下册 9.2 第二型曲线积分 第9题

\section*{解题过程：} （1）如图9．56所示， $$ \begin{aligned} & \int_{L}(\sin x+y)^{2} \mathrm{~d} x+\left(x^{2}+y^{2} \cos y\right) \mathrm{d} y \\ & =\int_{-1}^{1}\left[\left(\sin x+x^{2}\right)^{2}+\left(x^{2}+x^{4} \cos x^{2}\right) 2 x\right] \mathrm{d} x=\int_{-1}^{1}\left(\sin x+x^{2}\right)^{2} \mathrm{~d} x \\ & =\int_{-1}^{1}\left(\sin ^{2} x+2 x^{2} \sin x+x^{4}\right) \mathrm{d} x=2 \int_{0}^{1}\left(\sin ^{2} x+x^{4}\right) \mathrm{d} x=2 \int_{0}^{1}\left(\frac{1-\cos 2 x}{2}+x^{4}\right) \mathrm{d} x \\ & =\left.2\left(\frac{1}{2} x-\frac{1}{4} \sin 2 x+\frac{1}{5} x^{5}\right)\right|_{0} ^{1}=\frac{7}{5}-\frac{1}{2} \sin 2 \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-264.jpg?height=830&width=1251&top_left_y=7362&top_left_x=1277} \captionsetup{labelformat=empty} \caption{图 9.56} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-264.jpg?height=1223&width=796&top_left_y=6976&top_left_x=3729} \captionsetup{labelformat=empty} \caption{图 9.57} \end{figure} （2）如图 9.57 所示，补直线段 $L_{1}: x=0, y:-b \rightarrow b$ ，则 $L+L_{1}$ 为闭曲线，逆时针方向，记围成的区域为 $D$ ．由格林公式得 $$ \int_{L+L_{1}}\left(x^{2}+x y\right) \mathrm{d} x+\left(x y+y^{2}\right) \mathrm{d} y=\iint_{D}(y-x) \mathrm{d} x \mathrm{~d} y=a b \int_{\frac{\pi}{2}}^{\frac{3}{2} \pi} \mathrm{~d} \theta \int_{0}^{1} r^{2}(b \sin \theta-a \cos \theta) \mathrm{d} r=\frac{2}{3} \pi a^{2} b . $$ 又 $$ \int_{L_{1}}\left(x^{2}+x y\right) \mathrm{d} x+\left(x y+y^{2}\right) \mathrm{d} y=\int_{L_{1}} y^{2} \mathrm{~d} y=\int_{-b}^{b} y^{2} \mathrm{~d} y=\frac{2}{3} b^{3} . $$ 于是 $$ \int_{L}\left(x^{2}+x y\right) \mathrm{d} x+\left(x y+y^{2}\right) \mathrm{d} y=\frac{2}{3} \pi a^{2} b-\int_{L_{1}}\left(x^{2}+x y\right) \mathrm{d} x+\left(x y+y^{2}\right) \mathrm{d} y=\frac{2}{3} \pi a^{2} b-\frac{2}{3} b^{3} . $$ （3）令 $x=\cos \theta, y=\sin \theta$ ，则 $$ \begin{aligned} \int_{C} \frac{3 x}{3 x^{2}+4 y^{2}} \mathrm{~d} x-\frac{4 y}{3 x^{2}+4 y^{2}} \mathrm{~d} y & =\int_{0}^{2 \pi}\left(-\frac{3 \cos \theta \sin \theta}{3 \cos ^{2} \theta+4 \sin ^{2} \theta}-\frac{4 \cos \theta \sin \theta}{3 \cos ^{2} \theta+4 \sin ^{2} \theta}\right) \mathrm{d} \theta \\ & =-7 \int_{0}^{2 \pi}\left(\frac{\cos \theta \sin \theta}{3 \cos ^{2} \theta+4 \sin ^{2} \theta}\right) \mathrm{d} \theta=-\frac{7}{2} \int_{0}^{2 \pi}\left(\frac{1}{3+\sin ^{2} \theta}\right) \mathrm{dsin}^{2} \theta \\ & =-\left.\frac{7}{2} \ln \left(3+\sin ^{2} \theta\right)\right|_{0} ^{2 \pi}=0 \end{aligned} $$ （4）令 $x=\cos \theta, y=\sin \theta$ ，则 $$ \begin{aligned} \int_{C} \frac{x^{3} \mathrm{~d} x-y^{3} \mathrm{~d} y}{x^{4}+y^{4}} & =\int_{0}^{2 \pi} \frac{\cos ^{3} \theta(-\sin \theta)-\sin ^{3} \theta \cos \theta}{\cos ^{4} \theta+\sin ^{4} \theta} \mathrm{~d} \theta=-\int_{0}^{2 \pi} \frac{\sin \theta \cos \theta}{\cos ^{4} \theta+\sin ^{4} \theta} \mathrm{~d} \theta \\ & =-\int_{-\pi}^{\pi} \frac{\sin \theta \cos \theta}{\cos ^{4} \theta+\sin ^{4} \theta} \mathrm{~d} \theta=0 \end{aligned} $$