二重积分计算
求$$\iint_D \sin x\sin y(\sin x+\sin y)e^{\sin x\sin y}d\sigma,$$其中$D:0<x<\pi/2,0<y<\pi/2$.
解.首先把待求积分写成
\begin{align*}&\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin x\sin y\left( \sin x+\sin y \right) \text{e}^{\sin x\sin y}\text{d}x\text{d}y}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\sin y\text{e}^{\sin x\sin y}\text{d}x\text{d}y}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\cos y\text{e}^{\sin x\cos y}\text{d}x\text{d}y}}\\=&2\iint_S{x\text{e}^x\text{d}x\text{d}y},\end{align*}
其曲面$S$是单位球面在第一象限的部分.
考虑到平面$x=0$处距离为$x$的宽度为$\mathrm{d}x$的球面窄条的面积，相当于是底边长为$y=\sqrt{1-x^2}$，宽为$\mathrm{d}s=\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x=\frac{\mathrm{d}x}{\sqrt{1-x^2}}$,
因此
$$I=2\int_0^1{x\text{e}^x\frac{\pi}{2}\sqrt{1-x^2}\frac{\text{d}x}{\sqrt{1-x^2}}}=\pi \int_0^1{x\text{e}^x\text{d}x}=\pi.$$
解法二.利用分部积分及对称性可知
\begin{align*}&\iint_D{\sin x\sin y\left( \sin x+\sin y \right) e^{\sin x\sin y}d\sigma}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\sin ye^{\sin x\sin y}dxdy}}=2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\left( 1-\cos ^2x \right) \sin ye^{\sin x\sin y}dxdy}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\cos x\frac{\partial e^{\sin x\sin y}}{\partial x}dxdy}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\left[ \cos x\left. e^{\sin x\sin y} \right|_{0}^{\pi /2}-\left( \int_0^{\frac{\pi}{2}}{\left( -\sin x \right) e^{\sin x\sin y}dx} \right) \right] dy}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\left( -1 \right) dy}-2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin xe^{\sin x\sin y}dxdy}}\\=&\pi .\end{align*}
2022年9月16日 20:59
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