习题3.3
第 1 题
用二重积分的几何意义求下列二重积分的值.
(1) \(\iint\limits_{D}\sqrt{R^2 - x^2 - y^2}\text{d}{x}\text{d}{y}, D = \{(x, y) | x^2 + y^2 \le R^2\}\);
(2) \(\iint\limits_{D}\sqrt{x^2 + y^2}\text{d}{x}\text{d}{y}, D = \{(x, y) | x^2 + y^2 \le R^2\}\);
(3) \(\iint\limits_{D}\text{d}{x}\text{d}{y}, D = \{(x, y) | \left|x\right| + \left|y\right| \le 1\}\);
解
(1) 这个积分是半径为 \(R\) 的上半球的体积, 因此 \(\iint\limits_{D}\sqrt{R^2 - x^2 - y^2}\text{d}{x}\text{d}{y} = \dfrac{2}{3}\pi R^3\).
(2) 这个积分是底面半径为 \(R\), 高度为 \(R\) 的圆锥的体积, 因此 \(\iint\limits_{D}\sqrt{x^2 + y^2}\text{d}{x}\text{d}{y} = \dfrac{2}{3}\pi R^3\).
(3) 这个积分是一个对角线长度为 \(2\) 的正方形的面积, 因此 \(\iint\limits_{D}\text{d}{x}\text{d}{y} = 2\).
第 2 题
计算下列二重积分.
(1) \(\iint\limits_{I}\dfrac{x^2}{1 + y^2}\text{d}{x}\text{d}{y}, I = [0, 1]^2\);
(2) \(\iint\limits_{I}x\cos(xy)\text{d}{x}\text{d}{y}, I = \left[0, \dfrac{\pi}{2}\right] \times [0, 1]\);
(3) \(\iint\limits_{I}\sin(x + y)\text{d}{x}\text{d}{y}, I = [0, \pi]^2\).
解
(1)
\[
\begin{aligned}
\iint\limits_{I}\dfrac{x^2}{1 + y^2}\text{d}{x}\text{d}{y} & = \int_{0}^{1}\text{d}{x}\int_{0}^{1}\dfrac{x^2}{1 + y^2}\text{d}{y} \\
& = \int_{0}^{1}x^2\arctan y\bigg\vert_{y=0}^{y=1}\text{d}{x} \\
& = \dfrac{\pi}{4}\int_{0}^{1}x^2\text{d}{x} \\
& = \dfrac{\pi}{4}\cdot \dfrac{1}{3}x^3\bigg\vert_{0}^{1} \\
& = \dfrac{\pi}{12}
\end{aligned}
\]
(2)
\[
\begin{aligned}
\iint\limits_{I}x\cos(xy)\text{d}{x}\text{d}{y} & = \int_{0}^{\frac{\pi}{2}}\text{d}{x}\int_{0}^{1}x\cos(xy)\text{d}{y} \\
& = \int_{0}^{\frac{\pi}{2}}\sin(xy)\bigg\vert_{y=0}^{y=1}\text{d}{x} \\
& = \int_{0}^{\frac{\pi}{2}}\sin x\text{d}{x} \\
& = -\cos x\bigg\vert_{0}^{\frac{\pi}{2}} \\
& = 1
\end{aligned}
\]
(3)
\[
\begin{aligned}
\iint\limits_{I}\sin(x + y)\text{d}{x}\text{d}{y} & = \int_{0}^{\pi}\text{d}{x}\int_{0}^{\pi}\sin(x + y)\text{d}{y} \\
& = \int_{0}^{\pi}-\cos(x + y)\bigg\vert_{y=0}^{y=\pi}\text{d}{x} \\
& = 0
\end{aligned}
\]
第 3 题
设函数 \(f(x, y)\) 在 \(I = [a, b] \times [c, d]\) 上有连续的二阶偏导数, 计算\(\iint\limits_{I}\dfrac{\partial ^2 f}{\partial x\partial y}\text{d}{x}\text{d}{y}\).
解
\[
\begin{aligned}
\iint\limits_{I}\dfrac{\partial ^2 f}{\partial x\partial y}\text{d}{x}\text{d}{y} & = \int_{a}^{b}\text{d}{x}\int_{c}^{d}\dfrac{\partial ^2 f}{\partial x\partial y}\text{d}{y} \\
& = \int_{a}^{b}\dfrac{\partial f}{\partial x}(x, y)\bigg\vert_{y=c}^{y=d}\text{d}{x} \\
& = \int_{a}^{b}\left(\dfrac{\partial f(x,d)}{\partial x} - \dfrac{\partial f(x, c)}{\partial x}\right)\text{d}{x} \\
& = f(x, d)\bigg\vert_{a}^{b} - f(x, c)\bigg\vert_{a}^{b} \\
& = f(b, d) - f(a, d) - f(b, c) + f(a, c)
\end{aligned}
\]
第 4 题
将二重积分 \(\iint\limits_{D}f(x, y)\text{d}{x}\text{d}{y}\) 化为累次积分.
(1) \(D = \{(x, y) | x + y \le 1, y - x \le 1, y \ge 0\}\);
(2) \(D = \{(x, y) | y \ge x - 2, x \ge y^2\}\);
(3) \(D = \{(x, y) | \dfrac{2}{x} \le y \le 2x, 1 \le x \le 2\}\).
解
(1) 由题知 \(y-1\le x \le 1-y\), 因此 \(y \in [0, 1]\). 故 \(\iint\limits_{D}f(x, y)\text{d}{x}\text{d}{y} = \int_{0}^{1}\text{d}{y}\int_{y-1}^{1-y}f(x, y)\text{d}{x}\).
(2) 由题知 \(y^2 \le x \le y + 2\), 因此 \(y \in [-1, 2]\). 故 \(\iint\limits_{D}f(x, y)\text{d}{x}\text{d}{y} = \int_{-1}^{2}\text{d}{y}\int_{y^2}^{y+2}f(x, y)\text{d}{x}\).
(3) 按题目所给约束, 有 \(\iint\limits_{D}f(x, y)\text{d}{x}\text{d}{y} = \int_{1}^{2}\text{d}{x}\int_{\frac{2}{x}}^{2x}f(x, y)\text{d}{y}\).
第 5 题
在直角坐标系中画出下列积分的积分区域, 并交换积分次序.
(1) \(\int_{-1}^{0}\text{d}{x}\int_{0}^{1 + x}f(x, y)\text{d}{y} + \int_{0}^{1}\text{d}{x}\int_{0}^{1 - x}f(x, y)\text{d}{y}\);
(2) \(\int_{0}^{1}\text{d}{x}\int_{2\sqrt{1-x}}^{\sqrt{4-x^2}}f(x, y)\text{d}{y} + \int_{1}^{2}\text{d}{x}\int_{0}^{\sqrt{4-x^2}}f(x, y)\text{d}{y}\);
(3) \(\int_{0}^{1}\text{d}{x}\int_{0}^{\sqrt{2x-x^2}}f(x, y)\text{d}{y} + \int_{1}^{2}\text{d}{x}\int_{0}^{2-x}f(x, y)\text{d}{y}\);
解
(1)
\[\begin{figure}[!htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
samples=100,
%grid,
thick,
domain=-1:1,
%legend pos=outer north east,
smooth,
xmin=-1.1, xmax=1.1, ymin=0, ymax=1.1,
xlabel={$x$},
ylabel={$y$},
mark=halfcircle*,
mark size=2.9pt,
]
\addplot+[no marks,domain=0:1]{1-x};
\addplot+[no marks,domain=-1:0]{1+x};
%\addplot+[sharp plot] coordinates {(3, 5.6569)};
%\addplot+[sharp plot] coordinates {(3, -5.6569)};
\end{axis}
\end{tikzpicture}
%\put(0, 0){O}
\caption{习题3.3第5题(1)}
%\label{}
\end{figure}\]
由图知 \(\int_{-1}^{0}\text{d}{x}\int_{0}^{1 + x}f(x, y)\text{d}{y} + \int_{0}^{1}\text{d}{x}\int_{0}^{1 - x}f(x, y)\text{d}{y} = \int_{0}^{1}\text{d}{y}\int_{y-1}^{1-y}f(x, y)\text{d}{x}\).
(2)
\[\begin{figure}[!htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
samples=100,
%grid,
thick,
domain=-0.1:2.1,
%legend pos=outer north east,
smooth,
xmin=0, xmax=2.1, ymin=0, ymax=2.1,
xlabel={$x$},
ylabel={$y$},
mark=halfcircle*,
mark size=2.9pt,
]
\addplot+[no marks,domain=0:2]{sqrt(4-x^2)};
\addplot+[no marks,domain=0:1]{2*sqrt(1-x)};
%\addplot+[sharp plot] coordinates {(3, 5.6569)};
%\addplot+[sharp plot] coordinates {(3, -5.6569)};
\end{axis}
\end{tikzpicture}
%\put(0, 0){O}
\caption{习题3.3第5题(2)}
%\label{}
\end{figure}\]
由图知 \(\int_{0}^{1}\text{d}{x}\int_{2\sqrt{1-x}}^{\sqrt{4-x^2}}f(x, y)\text{d}{y} + \int_{1}^{2}\text{d}{x}\int_{0}^{\sqrt{4-x^2}}f(x, y)\text{d}{y} = \int_{0}^{2}\text{d}{y}\int_{1-\frac{1}{4}y^2}^{\sqrt{4-y^2}}f(x,y)\text{d}{x}\).
(3)
\[\begin{figure}[!htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
samples=100,
%grid,
thick,
domain=0:2,
%legend pos=outer north east,
smooth,
xmin=0, xmax=2.1, ymin=0, ymax=1.1,
xlabel={$x$},
ylabel={$y$},
mark=halfcircle*,
mark size=2.9pt,
]
\addplot+[no marks,domain=1:2]{2-x};
\addplot+[no marks,domain=0:1]{sqrt(2*x-x^2)};
%\addplot+[sharp plot] coordinates {(3, 5.6569)};
%\addplot+[sharp plot] coordinates {(3, -5.6569)};
\end{axis}
\end{tikzpicture}
%\put(0, 0){O}
\caption{习题3.3第5题(3)}
%\label{}
\end{figure}\]
由图知 \(\int_{0}^{1}\text{d}{x}\int_{0}^{\sqrt{2x-x^2}}f(x, y)\text{d}{y} + \int_{1}^{2}\text{d}{x}\int_{0}^{2-x}f(x, y)\text{d}{y} = \int_{0}^{1}\text{d}{y}\int_{1-\sqrt{1-y^2}}^{2-y}f(x,y)\text{d}{x}\).
第 6 题
计算下列二重积分.
(1) \(\iint\limits_{D}xy^2\text{d}{x}\text{d}{y}, D = \{(x, y) | 4x \ge y^2, x \le 1\}\);
(3) \(\iint\limits_{D}\left|xy\right|\text{d}{x}\text{d}{y}, D = \{(x, y) | x^2+y^2\le R^2\}\);
(5) \(\iint\limits_{D}(x^2 + y^2)\text{d}{x}\text{d}{y}\), \(D\) 是以 \(y=x, y=x+1, y=1, y=4\) 为边的平行四边形区域;
(7) \(\iint\limits_{D}\cos(x+y)\text{d}{x}\text{d}{y}, D = \{(x, y) | 0 \le x, y \le \pi\}\);
(9) \(\iint\limits_{D}y^2\text{d}{x}\text{d}{y}\), \(D\) 由 \(\begin{cases}x=a(t-\sin t), \\ y = a(1-\cos t),\end{cases}0\le t \le 2\pi\) 以及 \(x\) 轴围成;
解
(1)
\[
\begin{aligned}
\iint\limits_{D}xy^2\text{d}{x}\text{d}{y} & = \int_{-2}^{2}\text{d}{y}\int_{\frac{1}{4}y^2}^{1}xy^2\text{d}{x} \\
& = \int_{-2}^{2}\dfrac{1}{2}x^2y^2\bigg\vert_{x=\frac{1}{4}y^2}^{x=1}\text{d}{y} \\
& = \int_{-2}^{2}\left(\dfrac{1}{2}y^2-\dfrac{1}{32}y^6\right)\text{d}{y} \\
& = \left(\dfrac{1}{6}y^3 - \dfrac{1}{224}y^7\right)\bigg\vert_{-2}^2 \\
& = \dfrac{32}{21}
\end{aligned}
\]
(2) 考虑 \(D_{\rho, \varphi} = \{(\rho, \varphi) | 0 \le \rho \le R, 0 \le \phi \le \dfrac{\pi}{2}\}\), 则由对称性可知 \(\iint\limits_{D}\left|xy\right|\text{d}{x}\text{d}{y} = 4\iint\limits_{D_{\rho,\varphi}}\rho^2\sin\varphi\cos\varphi\rho\text{d}{\rho}\text{d}{\varphi}\).
\[
\begin{aligned}
\iint\limits_{D_{\rho,\varphi}}\rho^2\sin\varphi\cos\varphi\rho\text{d}{\rho}\text{d}{\varphi} & = \int_{0}^{\frac{\pi}{2}}\dfrac{1}{8}\sin2\varphi\rho^4\bigg\vert_{\rho = 0}^{\rho = R}\text{d}{\varphi} \\
& = \dfrac{R^4}{8}\int_{0}^{\frac{\pi}{2}}\sin2\varphi\text{d}{\varphi} \\
& = \dfrac{R^4}{16}(-\cos2\varphi)\bigg\vert_{0}^{\frac{\pi}{2}} \\
& = \dfrac{R^4}{8}
\end{aligned}
\]
故 \(\iint\limits_{D}\left|xy\right|\text{d}{x}\text{d}{y} = 4\iint\limits_{D_{\rho,\varphi}}\rho^2\sin\varphi\cos\varphi\rho\text{d}{\rho}\text{d}{\varphi} = \dfrac{R^4}{2}\).
(5)
\[
\begin{aligned}
\iint\limits_{D}(x^2 + y^2)\text{d}{x}\text{d}{y} & = \int_{1}^{4}text{d}{y}\int_{y-1}^{y}(x^2+y^2)\text{d}{x} \\
& = \int_{1}^{4}\left(\dfrac{1}{3}x^3+xy^2\right)\bigg\vert_{x=y-1}^{x=y}\text{d}{y} \\
& = \int_{1}^{4}\left(2y^2-y + \dfrac{1}{3}\right)\text{d}{y} \\
& = \left(\dfrac{2}{3}y^3 - \dfrac{1}{2}y^2 + \dfrac{1}{3}y\right)\bigg\vert_{1}^{4} \\
& = \dfrac{71}{2}
\end{aligned}
\]
(7)
\[
\begin{aligned}
\iint\limits_{D}\cos(x+y)\text{d}{x}\text{d}{y} & = \int_{0}^{\pi}\text{d}{x}\int_{0}^{\pi}\cos(x + y)\text{d}{y} \\
& = \int_{0}^{\pi}\sin(x + y)\bigg\vert_{y=0}^{y=\pi}\text{d}{x} \\
& = \int_{0}^{\pi}2\sin x\text{d}{x} \\
& = 2\cos x\bigg\vert_{0}^{\pi} \\
& = -4
\end{aligned}
\]
(9)
\[
\begin{aligned}
\iint\limits_{D}y^2\text{d}{x}\text{d}{y} & = \int_{0}^{2\pi a}\text{d}{x}\int_{0}^{y(x)}y^2\text{d}{y} \\
& = \int_{0}^{2\pi a}\dfrac{1}{3}{y^3}\bigg\vert_{y=0}^{y=y(x)}\text{d}{x} \\
& = \dfrac{1}{3}\int_{0}^{2\pi a}y^3(x)\text{d}{x} \\
& = \dfrac{1}{3}\int_{0}^{2\pi}a^3(1-\cos t)^3a(1-\cos t)\text{d}{t} \\
& = \dfrac{16a^4}{3}\int_{0}^{2\pi}\sin^8\dfrac{t}{2}\text{d}{t} \\
& = \dfrac{32a^4}{3}\int_{0}^{2\pi}\sin^8 u\text{d}{u} \\
& = \dfrac{64a^4}{3}\int_{0}^{\frac{\pi}{2}}\sin^8 u\text{d}{u} \\
& = \dfrac{64a^4}{3}\cdot\dfrac{\pi}{2}\cdot\dfrac{7!!}{8!!} \\
& = \dfrac{35}{12}\pi a^4
\end{aligned}
\]
第 8 题
利用第7题结论, 计算积分
\(\iint\limits_{D}x^2y^3\text{d}{x}\text{d}{y}\), \(\iint\limits_{D}\sqrt{R^2-x^2}\sin y\text{d}{x}\text{d}{y}\), \(D = \{(x, y) | x^2+y^2\le R^2\}\).
解
由于 \(D\) 关于 \(x\) 轴对称且关于 \(y\) 轴对称, 且第一个被积函数关于 \(y\) 为奇函数, 所以第一个积分式为 \(0\). 同理, 第二个被积函数关于 \(y\) 也为奇函数, 所以第二个积分式也为 \(0\). 此即 \(\iint\limits_{D}x^2y^3\text{d}{x}\text{d}{y} = \iint\limits_{D}\sqrt{R^2-x^2}\sin y\text{d}{x}\text{d}{y} = 0\).
第 9 题
分别求出由平面 \(z=x-y,z=0\) 与圆柱面 \(x^2+y^2=2x\) 所围成的两个空间几何体的体积.
解
记 \(D_1 = \{x^2+y^2\le 2x, y \ge x\}, D_2 = \{0 \le x \le 1, 0 \le y \le x\}, D_3 = \{x^2+y^2\le 2x, x\ge1\lor y \le 0\}\). 容易验证总积分区域 \(D = D_1 + D_2 + D_3\).
\[
\begin{aligned}
\iint\limits_{D_2}z\text{d}{x}\text{d}{y} & = \int_{0}^{1}\text{d}{x}\int_{0}^{x}(x-y)\text{d}{y} \\
& = \dfrac{1}{2}\int_{0}^{1}x^2\text{d}{x} \\
& = \dfrac{1}{6}
\end{aligned}
\]
\[
\begin{aligned}
\iint\limits_{D_1+D_2}z\text{d}{x}\text{d}{y} & = \int_{\frac{\pi}{2}}^{\pi}\text{d}{\theta}\int_{0}^{1}r(1+r\cos\theta-r\sin\theta)\text{d}{r} \\
& = \int_{\frac{\pi}{2}}^{\pi}\left(\dfrac{1}{2} + \dfrac{1}{3}\cos\theta - \dfrac{1}{3}\sin\theta\right)\text{d}{\theta} \\
& = \dfrac{\pi}{4} -\dfrac{2}{3}
\end{aligned}
\]
\[
\begin{aligned}
\iint\limits_{D_3}z\text{d}{x}\text{d}{y} & = \int_{-\pi}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{1}r(1+r\cos\theta-r\sin\theta)\text{d}{r} \\
& = \int_{-\pi}^{\frac{\pi}{2}}\left(\dfrac{1}{2} + \dfrac{1}{3}\cos\theta - \dfrac{1}{3}\sin\theta\right)\text{d}{\theta} \\
& = \dfrac{2}{3} + \dfrac{3\pi}{4}
\end{aligned}
\]
因此 \(V_1 = \int\limits_{D_1}z\text{d}{x}\text{d}{y} = \int\limits_{D_1+D_2}z\text{d}{x}\text{d}{y} - \iint\limits_{D_2}z\text{d}{x}\text{d}{y} = \dfrac{5}{6} - \dfrac{\pi}{4}, V_2 = \int\limits_{D_2+D_3}z\text{d}{x}\text{d}{y} = \int\limits_{D_2}z\text{d}{x}\text{d}{y} + \iint\limits_{D_3}z\text{d}{x}\text{d}{y} = \dfrac{5}{6} + \dfrac{3\pi}{4}\).
第 10 题
求由旋转抛物面 \(z = x^2+y^2\), 柱面 \(y=x^2\) 及平面 \(y=1,z=0\) 围成的空间几何体的体积.
解
体积为
\[
\begin{aligned}
V & = \int_{-1}^{1}\text{d}{x}\int_{x^2}^{1}(x^2+y^2)\text{d}{y} \\
& = \int_{-1}^{1}\left(\dfrac{1}{3} + x^2 - x^4 - \dfrac{1}{3}x^6\right)\text{d}{x} \\
& = \dfrac{88}{105}
\end{aligned}
\]
第 11 题
画出下列积分区域的图形, 并将二重积分 \(\iint\limits_{D}f(x, y)\text{d}{x}\text{d}{y}\) 化为极坐标下的累次积分.
(1) \(D = \{(x, y) | x^2 + y^2 \le 1, x^2 + (y-1)^2 \le 1\}\);
(2) \(D = \{(x, y) | x^2 + (y-a)^2 \le a^2, (x - a)^2 + y^2 \le a^2\}\).
解
(1) 由图知 \(\iint\limits_{D}f(x, y)\text{d}{x}\text{d}{y} = \int_{0}^{\frac{\pi}{6}}\text{d}{\theta}\int_{0}^{2\sin\theta}f(r\cos\theta, r\sin\theta)r\text{d}{r} + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{2}f(r\cos\theta, r\sin\theta)r\text{d}{r}\).
\[\begin{figure}[!htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
samples=100,
%grid,
thick,
domain=-0.1:1.1,
%legend pos=outer north east,
smooth,
xmin=0, xmax=1.1, ymin=0, ymax=1.1,
xlabel={$x$},
ylabel={$y$},
mark=halfcircle*,
mark size=2.9pt,
]
\addplot+[no marks,domain=0:sqrt(3)/2]{sqrt(1-x^2)};
\addplot+[no marks,domain=0:sqrt(3)/2]{1-sqrt(1-x^2)};
%\addplot+[sharp plot] coordinates {(3, 5.6569)};
%\addplot+[sharp plot] coordinates {(3, -5.6569)};
\end{axis}
\end{tikzpicture}
%\put(0, 0){O}
\caption{习题3.3第11题(1)}
%\label{}
\end{figure}\]
(2) 由图知 \(\iint\limits_{D}f(x, y)\text{d}{x}\text{d}{y} = \int_{0}^{\frac{\pi}{4}}\text{d}{\theta}\int_{0}^{2a\sin\theta}f(r\cos\theta, r\sin\theta)r\text{d}{r} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{2\cos\theta}f(r\cos\theta, r\sin\theta)r\text{d}{r}\).
\[\begin{figure}[!htbp]
\centering
\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
samples=100,
%grid,
thick,
domain=-0.1:1.1,
%legend pos=outer north east,
smooth,
xmin=0, xmax=1.1, ymin=0, ymax=1.1,
xlabel={$x$},
ylabel={$y$},
mark=halfcircle*,
mark size=2.9pt,
]
\addplot+[no marks,domain=0:1]{1-sqrt(1-x^2)};
\addplot+[no marks,domain=0:1]{sqrt(1-(x-1)^2)};
%\addplot+[sharp plot] coordinates {(3, 5.6569)};
%\addplot+[sharp plot] coordinates {(3, -5.6569)};
\end{axis}
\end{tikzpicture}
%\put(0, 0){O}
\caption{习题3.3第11题(2), 此时取 a=1}
%\label{}
\end{figure}\]
第 12 题
计算下列二重积分.
(1) \(\iint\limits_{D}(x^2 + y^2)\text{d}{x}\text{d}{y}, D = \{(x, y) | 2x \le x^2 + y^2 \le 4x\}\);
(3) \(\iint\limits_{D}(x+y)\text{d}{x}\text{d}{y}\), \(D\) 是由 \(x^2+y^2=x+y\) 围成的平面区域;
(5) \(\iint\limits_{D}\arctan\dfrac{y}{x}\text{d}{x}\text{d}{y}, D = \{(x, y) | x^2 + y^2 \le 1, x \le 0, y \le 0\}\);
解
(1) 代换 \(x = r\cos\theta, y = r\sin\theta\), 则区域变为 \(D_{r, \theta} = \{2\cos\theta \le r \le 4\cos\theta, \theta \in [0, \pi]\}\). 则
\[
\begin{aligned}
\iint\limits_{D}(x^2 + y^2)\text{d}{x}\text{d}{y} & = \iint\limits_{D_{r,\theta}}r^2\cdot r\text{d}{r}\text{d}{\theta} \\
& = \int_{0}^{\pi}\text{d}{\theta}\int_{2\cos\theta}^{4\cos\theta}r^3\text{d}{r} \\
& = \int_{0}^{\pi}60\cos^4\theta\text{d}{\theta} \\
& = 120 \int_{0}^{\frac{\pi}{2}}\cos^4\theta\text{d}{\theta} \\
& = 120\cdot\dfrac{\pi}{2}\cdot\dfrac{3\cdot 1}{4\cdot 2} \\
& = \dfrac{45}{2}\pi
\end{aligned}
\]
(3) 做坐标变换 \(\begin{cases}x = r\cos\theta, \\ y = r\sin\theta,\end{cases}\) 则有
\[
\begin{aligned}
\iint\limits_{D}(x+y)\text{d}{x}\text{d}{y} & = \int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_{0}^{\sin\theta + \cos\theta}r^2(\sin\theta + \cos\theta)\text{d}{r} \\
& = \dfrac{1}{3}\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}(\sin\theta + \cos\theta)^4\text{d}{\theta} \\
& = \dfrac{1}{3}\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\left(\dfrac{3-\cos4\theta}{2} + 2\sin2\theta\right)\text{d}{\theta} \\
& = \dfrac{1}{3}\cdot\dfrac{3\pi}{2} \\
& = \dfrac{\pi}{2}
\end{aligned}
\]
(5) 做坐标变换 \(\begin{cases}x = r\cos\theta, \\ y = r\sin\theta,\end{cases}\) 则有
\[
\begin{aligned}
\iint\limits_{D}\arctan\dfrac{y}{x}\text{d}{x}\text{d}{y} & = \int_{0}^{1}r\text{d}{r}\int_{0}^{\frac{\pi}{2}}\theta\text{d}{\theta} \\
& = \dfrac{1}{2}\cdot\dfrac{\pi^2}{8} \\
& = \dfrac{\pi^2}{16}
\end{aligned}
\]
第 13 题
求下列曲线所围图形的面积.
(1) 双纽线 \((x^2 + y^2)^2 = 2a^2(x^2-y^2)\) 与圆 \(x^2 + y^2 = a^2\) 所围图形 (圆外部分) 的面积 \((a > 0)\);
(2) 心脏线 \(r = a(1 + \cos\theta)\) 与圆 \(x^2 + y^2 = \sqrt{3}ay\) 所围图形 (心脏线内部) 的面积 \((a > 0)\).
解
(1) 考虑 \(D = \{a \le \rho \le a\sqrt{2\cos 2\theta}, -\dfrac{\pi}{6} \le \theta \le \dfrac{\pi}{6} \lor \dfrac{5\pi}{6} \le \theta \le \dfrac{7\pi}{6}\}\). 则面积
\[
\begin{aligned}
S & = \iint\limits_{D}\rho\text{d}{\rho}\text{d}{\theta} \\
& = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\text{d}{\theta}\int_{a}^{a\sqrt{2\cos2\theta}}\rho \text{d}{\rho}\text{d}{\theta} + \int_{\frac{5\pi}{6}}^{\frac{7\pi}{6}}\text{d}{\theta}\int_{a}^{a\sqrt{2\cos2\theta}}\rho \text{d}{\rho}\text{d}{\theta} \\
& = \left(\sqrt{3} - \dfrac{\pi}{3}\right)a^2
\end{aligned}
\]
(2) 考虑 \(D = \{0 \le \rho \le \sqrt{3}a\sin\theta, 0 \le \theta \le \dfrac{\pi}{3}\} \bigcup \{0 \le \rho \le a(1 + \cos\theta), \dfrac{\pi}{3} \le \theta \le \pi\}\). 则面积
\[
\begin{aligned}
S & = \iint\limits_{D}\rho\text{d}{\rho}\text{d}{\theta} \\
& = \int_{0}^{\frac{\pi}{3}}\text{d}{\theta}\int_{0}^{\sqrt{3}a\sin\theta}\rho \text{d}{\rho}\text{d}{\theta} + \int_{\frac{\pi}{3}}^{\pi}\text{d}{\theta}\int_{0}^{a(1+\cos\theta)}\rho \text{d}{\rho}\text{d}{\theta} \\
& = \dfrac{3\pi - 3\sqrt{3}}{4}a^2
\end{aligned}
\]
第 14 题
通过恰当的变量代换, 计算下列二重积分.
(1) \(\iint\limits_{D}x^2y^2\text{d}{x}\text{d}{y}\), \(D\) 是由 \(xy = 2, xy = 4, y=x, y=3x\) 在第一象限所围成的平面区域;
(2) \(\iint\limits_{D}(x^2 + y^2)\text{d}{x}\text{d}{y}\), \(D\) 是由 \(xy=1, xy=2, x^2-y^2=1, x^2-y^2=2\) 所围成的平面区域;
解
(1) 做坐标变换 \((u, v) = (xy, \dfrac{y}{x})\), 则 \(\dfrac{D(x, y)}{D(u, v)} = \dfrac{1}{4v}\). 因此
\[
\begin{aligned}
\iint\limits_{D}x^2y^2\text{d}{x}\text{d}{y} & = \int_{1}^{3}\text{d}{v}\int_{2}^{4}\dfrac{D(x, y)}{D(u, v)}u^2\text{d}{u} \\
& = \int_{1}^{3}\dfrac{1}{4v}\text{d}{v}\int_{2}^{4}u^2\text{d}{u} \\
& = \dfrac{\ln3}{4}\cdot\dfrac{56}{3} \\
& = \dfrac{28\ln3}{3}
\end{aligned}
\]
(2) 做坐标变换 \((u, v) = (x^2-y^2, xy)\), 则 \(\dfrac{D(x, y)}{D(u, v)} = \dfrac{1}{2\sqrt{u^2+4v^2}}\). 因此
\[
\begin{aligned}
\iint\limits_{D}(x^2 + y^2)\text{d}{x}\text{d}{y} & = \int_{1}^{2}\text{d}{v}\int_{1}^{2}\dfrac{D(x, y)}{D(u, v)}\sqrt{u^2+4v^2}\text{d}{u} \\
& = \dfrac{1}{2}\int_{1}^{2}\text{d}{v}\int_{1}^{2}\text{d}{u} \\
& = \dfrac{1}{2}
\end{aligned}
\]
第 15 题
求下列图形围成区域的面积.
(1) \((a_1x + b_1y + c_1)^2 + (a_2x + b_2y + c_2)^2 = 1\), 其中 \(a_1b_2 \neq a_2b_1\);
(2) \(\sqrt{x} + \sqrt{y} = \sqrt{a}\) 与 \(x = 0, y = 0\).
解
(1) 做坐标变换 \((u, v) = (a_1x + b_1y + c_1, a_2x, b_2y, c_2)\), 则 \(\dfrac{D(x, y)}{D(u, v)} = \dfrac{1}{a_1b_2 - a_2b_1}\). 因此 \(S = \iint\limits_{u^2 + v^2 \le 1}\left|\dfrac{D(x, y)}{D(u, v)}\right|\text{d}{u}\text{d}{v} = \dfrac{\pi}{\left|a_1b_2 - a_2b_1\right|}\).
(2) 做坐标变换 \((u, v) = (\sqrt{x}, \sqrt{y})\), 则 \(\dfrac{D(x, y)}{D(u, v)} = 4uv\). 因此 \(S = \iint\limits_{u+v\le \sqrt{a}, u,v\ge 0}\left|\dfrac{D(x, y)}{D(u, v)}\right|\text{d}{u}\text{d}{v} = \iint\limits_{u+v\le\sqrt{a}, u, v\ge 0}4uv\text{d}{u}\text{d}{v} = \dfrac{1}{6}a^2\).
第 16 题
设函数 \(f(t)\) 连续, 证明:
(1) \(\iint\limits_{\left|x\right| + \left|y\right| \le 1}f(x+y)\text{d}{x}\text{d}{y} = \int_{-1}^{1}f(t)\text{d}{t}\).
(2) \(\iint\limits_{D}f(xy)\text{d}{x}\text{d}{y} = \ln 2\int_{1}^{2}f(t)\text{d}{t}\), \(D\) 是由 \(xy = 1, xy = 2, y = x, y = 4x\) 所围成的第一象限的区域.
证明
(1) 做坐标变换 \((u, v) = (x+y, x-y)\), 则 \(\dfrac{D(x, y)}{D(u, v)} = -\dfrac{1}{2}\). 因此
\[
\begin{aligned}
\iint\limits_{\left|x\right| + \left|y\right| \le 1}f(x+y)\text{d}{x}\text{d}{y} & = \int_{-1}^{1}\text{d}{v}\int_{-1}^{1}\left|\dfrac{D(x, y)}{D(u, v)}\right|f(u)\text{d}{u} \\
& = \int_{-1}^{1}f(u)\text{d}{u} \\
& = \int_{-1}^{1}f(t)\text{d}{t}
\end{aligned}
\]
(2) 做坐标变换 \((u, v) = (xy, \dfrac{y}{x})\), 则 \(\dfrac{D(x, y)}{D(u, v)} = \dfrac{1}{4v}\). 因此
\[
\begin{aligned}
\iint\limits_{D}f(xy)\text{d}{x}\text{d}{y} & = \int_{1}^{4}\dfrac{D(x, y)}{D(u, v)}\text{d}{v}\int_{1}^{2}f(u)\text{d}{u} \\
& = \int_{1}^{4}\dfrac{1}{4v}\text{d}{v}\int_{1}^{2}f(u)\text{d}{u} \\
& = \ln2\int_{1}^{2}f(t)\text{d}{t}
\end{aligned}
\]
第 17 题
设函数 \(f(t)\) 连续, \(f(t) > 0\), 求积分 \(\iint\limits_{x^2 + y^2 \le R^2}\dfrac{af(x) + bf(y)}{f(x) + f(y)}\text{d}{x}\text{d}{y}\).
解
由对称性, 记 \(S = \iint\limits_{x^2 + y^2 \le R^2}\dfrac{f(x)}{f(x) + f(y)}\text{d}{x}\text{d}{y} = \iint\limits_{x^2 + y^2 \le R^2}\dfrac{f(y)}{f(x) + f(y)}\text{d}{x}\text{d}{y}\), 则 \(2S = \iint\limits_{x^2 + y^2 \le R^2}\dfrac{f(x) + f(y)}{f(x) + f(y)}\text{d}{x}\text{d}{y} = \pi R^2\). 故 \(S = \dfrac{1}{2}\pi R^2\). 因此 \(\iint\limits_{x^2 + y^2 \le R^2}\dfrac{af(x) + bf(y)}{f(x) + f(y)}\text{d}{x}\text{d}{y} = aS + bS = \dfrac{a+b}{2}\pi R^2\).
第 18 题
设函数 \(f(t, s)\) 连续, 求 \(F(x) = \int_{0}^{x}\int_{t^2}^{x^2}f(t, s)\text{d}{s}\text{d}{t}\) 的导函数.
解
记 \(g(x, t) = \int_{t^2}^{x^2}f(t, s)\text{d}{s}\), 则 \(F(x) = \int_{0}^{x}g(x, t)\text{d}{t}\).
\[
\begin{aligned}
F'(x) & = \int_{0}^{x}\dfrac{\partial g}{\partial x}(x, t)\text{d}{t} + g(x, x) \\
& = \int_{0}^{x} f(t, x^2)\cdot 2x\text{d}{t} + \int_{x^2}^{x^2}f(x, s)\text{d}{s} \\
& = 2x\int_{0}^{x}f(t, x^2)\text{d}{t}
\end{aligned}
\]