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Last edited by hbghlyj 2025-3-31 00:32这是书上的解答！
结果就是-8，现在想想，我也不晓得为什么
计算 $\iint_D\left(\frac{y}{x}\right)^2, D$ 乃是 $y=x, y=3 x, x y=1, x y=5$ 围成之第一象限区域
Solution：
\[
\begin{aligned}
& \left\{\begin{array}{l}
\frac{y}{x}=u \\
x y=v
\end{array}\right\} \Rightarrow\left\{\begin{array}{l}
x=\sqrt{\frac{v}{u}} \\
y=\sqrt{u v}
\end{array} \right.\Rightarrow\binom{d x}{d y}=\left[\begin{array}{ll}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\partial v & \frac{\partial v}{\partial x} \\
\partial x & \partial y
\end{array}\right]\left[\begin{array}{l}
d u \\
d v
\end{array}\right]=\left[\begin{array}{ll}
\frac{-1}{2 u} \sqrt{\frac{v}{u}} & \frac{1}{2 \sqrt{u v}} \\
\frac{1}{2} \sqrt{\frac{v}{u}} & \frac{1}{2} \sqrt{\frac{u}{v}}
\end{array}\right]\binom{d u}{d v} \\
& \Rightarrow \text { Jacobian determinant }=|J|=\left|\begin{array}{ll}
\frac{-1}{2 u} \sqrt{\frac{v}{u}} & \frac{1}{2 \sqrt{u v}} \\
\frac{1}{2} \sqrt{\frac{v}{u}} & \frac{1}{2} \sqrt{\frac{u}{v}}\end{array}\right| =\frac{1}{-2 u}
\end{aligned}
\]
It's obvious that： $1 \leq u \leq 3,1 \leq v \leq 5$\begin{aligned}
& \Rightarrow \iint_D\left(\frac{y}{x}\right)^2=\iint_{D^{\prime}} \left[\frac{y(u, v)}{x(u, v)}\right]^2 \cdot|J| \cdot d u \cdot d v \\
& =\iint_{D^{\prime}} u^2\left(\frac{1}{-2 u}\right) d u \bullet d v \\
& =-\int_1^3 \frac{u d u}{2} \int_1^5 d v \\
& =-8
\end{aligned} |
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战巡
Posted 2016-6-27 15:58
чшучшу
