题目
利用极坐标计算下列 各 题:-|||-(1) int (int )_({e)^(x^2)+(y)^2}dx, 其中D是由圆周 ^2+(y)^2=4 所围成的闭区域;-|||-(2) iint ln (1+(x)^2+(y)^2)dtheta , 其中D是由圆周 ^2+(y)^2=1 及坐标轴所围成的在第一象限内-|||-的闭区域;-|||-(3) int arctan dfrac (y)(x)dsigma , 其中D是由圆周 ^2+(y)^2=4 ^2+(y)^2=1 及直线 =0, y=x 所围-|||-成的在第一象限内的闭区域.
题目解答
答案
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解析
步骤 1：转换为极坐标
将直角坐标系下的积分转换为极坐标系下的积分。在极坐标系中，$x = r\cos\theta$，$y = r\sin\theta$，$d\sigma = r dr d\theta$。因此，积分可以表示为：
(1) $\iint {e}^{{x}^{2}+{y}^{2}}d\sigma = \int_{0}^{2\pi} \int_{0}^{2} {e}^{r^2} r dr d\theta$
(2) $\iint \ln (1+{x}^{2}+{y}^{2})d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \ln (1+r^2) r dr d\theta$
(3) $\int \arctan \dfrac {y}{x}d\sigma = \int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \arctan \dfrac {r\sin\theta}{r\cos\theta} r dr d\theta$
步骤 2：计算积分
(1) $\int_{0}^{2\pi} \int_{0}^{2} {e}^{r^2} r dr d\theta = \int_{0}^{2\pi} d\theta \int_{0}^{2} {e}^{r^2} r dr = 2\pi \int_{0}^{2} {e}^{r^2} r dr$
令 $u = r^2$，则 $du = 2r dr$，因此 $\int_{0}^{2} {e}^{r^2} r dr = \frac{1}{2} \int_{0}^{4} {e}^{u} du = \frac{1}{2} ({e}^{4} - 1)$
所以，$\int_{0}^{2\pi} \int_{0}^{2} {e}^{r^2} r dr d\theta = \pi ({e}^{4} - 1)$
(2) $\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \ln (1+r^2) r dr d\theta = \int_{0}^{\frac{\pi}{2}} d\theta \int_{0}^{1} \ln (1+r^2) r dr$
令 $u = 1+r^2$，则 $du = 2r dr$，因此 $\int_{0}^{1} \ln (1+r^2) r dr = \frac{1}{2} \int_{1}^{2} \ln u du = \frac{1}{2} (2\ln 2 - 1)$
所以，$\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \ln (1+r^2) r dr d\theta = \frac{\pi}{4} (2\ln 2 - 1)$
(3) $\int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \arctan \dfrac {r\sin\theta}{r\cos\theta} r dr d\theta = \int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \arctan \tan\theta r dr d\theta = \int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \theta r dr d\theta$
$\int_{1}^{2} \theta r dr = \theta \int_{1}^{2} r dr = \theta \left[\frac{r^2}{2}\right]_{1}^{2} = \frac{3}{2}\theta$
所以，$\int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \theta r dr d\theta = \int_{0}^{\frac{\pi}{4}} \frac{3}{2}\theta d\theta = \frac{3}{2} \left[\frac{\theta^2}{2}\right]_{0}^{\frac{\pi}{4}} = \frac{3}{64}{\pi}^{2}$
将直角坐标系下的积分转换为极坐标系下的积分。在极坐标系中，$x = r\cos\theta$，$y = r\sin\theta$，$d\sigma = r dr d\theta$。因此，积分可以表示为：
(1) $\iint {e}^{{x}^{2}+{y}^{2}}d\sigma = \int_{0}^{2\pi} \int_{0}^{2} {e}^{r^2} r dr d\theta$
(2) $\iint \ln (1+{x}^{2}+{y}^{2})d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \ln (1+r^2) r dr d\theta$
(3) $\int \arctan \dfrac {y}{x}d\sigma = \int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \arctan \dfrac {r\sin\theta}{r\cos\theta} r dr d\theta$
步骤 2：计算积分
(1) $\int_{0}^{2\pi} \int_{0}^{2} {e}^{r^2} r dr d\theta = \int_{0}^{2\pi} d\theta \int_{0}^{2} {e}^{r^2} r dr = 2\pi \int_{0}^{2} {e}^{r^2} r dr$
令 $u = r^2$，则 $du = 2r dr$，因此 $\int_{0}^{2} {e}^{r^2} r dr = \frac{1}{2} \int_{0}^{4} {e}^{u} du = \frac{1}{2} ({e}^{4} - 1)$
所以，$\int_{0}^{2\pi} \int_{0}^{2} {e}^{r^2} r dr d\theta = \pi ({e}^{4} - 1)$
(2) $\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \ln (1+r^2) r dr d\theta = \int_{0}^{\frac{\pi}{2}} d\theta \int_{0}^{1} \ln (1+r^2) r dr$
令 $u = 1+r^2$，则 $du = 2r dr$，因此 $\int_{0}^{1} \ln (1+r^2) r dr = \frac{1}{2} \int_{1}^{2} \ln u du = \frac{1}{2} (2\ln 2 - 1)$
所以，$\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \ln (1+r^2) r dr d\theta = \frac{\pi}{4} (2\ln 2 - 1)$
(3) $\int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \arctan \dfrac {r\sin\theta}{r\cos\theta} r dr d\theta = \int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \arctan \tan\theta r dr d\theta = \int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \theta r dr d\theta$
$\int_{1}^{2} \theta r dr = \theta \int_{1}^{2} r dr = \theta \left[\frac{r^2}{2}\right]_{1}^{2} = \frac{3}{2}\theta$
所以，$\int_{0}^{\frac{\pi}{4}} \int_{1}^{2} \theta r dr d\theta = \int_{0}^{\frac{\pi}{4}} \frac{3}{2}\theta d\theta = \frac{3}{2} \left[\frac{\theta^2}{2}\right]_{0}^{\frac{\pi}{4}} = \frac{3}{64}{\pi}^{2}$