题目

某工厂用一种面密度为=√(x^2+y^2(0≤z≤3),=2的金属板做一个圆锥面壳,圆锥面壳的方程为=√(x^2+y^2(0≤z≤3),=2求此圆锥面壳的质量.

某工厂用一种面密度为 $\rho=z$ 的金属板做一个圆锥面壳,圆锥面壳的方程为 $z= \sqrt {x^2+y^2} (0 \le z \le 3),$ 求此圆锥面壳的质量.

题目解答

答案

$\begin{aligned}&本题考查第一型曲面积分的计算：\\ &若曲面方程可以表示为z=z(x,y)\\ &将曲面向xOy平面投影，投影区域为D_{xy}\\ &其中空间曲面微元dS与平面面积微元dxdy\\ &法向量夹角的余弦值为：\\ &\mid cos(\gamma )\mid=\frac{1}{\sqrt{1+z^2_x(x,y)+z^2_y(x,y)}}\\ &则:\iint_\limits{\sum}f(x,y,z)dS\\ &=\iint_\limits{D_{xy}}f[x,y,z(x,y)]\sqrt{1+z^2_x(x,y)+z^2_y(x,y)}dxdy\\ &该题首先求函数的偏导数：\\ &\frac{\partial z}{\partial x}=\frac{x}{\sqrt{x^2 + y^2}}\\ &\frac{\partial z}{\partial y}=\frac{y}{\sqrt{x^2 + y^2}}\\ &所以：\sqrt{1+{(\frac{\partial z}{\partial x})}^2+{(\frac{\partial z}{\partial y})}^2}\\ &=\sqrt{2}\\ &曲面积分可以转换为二重积分：\\ \end{aligned}$ $\begin{aligned}\\ &\iint_\limits{\sum}\left(z\right)dS\\ &=\iint_\limits{D_{ x y}}(\sqrt{2}\, \sqrt{x^2 + y^2})dxdy\\ &其中投影区域D=\{(\mathrm{\rho },\mathrm{\theta })|0\leq \mathrm{\rho }\leq 3,0\leq \mathrm{\theta }\leq 2\, \pi\}\\ &本题可以看作\rho 型二重积分：\\ &积分区域的极角\theta 区间为[0,2\, \pi],\\ &\mathrm{\rho }的上下边界为\mathrm{\rho }_2(\mathrm{\theta })=3,\mathrm{\rho }_1(\mathrm{\theta })=0\\ &\int_{0}^{2\, \pi}d\mathrm{\theta }\int_{0}^{3}\left[\left(\sqrt{2}\, \sqrt{{\mathrm{\rho }}^2}\right)\, \mathrm{\rho }\right]d\mathrm{\rho }\\&=\int_{0}^{2\, \pi}\left[\left(\frac{\sqrt{2}\, \sqrt{{\mathrm{\rho }}^6}}{3}\right)\bigg|^{3}_{0}\right]d\mathrm{\theta }\\&=\int_{0}^{2\, \pi}\left(9\, \sqrt{2}\right)d\mathrm{\theta }\\&=\left(9\, \sqrt{2}\, \mathrm{\theta }\right)\bigg|^{2\, \pi}_{0}\\&=18\, \pi\, \sqrt{2}\end{aligned}$