空间立体的质心坐标(B020)

问题

已知空间立体 $\Omega$ 的体密度为 $\rho(x, y, z)$, 且 $\rho(x, y, z)$ 在空间立体 $\Omega$ 上连续,则,该立体的质心坐标 $(\bar{x}, \bar{y}, \bar{z})$ 为多少?

选项

[A].   $\bar{x}$ $=$ $\frac{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} x \rho(x, y, z) \mathrm{d} v}$, $\bar{y}$ $=$ $\frac{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} y \rho(x, y, z) \mathrm{d} v}$, $\bar{z}$ $=$ $\frac{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} z \rho(x, y, z) \mathrm{d} v}$

[B].   $\bar{x}$ $=$ $\frac{\iiint_{\Omega} x \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{y}$ $=$ $\frac{\iiint_{\Omega} y^{2} \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{z}$ $=$ $\frac{\iiint_{\Omega} z^{2} \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$

[C].   $\bar{x}$ $=$ $\frac{\iiint_{\Omega} x \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{y}$ $=$ $\frac{\iiint_{\Omega} y \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{z}$ $=$ $\frac{\iiint_{\Omega} z \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$

[D].   $\bar{x}$ $=$ $\frac{\iiint_{\Omega} x \rho^{\prime}(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{y}$ $=$ $\frac{\iiint_{\Omega} y \rho^{\prime}(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{z}$ $=$ $\frac{\iiint_{\Omega} z \rho^{\prime}(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$


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$\bar{x}$ $=$ $\frac{\iiint_{\Omega} x \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{y}$ $=$ $\frac{\iiint_{\Omega} y \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$, $\bar{z}$ $=$ $\frac{\iiint_{\Omega} z \rho(x, y, z) \mathrm{d} v}{\iiint_{\Omega} \rho(x, y, z) \mathrm{d} v}$