题号:1230    题型:填空题    来源:2001年全国硕士研究生招生考试试题
类型:考研真题
设 $r=\sqrt{x^{2}+y^{2}+z^{2}}$, 则 $\left.\operatorname{div}(\operatorname{grad} r)\right|_{(1,-2,2)}=$
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答案:
$frac{2}{3} $

解析:

本题实际上是计算 $\frac{\partial^{2} r}{\partial x^{2}}+\frac{\partial^{2} r}{\partial y^{2}}+\frac{\partial^{2} r}{\partial z^{2}}$
$$
\frac{\partial r}{\partial x}=\frac{\partial \sqrt{x^{2}+y^{2}+z^{2}}}{\partial x}=\frac{2 x}{2 \sqrt{x^{2}+y^{2}+z^{2}}}=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}=\frac{x}{r}
$$

$$
\frac{\partial^{2} r}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{x}{r}\right)=\frac{r-x \frac{\partial r}{\partial x}}{r^{2}} \underline{\underline{\frac{\partial r}{\partial x}}=\frac{x}{r}} \frac{r-x \frac{x}{r}}{r^{2}}=\frac{r^{2}-x^{2}}{r^{3}}
$$
类似可得 $\quad \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial^{2} r}{\partial y^{2}}=\frac{r^{2}-y^{2}}{r^{3}} ; \frac{\partial r}{\partial z}=\frac{z}{r}, \frac{\partial^{2} r}{\partial z^{2}}=\frac{r^{2}-z^{2}}{r^{3}}$
根据定义有 $\operatorname{div}(\mathrm{gradr})=\frac{\partial^{2} r}{\partial x^{2}}+\frac{\partial^{2} r}{\partial y^{2}}+\frac{\partial^{2} r}{\partial z^{2}}=\frac{r^{2}-x^{2}}{r^{3}}+\frac{r^{2}-y^{2}}{r^{3}}+\frac{r^{2}-z^{2}}{r^{3}}$
$$
=\frac{3 r^{2}-x^{2}-y^{2}-z^{2}}{r^{3}}=\frac{3 r^{2}-r^{2}}{r^{3}}=\frac{2 r^{2}}{r^{3}}=\frac{2}{r}=\frac{2}{\sqrt{x^{2}+y^{2}+z^{2}}}
$$
于是
$$
\left.\operatorname{div}(\operatorname{gradr})\right|_{(1,-2,2)}=\left.\frac{2}{\sqrt{x^{2}+y^{2}+z^{2}}}\right|_{(1,-2,2)}=\frac{2}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}=\frac{2}{3}
$$

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