📝 题目
例 3 考察抛物线 ${y}^{2} = {\alpha x},{y}^{2} = {\beta x},{x}^{2} = {\gamma y}$ 和 ${x}^{2} = {\delta y}$ $\left( {0 < \alpha < \beta ,0 < \gamma < \delta }\right)$ . 设 $D$ 是由这四条抛物线围成的闭区域,试计算:
(1) $D$ 的面积 $\sigma \left( D\right)$ ；
(2) $I = {\iint }_{D}{xy}\mathrm{\;d}\left( {x,y}\right)$ ;
(3) $J = {\iint }_{D}\frac{1}{xy}\mathrm{\;d}\left( {x,y}\right)$ .
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图 13-9
💡 答案与解析
解 闭区域 $D$ 的形状 (参看图 13-9) 提示我们做变换
$$ \psi : \left\{ \begin{array}{l} u = \frac{{y}^{2}}{x}, \\ v = \frac{{x}^{2}}{y}. \end{array}\right. $$
计算变换的雅可比行列式得
$$ \frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) } = \left| \begin{matrix} - \frac{{y}^{2}}{{x}^{2}} & \frac{2y}{x} \\ \frac{2x}{y} & - \frac{{x}^{2}}{{y}^{2}} \end{matrix}\right| = - 3, $$
$$ \frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) } = - \frac{1}{3}. $$
通过变元替换, 我们得到
$$ \sigma \left( D\right) = {\iint }_{D}\mathrm{\;d}\left( {x,y}\right) = \frac{1}{3}{\int }_{a}^{\beta }\mathrm{d}u{\int }_{\gamma }^{\delta }\mathrm{d}v $$
$$ = \frac{1}{3}\left( {\beta - \alpha }\right) \left( {\delta - \gamma }\right) . $$
$$ I = {\iint }_{D}{xy}\mathrm{\;d}\left( {x,y}\right) = \frac{1}{3}{\int }_{\alpha }^{\beta }\mathrm{d}u{\int }_{\gamma }^{\delta }{uv}\mathrm{\;d}v $$
$$ = \frac{1}{12}\left( {{\beta }^{2} - {\alpha }^{2}}\right) \left( {{\delta }^{2} - {\gamma }^{2}}\right) . $$
$$ J = {\iint }_{D}\frac{1}{xy}\mathrm{\;d}\left( {x,y}\right) $$
$$ = \frac{1}{3}{\int }_{a}^{\beta }\mathrm{d}u{\int }_{\gamma }^{\delta }\frac{1}{uv}\mathrm{\;d}v $$
$$ = \frac{1}{3}\ln \frac{\beta }{\alpha } \cdot \ln \frac{\delta }{\gamma }. $$