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5楼: Originally posted by clcfang at 2020-08-20 09:47:19
感谢你的回答！仍有不懂。我举个例子，比如两个圆柱相交（从侧面）求公共部分体积。（1）在两圆柱底面半径相等且正对着垂直相交（2）底面半径不等或斜着相交时
你能教下不画图怎么计算吗？...

[latex]\left\{x,y,z\vert x^2+y^2\leq r^2,\;{(x-a)}^2+y^2\leq r^2,\;0<z<h\right\},\;a<2r[/latex]
[latex]x=\rho\cos\theta,\;y=\rho\sin\theta[/latex]
[latex]\left\{\begin{array}{l}0\leq\rho\leq r,\;\\\rho^2-2a\rho\cos\theta+a^2-r^2\leq0\end{array}\right.[/latex]
这里只写r<a<2r的情况，[latex]4a^2\cos^2\theta-4(a^2-r^2)>0,\;-arc\sin\frac ra<\theta<arc\sin\frac ra[/latex],
[latex]\left\{\begin{array}{l}0\leq\rho\leq r,\;\\a\cos\theta-\sqrt{r^2-a^2\sin^2\theta}\leq\rho\leq a\cos\theta+\sqrt{r^2-a^2\sin^2\theta}\end{array}\right.[/latex]
r<a, 只需比较端点r和[latex]a\cos\theta+\sqrt{r^2-a^2\sin^2\theta}[/latex]，
若[latex]arc\cos\frac a{2r}<arc\sin\frac ar[/latex],
[latex]-arc\cos\frac a{2r}\leq\theta\leq arc\cos\frac a{2r}[/latex]时，
[latex]a\cos\theta+\sqrt{r^2-a^2\sin^2\theta}\leq\rho\leq r[/latex].
[latex]arc\cos\frac a{2r}\leq\theta\leq arc\sin\frac ar[/latex]或[latex]-arc\sin\frac ar\leq\theta\leq-arc\cos\frac a{2r}[/latex]时，
[latex]a\cos\theta-\sqrt{r^2-a^2\sin^2\theta}\leq\rho\leq a\cos\theta+\sqrt{r^2-a^2\sin^2\theta}[/latex].
若[latex]arc\cos\frac a{2r}<arc\sin\frac ar[/latex],
则[latex]-arc\sin\frac ar<\theta<arc\sin\frac ar[/latex]时[latex]a\cos\theta+\sqrt{r^2-a^2\sin^2\theta}>r[/latex], 只有[latex]a\cos\theta-\sqrt{r^2-a^2\sin^2\theta}\leq\rho\leq r[/latex]
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尽可能地会画图吧，现在也有Geogebra软件的