\theta)} = \rho^2\sin\varphi\end{aligned}\)

\(\begin{aligned} \iiint_{\Omega}\text{d}V = \iiint_{\Omega'}|\textbf{J}|\text{d}\rho\text{d}\varphi\text{d}\theta = \int_0^{2\pi}\text{d}\theta\int_0^{\pi}\sin\varphi\text{d}\varphi\int_0^{R}\rho^2\text{d}\rho = \frac{4\pi R^3}{3} \end{aligned}\)

\(T_4 = \{(x_1,x_2,x_3,x_4)\mid x_1^2+x_2^2+x_3^2+x_4^2\leq R^2\}\)

\(\begin{cases}x_1 = \rho\sin\psi\sin\varphi\cos\theta\newline x_2 = \rho\sin\psi\sin\varphi\sin\theta\newline x_3 = \rho\sin\psi\cos\varphi\newline x_4 = \rho\cos\psi \end{cases} \quad \textbf{J} = \begin{aligned}\frac{\partial (x_1,x_2,x_3,x_4)}{\partial (\rho,\psi,\varphi,\theta)} = \rho^3\sin\psi^2\sin\varphi\end{aligned}\)

\(\begin{aligned} \int_{\Omega}\text{d}V = \int_{\Omega'}|\textbf{J}|\text{d}\rho\text{d}\varphi\text{d}\theta = \int_0^{2\pi}\text{d}\theta\int_0^{\pi}\sin^2\psi\text{d}\psi\int_0^{\pi}\sin\varphi\text{d}\varphi\int_0^{R}\rho^3\text{d}\rho = \frac{\pi^2 R^4}{2} \end{aligned}\)

\(T_n = \{(x_1,x_2,\dots,x_n)\mid x_1^2+x_2^2+\cdots+x_n^2\leq R^2\}\)

\(\begin{cases}x_1 = \rho\sin\theta_1\sin\theta_2\cdots\cos\theta_{n-1}\newline x_2 = \rho\sin\theta_1\sin\theta_2\cdots\sin\theta_{n-1}\newline \ \vdots\newline x_n = \rho\cos\theta_1 \end{cases} \displaystyle \textbf{J} = \frac{\partial (x_1,x_2,\dots,x_n)}{\partial (\rho,\theta_1,\theta_2,\dots,\theta_{n-1})} = \rho^n\sin^{n-1}\theta_1\cdots\sin\theta_{n-1}\)

\(\begin{aligned} &\int_{\Omega}\text{d}V = \int_{\Omega'}|\textbf{J}|\text{d}\rho\text{d}\theta_1\cdots\text{d}\theta_{n-1}\\ =& \int_0^{2\pi}\text{d}\theta_{n-1}\int_0^{\pi}\sin^{n-1}\theta_1\text{d}\theta_1\cdots\int_0^{\pi}\sin\theta_{n-2}\text{d}\theta_{n-2}\int_0^{R}\rho^{n-1}\text{d}\rho\newline =& \begin{cases}\begin{aligned} &2\pi\cdot 2\cdot \frac{\pi}{2}\cdots\frac{2(n-1)!!}{n!!}\cdot \frac{\pi}{2}\cdot \frac{R^{n}}{n}, &n\text{为偶数}\\ &2\pi\cdot 2\cdot \frac{\pi}{2}\cdots\frac{2(n-1)!!}{n!!}\cdot \frac{R^{n}}{n}, &n\text{为奇数} \end{aligned}\end{cases} \end{aligned}\)

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