T19 斜棱锥 Jacobian 换元

设 $\Omega$ 为锥面 $x^2+(y-z{)}^2=(1-z{)}^2$ 与平面 $z=0$ 所围成的斜棱锥体，求 $\Omega$ 的形心坐标。

• 由对称性有 $\bar{x}=0$
• 引入柱坐标变换 $\{\begin{array}{l}x=r\cos \theta \\ y-z=r\sin \theta \\ z=z\end{array}$，则锥面方程化为 $r^2=(1-z{)}^2$
  • 计算 Jacobian 行列式 $J=\left|\begin{array}{ccc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }& \frac{\partial x}{\partial z}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }& \frac{\partial y}{\partial z}\\ \frac{\partial z}{\partial r}& \frac{\partial z}{\partial \theta }& \frac{\partial z}{\partial z}\end{array}\right|=\left|\begin{array}{ccc}\cos \theta & -r\sin \theta & 0\\ \sin \theta & r\cos \theta & -1\\ 0& 0& 1\end{array}\right|=r$
  • 故体积微元 $\mathrm{d}V=|J|\mathrm{d}r\mathrm{d}\theta \mathrm{d}z=r\mathrm{d}r\mathrm{d}\theta \mathrm{d}z$
• 计算体积 $V={\iiint }_{\Omega }\mathrm{d}V={\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^1\mathrm{d}z{\int }_0^{1-z}r\mathrm{d}r=\frac{\pi }{3}$
• 计算 $\bar{y}=\frac{1}{V}{\iiint }_{\Omega }y\mathrm{d}V=\frac{1}{V}{\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^1\mathrm{d}z{\int }_0^{1-z}(r\sin \theta +z)r\mathrm{d}r=\frac{1}{4}$
  • 注意 ${\int }_0^{2\pi }\sin \theta \mathrm{d}\theta =0$
• 计算 $\bar{z}=\frac{1}{V}{\iiint }_{\Omega }z\mathrm{d}V=\frac{1}{V}{\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^1\mathrm{d}z{\int }_0^{1-z}zr\mathrm{d}r=\frac{1}{4}$