下册 8.1 二重积分 第31题

\section*{解题过程：} （1）利用极坐标变换得 $$ \begin{aligned} \frac{1}{\pi h^{2}} \iint_{B\left(z_{0}, h\right)} f(x, y) \mathrm{d} x \mathrm{~d} y-f\left(x_{0}, y_{0}\right) & =\frac{1}{\pi h^{2}} \iint_{B\left(z_{0}, h\right)}\left(f(x, y)-f\left(x_{0}, y_{0}\right)\right) \mathrm{d} x \mathrm{~d} y \\ & =\frac{1}{\pi h^{2}} \int_{0}^{h} \mathrm{~d} r \int_{0}^{2 \pi}\left(f\left(x_{0}+r \cos \theta, y_{0}+r \sin \theta\right)-f\left(x_{0}, y_{0}\right)\right) r \mathrm{~d} \theta \end{aligned} $$ 于是 $$ \begin{aligned} & \lim _{h \rightarrow 0^{+}} \frac{1}{\pi h^{4}} \int_{0}^{h} \mathrm{~d} r \int_{0}^{2 \pi}\left(f\left(x_{0}+r \cos \theta, y_{0}+r \sin \theta\right)-f\left(x_{0}, y_{0}\right)\right) r \mathrm{~d} \theta \\ & =\lim _{h \rightarrow 0^{+}} \frac{1}{4 \pi h^{2}} \int_{0}^{2 \pi}\left(f\left(x_{0}+h \cos \theta, y_{0}+h \sin \theta\right)-f\left(x_{0}, y_{0}\right)\right) \mathrm{d} \theta \\ & =\lim _{h \rightarrow 0^{+}} \frac{1}{8 \pi h} \int_{0}^{2 \pi}\left(\frac{\partial f}{\partial x} \cos \theta+\frac{\partial f}{\partial y} \sin \theta\right) \mathrm{d} \theta \\ & =\frac{1}{8 \pi} \lim _{h \rightarrow 0^{+}} \int_{0}^{2 \pi}\left[\left(\frac{\partial^{2} f}{\partial x^{2}} \cos \theta+\frac{\partial^{2} f}{\partial y \partial x} \sin \theta\right) \cos \theta+\left(\frac{\partial^{2} f}{\partial x \partial y} \cos \theta+\frac{\partial^{2} f}{\partial y^{2}} \sin \theta\right) \sin \theta\right] \mathrm{d} \theta \\ & =\frac{1}{8 \pi} \lim _{h \rightarrow 0^{+}} \int_{0}^{2 \pi}\left(\frac{\partial^{2} f}{\partial x^{2}}\left(z_{0}\right) \cos ^{2} \theta+2 \frac{\partial^{2} f}{\partial x \partial y}\left(z_{0}\right) \sin \theta \cos \theta+\frac{\partial^{2} f}{\partial y^{2}}\left(z_{0}\right) \sin ^{2} \theta\right) \mathrm{d} \theta \\ & =\frac{1}{8 \pi} \cdot \pi\left(\frac{\partial^{2} f}{\partial x^{2}}\left(z_{0}\right)+\frac{\partial^{2} f}{\partial y^{2}}\left(z_{0}\right)\right)=\frac{1}{8}\left(\frac{\partial^{2} f}{\partial x^{2}}\left(z_{0}\right)+\frac{\partial^{2} f}{\partial y^{2}}\left(z_{0}\right)\right) . \end{aligned} $$ （2）利用极坐标变换得 $$ F(t)=\iint_{x^{2}+y^{2}