题目

求由曲面=(x)^2+(y)^2 和 =6-sqrt ({x)^2+(y)^2}所围的立体的质心坐标(假设密度为1).

求由曲面 $z=x^2+y^2和z=6- \sqrt {x^2+y^2}$ 所围的立体的质心坐标(假设密度为1).

题目解答

答案

$\begin{align*}&首先，联立方程组\begin{cases}z = x^2 + y^2 \\ z = 6 - \sqrt{x^2 + y^2}\end{cases}\\&可得x^2 + y^2 = 6 - \sqrt{x^2 + y^2}\\&令r^2 = x^2 + y^2，则r^2 = 6 - r，r^2 + r - 6 = 0，(r + 3)(r - 2) = 0\\&解得r = 2（r = -3舍去）\\&所以两曲面的交线在xOy平面上的投影为x^2 + y^2 = 4\\&所求立体在xOy平面上的投影区域为D : x^2 + y^2 \leq 4\\&体积V为：\\ V&=\iint_D [(6 - \sqrt{x^2 + y^2}) - (x^2 + y^2)]dxdy\\ &=\int_0^{2\pi}d\theta\int_0^2 [(6 - r) - r^2]r dr\\ &=2\pi\int_0^2 (6r - r^2 - r^3)dr\\ &=2\pi\left[3r^2 - \frac{1}{3}r^3 - \frac{1}{4}r^4\right]_0^2\\ &=2\pi\left(12 - \frac{8}{3} - 4\right)\\ &=2\pi\left(8 - \frac{8}{3}\right)\\ &=\frac{32\pi}{3} \\&设质心坐标为(\bar{x}, \bar{y}, \bar{z})\\&\bar{z}为：\\ \bar{z}&=\frac{1}{V}\iint_D \left[\int_{x^2 + y^2}^{6 - \sqrt{x^2 + y^2}} z dz\right]dxdy\\ &=\frac{3}{32\pi}\int_0^{2\pi}d\theta\int_0^2 \left[\frac{1}{2}((6 - r)^2 - (r^2)^2)\right]r dr\\ &=\frac{3}{32\pi}\int_0^{2\pi}d\theta\int_0^2 \left(\frac{1}{2}(36 - 12r + r^2 - r^4)\right)r dr\\ &=\frac{3}{32\pi}\times 2\pi\int_0^2 (18r - 6r^2 + \frac{1}{2}r^3 - \frac{1}{2}r^5)dr\\ &=\frac{3}{16}\left[9r^2 - 2r^3 + \frac{1}{8}r^4 - \frac{1}{12}r^6\right]_0^2\\ &=\frac{3}{16}\left(36 - 16 + 2 - \frac{32}{3}\right)\\ &=\frac{3}{16}\left(22 - \frac{32}{3}\right)\\ &=\frac{3}{16}\times\frac{34}{3}\\ &=\frac{17}{8} \\&由于立体关于z轴的对称性，\bar{x} = \bar{y} = 0\\&所以质心坐标为(0, 0, \frac{17}{8}) \\&\end{align*}$