# 多重积分计算的一些题

Eufisky posted @ 2015年8月29日 04:53 in 数学分析 with tags 积分计算 多元 , 1045 阅读

（1）设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且$\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=x^2+y^2,$求$\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.$

\begin{align*}&\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\&=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}s\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\frac{\partial f}{\partial n}\mathrm{d}s=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}\Delta f\mathrm{d}x\mathrm{d}y\\&=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}(x^2+y^2)\mathrm{d}x\mathrm{d}y=\int_0^1\frac{\pi r^4}{2}\mathrm{d}r =\frac{\pi}{10} .\end{align*}

（2）设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且$\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\exp{(-x^2-y^2)},$求

$\iint\limits_D\left(x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.$

\begin{align*}&\iint\limits_D\left(x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\left(x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}\right)\mathrm{d}s\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 =r^2}r\frac{\partial f}{\partial n}\mathrm{d}s=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}r\Delta f\mathrm{d}x\mathrm{d}y\\ &=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}r\exp{(-x^2-y^2)}\mathrm{d}x\mathrm{d}y=\int_0^1\pi r(1-e^{-r^2})\mathrm{d}r =\frac{\pi }{2e} .\end{align*}

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