多重积分合元 · 余面积公式法

该方法由 Songby 提出.

余面积公式

\[\iint \limits_D g(x,y)\text dS=\int_a^b \int \limits_L g(x,y)\frac{\text dy}{f_x}\text dz \]

我们来证明这个定理.

画出 \(f(x,y)=z\) 和 \(f(x,y)=z+\Delta z\), 我们分别得到等值线 \(\overset{\LARGE{\frown}}{AC}\) 和 \(\overset{\LARGE{\frown}}{BD}\), 面积微元 \(\text dS\) 即为等值线间的面积.

考虑 \(\vec{AB}\) 与 \(\nabla f\) 的方向相同

\[\Delta r=|AB|=\frac{\Delta z}{|\nabla f|}=\frac{\Delta z}{\sqrt{f_x^2+f_y^2}}\\ \]

那么

\[\text dr=\frac{\text dz}{\sqrt{f_x^2+f_y^2}} \]

由弧长公式

\[\text dl=|\overset{\LARGE{\frown}}{AC}|=\sqrt{1+\left(\frac{f_x}{f_y}\right)^2}\text dx=\sqrt{1+\left(\frac{f_y}{f_x}\right)^2}\text dy \]

因此

\[\text dS=\int\limits_L \text dl\cdot\text dr=\int\limits_{L} \frac{\text dx}{f_y}\text dz=\int\limits_{L} \frac{\text dy}{f_x}\text dz \]

关于 \(z\) 的积分限由小到大积分

\[\iint \limits_D g(x,y)\text dS=\int_a^b \int \limits_L g(x,y)\frac{\text dx}{f_y}\text dz=\int_a^b \int \limits_L g(x,y)\frac{\text dy}{f_x}\text dz \]

证毕.

值得注意的是 \(L\) 的取向. 这点可由 \(\text dS, \text dz\) 恒为正值, \(\text dx, \text dy\) 是切向增量, \(f_x, f_y\) 是梯度分量简单得出.

总结之, 余面积公式是积分合元法的推广.

例题

\(1.\quad \displaystyle{\iint\limits_{\mathbb R^2}\text e^{-(x^2+y^2)}\ \text dx\text dy}\)

\[\begin{aligned} I=&\iint\limits_{\mathbb R^2}\text e^{-(x^2+y^2)}\ \text dx\text dy\\ =&\int_0^{+\infty}\text dz\oint\limits_L \frac{\text dy}{2x}\\ =&\frac{1}{2}\int_0^{+\infty}\text e^{-z}\text dz\int_0^{2\pi}\text d\theta\\ =&\pi \end{aligned} \]

一点补充：若设 \(z=-(x^2+y^2)\), 那么 \(z\) 的积分域是 \(\displaystyle\int_{-\infty}^0\), \(\displaystyle{\oint\limits_L \frac{\text dy}{f_x}}\) 应采取顺时针方向, \(\displaystyle{\oint\limits_L \frac{\text dx}{f_y}}\) 应采取逆时针方向.

\(2.\quad \displaystyle{\iint\limits_{D}\text e^{\frac{y}{x+y}}\ \text dx\text dy,\quad D=\left\{(x,y)\bigg|x+y\leq1,x\geq0,y\geq0\right\}}\)

\[\begin{aligned} I=&\iint\limits_{D}\text e^{\frac{y}{x+y}}\ \text dx\text dy\\ =&\int_0^1 \text dz\int_0^z \text e^{\frac{y}{z}} \text dy\\ =&(\text e-1)\int_0^1 z\text dz\\ =&\frac{\text e-1}{2} \end{aligned} \]

\(3.\quad \displaystyle{\iint\limits_{D}\frac{(x+y)\ln\left(1+\frac{y}{x}\right)}{\sqrt{1-x-y}}\ \text dx\text dy,\quad D=\left\{(x,y)\bigg|x+y\leq1,x\geq0,y\geq0\right\}}\)

\[\begin{aligned} I=&\iint\limits_{D}\frac{(x+y)\ln\left(1+\frac{y}{x}\right)}{\sqrt{1-x-y}}\ \text dx\text dy\\ =&\int_0^1\text dz\int_0^z \frac{z\ln\left(\frac{z}{x}\right)}{\sqrt{1-z}}\text dx\\ =&\int_0^1 \frac{z^2}{\sqrt{1-z}}\text dz\\ =&\frac{15}{16} \end{aligned} \]

\(4.\quad \displaystyle{\iiint\limits_{\Omega} \frac{\text dx\text dy\text dz}{(1+|x|+|y|+|z|)^3},\quad \Omega=\left\{ (x,y,z)\bigg| |x|+|y|+|z|\leq 1\right\}}\)

三重积分是类似的 \(\displaystyle{\text dV=\iint\limits_S \text dS\cdot\text dr=\iint\limits_S \frac{\text dx\text dy}{f_z}\text du}\), 方向判定同样是看 \(\text dx\text dy\) 和 \(f_z\)

\[\begin{aligned} I=&\iiint\limits_{\Omega} \frac{\text dx\text dy\text dz}{(1+|x|+|y|+|z|)^3}\\ =&8\iiint\limits_{\Omega} \frac{\text dx\text dy\text dz}{(1+x+y+z)^3}\\ =&8\int_0^1 \frac{1}{(1+u)^3}\text du\iint\limits_S \text dx\text dy\\ =&4\int_0^1\frac{u^2}{(1+u)^3}\text du\\ =&4\ln2-\frac52 \end{aligned} \]