📝 题目
例 5 设 $f\left( {x,y}\right)$ 在 ${\mathbf{R}}^{2}$ 上连续, $r$ 为定数,记 $D$ 是以(x, y)点为心,以 $r$ 为半径的圆,即 ${\left( \xi - x\right) }^{2} + {\left( \eta - y\right) }^{2} \leq {r}^{2},L$ 为 $D$ 的边界沿逆时针方向. 令
$$ F\left( {x,y}\right) = {\iint }_{D}f\left( {\xi ,\eta }\right) \mathrm{d}\xi \mathrm{d}\eta . $$
💡 答案与解析
证明:
(1) $F\left( {x,y}\right)$ 的偏导数存在,且
$$ \frac{\partial F}{\partial x} = {\int }_{L}f\left( {\xi ,\eta }\right) \mathrm{d}\eta ,\;\frac{\partial F}{\partial y} = - {\int }_{L}f\left( {\xi ,\eta }\right) \mathrm{d}\xi ; $$
(2)上述偏导数是 $x,y$ 的连续函数.
证 (1) 由重积分化累次积分公式, 得
$$ F\left( {x,y}\right) = {\int }_{y - r}^{y + r}\mathrm{\;d}\eta {\int }_{x - \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}}}^{x + \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}}}f\left( {\xi ,\eta }\right) \mathrm{d}\xi . $$
当 $y$ 固定时,作为 $\eta ,x$ 的函数,
$$ g\left( {\eta ,x}\right) = {\int }_{x - \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}}}^{x + \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}}}f\left( {\xi ,\eta }\right) \mathrm{d}\xi $$
在 $y - r \leq \eta \leq y + r,x \in \left( {-\infty , + \infty }\right)$ 上连续,且 $g\left( {\eta ,x}\right)$ 对 $x$ 的偏导数
$$ {g}_{x}^{\prime }\left( {\eta ,x}\right) = f\left( {x + \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}},\eta }\right) $$
$$ - f\left( {x - \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}},\eta }\right) $$
也在 $y - r \leq \eta \leq y + r,x \in \left( {-\infty , + \infty }\right)$ 上连续,故根据参变积分求导定理知
$$ \frac{\partial F}{\partial x} = {\int }_{y - r}^{y + r}\left\lbrack {f\left( {x + \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}},\eta }\right) }\right. $$
$$ \left. {-f\left( {x - \sqrt{{r}^{2} - {\left( \eta - y\right) }^{2}},\eta }\right) }\right\rbrack \mathrm{d}\eta . $$
再根据第二型曲线积分的计算, 即可看出
$$ \frac{\partial F}{\partial x} = {\int }_{{\left( \xi - x\right) }^{2} + {\left( \eta - y\right) }^{2} = {r}^{2}}f\left( {\xi ,\eta }\right) \mathrm{d}\eta = {\int }_{L}f\left( {\xi ,\eta }\right) \mathrm{d}\eta . $$
同理可证
$$ \frac{\partial F}{\partial y} = - {\int }_{L}f\left( {\xi ,\eta }\right) \mathrm{d}\xi . $$
(2)为了说明 $\frac{\partial F}{\partial x},\frac{\partial F}{\partial y}$ 是 $x,y$ 的连续函数,只需注意
$$ \frac{\partial F}{\partial x} = {\int }_{0}^{2\pi }f\left\lbrack {x + r\cos \theta ,y + r\sin \theta }\right\rbrack r\cos \theta \mathrm{d}\theta , $$
$$ \frac{\partial F}{\partial y} = {\int }_{0}^{2\pi }f\left\lbrack {x + r\cos \theta ,y + r\sin \theta }\right\rbrack r\sin \theta \mathrm{d}\theta . $$
这样根据参变积分中的连续性定理,即可看出 $\frac{\partial F}{\partial x},\frac{\partial F}{\partial y}$ 是 $x,y$ 的二元连续函数.