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5樓: Originally posted by clcfang at 2020-08-20 09:47:19
感謝你的回答！仍有不懂。我舉個(gè)例子，比如兩個(gè)圓柱相交（從側(cè)面）求公共部分體積。（1）在兩圓柱底面半徑相等且正對(duì)著垂直相交（2）底面半徑不等或斜著相交時(shí)
你能教下不畫圖怎么計(jì)算嗎？...

[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]
這里只寫r<a<2r的情況，[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, 只需比較端點(diǎn)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]時(shí)，
[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]時(shí)，
[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],
則[latex]-arc\sin\frac ar<\theta<arc\sin\frac ar[/latex]時(shí)[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]
，

盡可能地會(huì)畫圖吧，現(xiàn)在也有Geogebra軟件的