下册 8.1 二重积分 第7题

\section*{解题过程：} 如图 8.22 所示，在极坐标系中，闭区域 $D: x^{2}+y^{2} \leqslant a^{2}$ 可表示为： $0 \leqslant r \leqslant a, 0 \leqslant \theta \leqslant 2 \pi$ ． （1） $\iint_{D}\left(x^{3}+y^{3}\right) \mathrm{e}^{\left(x^{2}+y^{2}\right)^{2}} \mathrm{~d} x \mathrm{~d} y=\int_{0}^{2 \pi}\left(\sin ^{3} \theta+\cos ^{3} \theta\right) \mathrm{d} \theta \int_{0}^{1} r^{4} \mathrm{e}^{r^{4}} \mathrm{~d} r=0$ ． （2） $\iint_{D} x^{2} y^{2} \ln \left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y=\int_{0}^{2 \pi} \sin ^{2} \theta \cos ^{2} \theta \mathrm{~d} \theta \int_{0}^{1} r^{5} \ln r^{2} \mathrm{~d} r$ $$ =8 \int_{0}^{\frac{\pi}{2}} \sin ^{2} \theta \cos ^{2} \theta \mathrm{~d} \theta \int_{0}^{1} r^{6} \ln r \mathrm{~d} r=-\frac{\pi}{72} . $$ （3）令 $\displaystyle I=\iint_{D}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right) \mathrm{d} x \mathrm{~d} y$ ，根据对称性有 $\displaystyle I=\iint_{D}\left(\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}\right) \mathrm{d} x \mathrm{~d} y$ 。从而 \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-131.jpg?height=1016&width=1071&top_left_y=5028&top_left_x=4434} \captionsetup{labelformat=empty} \caption{图8．22} \end{figure} $$ 2 I=\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) \iint_{D}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y=\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} r^{3} \mathrm{~d} r=\frac{\pi}{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) . $$ 所以 $\displaystyle I=\frac{\pi}{4}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)$ ． （4） $\displaystyle \iint_{D}|x y| \mathrm{d} x \mathrm{~d} y=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{2}|r \sin \theta \cdot r \cos \theta| \cdot r \mathrm{~d} r=\int_{0}^{2 \pi} \frac{1}{2}|\sin 2 \theta| \mathrm{d} \theta \cdot \int_{0}^{2} r^{3} \mathrm{~d} r=2 \int_{0}^{2 \pi}|\sin 2 \theta| \mathrm{d} \theta=8$ ．