题目

∥1(x+1)^2+(y-1)^2]dxdydz= () ,Ω是由 ∥1(x+1)^2+(y-1)^2]dxdydz= () ,Ω 及平面 z=1 所围成的闭区域。∥1(x+1)^2+(y-1)^2]dxdydz= () ,Ω

$\iiint_{\Omega}\left[(x+1)^{2}+(y-1)^{2}\right] \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=(\quad), \Omega$ 是由 $3z=x^2+y^2$ 及平面 z=1 所围成的闭区域。

$A.\frac{7}{2}\pi B.\frac{9}{2}\pi C.\frac{5}{2}\pi D.\frac{11}{2}\pi$

题目解答

答案

解：

$\Omega$ 在xOy平面上的投影区域为半径等于 $\sqrt{3}$ 的圆形闭区域, 即 $D_{x y}=|(\rho, \theta)| 0 \leqslant p \leqslant \sqrt{3},0 \leqslant \theta \leqslant 2 \pi \mid 。$ 在投影区域内任一点作一与 z 轴同向的射线, 它由 $3z=x^2+y^2$ 穿入 $\Omega$ , 再由 z=1 穿出, 故积分区域可表示为

$\frac{1}{3} \rho^{2} \leqslant z \leqslant 1,0 \leqslant \rho \leqslant \sqrt{3},0 \leqslant \theta \leqslant 2 \pi$ , 再利用柱坐标计算, 得

$\begin{aligned} \iiint_{\Omega}\left[(x+1)^{2}+(y-1)^{2}\right] \mathrm{d} x \mathrm{~d} y \mathrm{~d} z &= \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{\frac{r^2}{3}}^{1} \left[ (r\cos\theta + 1)^{2} + (r\sin\theta - 1)^{2} \right] r \mathrm{d} z \mathrm{d} r \mathrm{d} \theta \\&= \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{\frac{r^2}{3}}^{1} (r^2\cos^2\theta + 2r\cos\theta + 1 + r^2\sin^2\theta - 2r\sin\theta + 1 ) r \mathrm{d} z \mathrm{d} r \mathrm{d} \theta \\&= \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{\frac{r^2}{3}}^{1} (r^2 + 2r\cos\theta - 2r\sin\theta + 2 ) r \mathrm{d} z \mathrm{d} r \mathrm{d} \theta \\&= \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} (r^2 + 2r\cos\theta - 2r\sin\theta + 2 ) r \left(1- \frac{r^2}{3} \right)\mathrm{d} r \mathrm{d} \theta \\&= \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} (-\frac{r^5}{3} - \frac{2r^4\cos\theta}{3} + \frac{2r^4\sin\theta}{3} + \frac{1}{3}r^3 + 2r^2\cos\theta - 2r^2\sin\theta + 2r ) \mathrm{d} r \mathrm{d} \theta \\&= \int_{0}^{2\pi} (\frac{9}{4} + \frac{4\sqrt{3}\cos\theta}{5} - \frac{4\sqrt{3}\sin\theta}{5} ) \mathrm{d} \theta \mathrm{d} \theta \\&= (\frac{9}{4}\theta + \frac{4\sqrt{3}\sin\theta}{5} + \frac{4\sqrt{3}\cos\theta}{5})| _{0}^{2\pi} \\&=\frac{9}{4}\bullet 2\pi \\&=\frac{9\pi}{2} \end{aligned}$

故本题正确答案为B