指数型曲面与坐标轴所围成的体积

青青子衿

在二维的情形中，可以求出曲线\(\,e^{x+y}=\,e^{x}+e^{y}\,\)与\(\,x\,\)轴、\(\,y\,\)轴所围成区域的面积。
\begin{align*}
&&e^{x+y}&=e^x+e^y\\
\Rightarrow&&y&=x-\ln\left(e^x-1\right)
\end{align*}
\begin{align*}
\int_0^{+\infty}\bigg(x-\ln\left(e^x-1\right)\bigg)
\mathrm{d}x=\dfrac{\pi^2}{6}
\end{align*}
那么，
在三维的情形中，如何求出曲面\(\,e^{x+y}+e^{y+z}+e^{z+x}=\,2(e^{x}+e^{y}+e^{z})\,\)与\(\,xOy\,\)平面、\(\,yOz\,\)平面、\(\,xOz\,\)平面所围成区域的体积。

1. NIntegrate[
2. Boole[E^x + E^y + 2 >= E^(x + y)]
3. Log[(2 E^x + 2 E^y - E^(x + y))/(E^x + E^y - 2)],
4. {x, 0, \[Infinity]}, {y, 0, \[Infinity]}]

复制代码

hbghlyj

Asymptote: HTML format
import graph3;
import contour3;

size(200);

currentprojection = perspective(2,2,2);

real f(real x,real y,real z) { return exp(x+y) + exp(y+z) + exp(z + x) - 2*(exp(x) + exp(y) + exp(z)); }

draw(
  surface(contour3(f,(0,0,0),(5,5,5),10)),
  blue+opacity(0.75)
);

draw((0,0,0)--(6,0,0),Arrow3(6));
draw((0,0,0)--(0,6,0),Arrow3(6));
draw((0,0,0)--(0,0,6),Arrow3(6));

label("$x$",(6,0,0),SW);
label("$y$",(0,6,0),E);
label("$z$",(0,0,6),N);

draw((0,0,0)--(0,5,0)--(5,5,0)--(5,0,0)--cycle, gray(0.7));
draw((0,0,0)--(0,0,5)--(5,0,5)--(5,0,0)--cycle, gray(0.7));
draw((0,0,0)--(0,5,0)--(0,5,5)--(0,0,5)--cycle, gray(0.7));