人邮高数 第7章 第7-1-21题

[AI解答]

[AI解答]

首先，观察积分区域 $D$ 由两个圆和两条直线围成。 圆方程： $$ x^2 + y^2 = 2y \quad\Rightarrow\quad x^2 + (y-1)^2 = 1 $$ $$ x^2 + y^2 = 4y \quad\Rightarrow\quad x^2 + (y-2)^2 = 4 $$ 直线方程： $$ x - \sqrt{3}y = 0 \quad\Rightarrow\quad \theta = \frac{\pi}{6} $$ $$ y - \sqrt{3}x = 0 \quad\Rightarrow\quad \tan\theta = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3} $$ 因此区域在极坐标下表示为： $$ r^2 = x^2 + y^2,\quad x = r\cos\theta,\ y = r\sin\theta $$ 圆方程变为： $$ r^2 = 2r\sin\theta \Rightarrow r = 2\sin\theta $$ $$ r^2 = 4r\sin\theta \Rightarrow r = 4\sin\theta $$ 角度范围：$\displaystyle \frac{\pi}{6} \le \theta \le \frac{\pi}{3}$。 于是积分化为： $$ \iint_D (x^2+y^2)\,dxdy = \int_{\theta=\frac{\pi}{6}}^{\frac{\pi}{3}} \int_{r=2\sin\theta}^{4\sin\theta} r^2 \cdot r\, dr d\theta $$ $$ = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \int_{2\sin\theta}^{4\sin\theta} r^3 dr \right) d\theta $$ 先计算内层积分： $$ \int_{2\sin\theta}^{4\sin\theta} r^3 dr = \left[ \frac{r^4}{4} \right]_{2\sin\theta}^{4\sin\theta} = \frac{1}{4}\left( (4\sin\theta)^4 - (2\sin\theta)^4 \right) $$ $$ = \frac{1}{4}\left( 256\sin^4\theta - 16\sin^4\theta \right) = \frac{1}{4} \cdot 240 \sin^4\theta = 60\sin^4\theta $$ 于是原积分： $$ = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 60\sin^4\theta \, d\theta $$ 利用倍角公式降幂： $$ \sin^4\theta = \left( \frac{1-\cos2\theta}{2} \right)^2 = \frac{1}{4}(1 - 2\cos2\theta + \cos^2 2\theta) $$ $$ = \frac{1}{4}\left(1 - 2\cos2\theta + \frac{1+\cos4\theta}{2}\right) = \frac{1}{4}\left( \frac{3}{2} - 2\cos2\theta + \frac{1}{2}\cos4\theta \right) $$ $$ = \frac{3}{8} - \frac{1}{2}\cos2\theta + \frac{1}{8}\cos4\theta $$ 因此： $$ 60\sin^4\theta = 60\left( \frac{3}{8} - \frac{1}{2}\cos2\theta + \frac{1}{8}\cos4\theta \right) = \frac{45}{2} - 30\cos2\theta + \frac{15}{2}\cos4\theta $$ 积分得： $$ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{45}{2} - 30\cos2\theta + \frac{15}{2}\cos4\theta \right) d\theta $$ $$ = \left[ \frac{45}{2}\theta - 15\sin2\theta + \frac{15}{8}\sin4\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} $$ 分别代入上下限： 在 $\theta = \frac{\pi}{3}$： $$ \frac{45}{2}\cdot\frac{\pi}{3} = \frac{15\pi}{2},\quad \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2},\quad \sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2} $$ 得： $$ \frac{15\pi}{2} - 15\cdot\frac{\sqrt{3}}{2} + \frac{15}{8}\left(-\frac{\sqrt{3}}{2}\right) = \frac{15\pi}{2} - \frac{15\sqrt{3}}{2} - \frac{15\sqrt{3}}{16} $$ 在 $\theta = \frac{\pi}{6}$： $$ \frac{45}{2}\cdot\frac{\pi}{6} = \frac{15\pi}{4},\quad \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2},\quad \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2} $$ 得： $$ \frac{15\pi}{4} - 15\cdot\frac{\sqrt{3}}{2} + \frac{15}{8}\cdot\frac{\sqrt{3}}{2} = \frac{15\pi}{4} - \frac{15\sqrt{3}}{2} + \frac{15\sqrt{3}}{16} $$ 上下限相减： $$ \left( \frac{15\pi}{2} - \frac{15\sqrt{3}}{2} - \frac{15\sqrt{3}}{16} \right) - \left( \frac{15\pi}{4} - \frac{15\sqrt{3}}{2} + \frac{15\sqrt{3}}{16} \right) $$ $$ = \frac{15\pi}{2} - \frac{15\pi}{4} + \left( -\frac{15\sqrt{3}}{2} + \frac{15\sqrt{3}}{2} \right) + \left( -\frac{15\sqrt{3}}{16} - \frac{15\sqrt{3}}{16} \right) $$ $$ = \frac{15\pi}{4} - \frac{30\sqrt{3}}{16} = \frac{15\pi}{4} - \frac{15\sqrt{3}}{8} $$ 因此最终结果为： $$ \boxed{\displaystyle \frac{15\pi}{4} - \frac{15\sqrt{3}}{8}} $$

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