题号:1679    题型:填空题    来源:2008年全国硕士研究生招生考试试题
类型:考研真题
设曲面 $\Sigma$ 是 $z=\sqrt{4-x^{2}-y^{2}}$ 的上侧, 则 $\iint_{\Sigma} x y \mathrm{~d} y \mathrm{~d} z+x \mathrm{~d} z \mathrm{~d} x+x^{2} \mathrm{~d} x \mathrm{~d} y=$
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答案:
$4 \pi$

解析:

$$
\begin{aligned}
& \iint_{\Sigma} x y \mathrm{~d} y \mathrm{~d} z+x \mathrm{~d} z \mathrm{~d} x+x^{2} \mathrm{~d} x \mathrm{~d} y \\
=& \iint_{\Sigma+D} x y \mathrm{~d} y \mathrm{~d} z+x \mathrm{~d} z \mathrm{~d} x+x^{2} \mathrm{~d} x \mathrm{~d} y+\iint_{D_{\pm}} x^{2} \mathrm{~d} x \mathrm{~d} y \\
=& \iint_{\Omega} y \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z+\iint_{D_{\pm}} x^{2} \mathrm{~d} x \mathrm{~d} y \\
=& 0+\frac{1}{2} \iint_{D_{\text {E. }}}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{1}{2} \int_{0}^{2 \pi} \mathrm{d} \theta \int_{0}^{2} r^{2} r \mathrm{~d} r \\
&=4 \pi .
\end{aligned}
$$

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