设 $D=\{(x,y)\in\mathbb{R}^2\mid 2\leqslant\frac{x^2}{x^5+y^2}\leqslant 5,\ 4\leqslant\frac{y}{x^5+y^2}\leqslant 7\}$, 求二重积分 $\displaystyle\iint_{D}\frac{1}{x^3 y}\mathrm{d}x\mathrm{d}y$.
1
令 $u=\dfrac{x^2}{x^5+y^2}$, $v=\dfrac{y}{x^5+y^2}$, 于是 $D$ 在这个变换下变为区域 $\Omega=\{(u,v)\mid 2\leqslant u\leqslant 5,\ 4\leqslant v\leqslant 7\}=[2,5]\times [4,7]$.
计算 $u,v$ 关于 $x,y$ 的偏导数如下:
\[
\begin{aligned}
\frac{\partial u}{\partial x}&=\frac{2x(x^5+y^2)-x^2\cdot 5x^4}{(x^5+y^2)^2}=\frac{-3x^6+2xy^2}{(x^5+y^2)^2},\\
\frac{\partial u}{\partial y}&=-\frac{x^2}{(x^5+y^2)^2}\cdot 2y=\frac{-2x^2 y}{(x^5+y^2)^2},\\
\frac{\partial v}{\partial x}&=-\frac{y}{(x^5+y^2)^2}\cdot 5x^4=\frac{-5x^4 y}{(x^5+y^2)^2},\\
\frac{\partial v}{\partial y}&=\frac{1\cdot(x^5+y^2)-y\cdot 2y}{(x^5+y^2)^2}=\frac{x^5-y^2}{(x^5+y^2)^2}.
\end{aligned}
\]
于是
\[
\begin{split}
\biggl|\frac{\partial(u,v)}{\partial(x,y)}\biggr|&=\begin{vmatrix}
\dfrac{-3x^6+2xy^2}{(x^5+y^2)^2} & \dfrac{-2x^2 y}{(x^5+y^2)^2}\\
\dfrac{-5x^4 y}{(x^5+y^2)^2} & \dfrac{x^5-y^2}{(x^5+y^2)^2}
\end{vmatrix}=\frac{1}{(x^5+y^2)^4}\begin{vmatrix}
-3x^6+2xy^2 & -2x^2 y\\
-5x^4 y & x^5-y^2
\end{vmatrix}\\
&=\frac{1}{(x^5+y^2)^4}\Bigl[(-3x^6+2xy^2)(x^5-y^2)-2x^2 y\cdot 5x^4y\Bigr]\\
&=\frac{-3x^{11}-5x^6 y^2-2xy^4}{(x^5+y^2)^4}\\
&=\frac{-x(3x^5+2y^2)}{(x^5+y^2)^3}.
\end{split}
\]
因此,
\[\dfrac{\partial(x,y)}{\partial(u,v)}=\frac{(x^5+y^2)^3}{-x(3x^5+2y^2)}.\]
原积分化为
\[
\begin{split}
&\int_{2}^{5}\int_{4}^{7}\frac{1}{x^3 y}\cdot\biggl|\dfrac{\partial(x,y)}{\partial(u,v)}\biggr|\mathrm{d}u\mathrm{d}v\\
=&\int_{2}^{5}\int_{4}^{7}\frac{1}{x^3 y}\cdot\frac{(x^5+y^2)^3}{x(3x^5+2y^2)}\mathrm{d}u\mathrm{d}v.
\end{split}
\]
其中
\[
\begin{split}
\frac{1}{x^3 y}\cdot\frac{(x^5+y^2)^3}{x(3x^5+2y^2)}&=\frac{(x^5+y^2)^3}{x^4 y(3x^5+2y^2)}=\frac{\bigl(\frac{x^2}{u}\bigr)^3}{x^4 y(3x^5+2y^2)}\\
&=\frac{x^2}{yu^3(3x^5+2y^2)}=\frac{u}{v}\cdot\frac{1}{u^3}\cdot\frac{1}{3x^5+2y^2}\\
&=\frac{1}{vu^2(3x^5+2y^2)}
\end{split}
\]