人邮高数 第7章 第7-3-9题

[AI解答]

[AI解答]

我们要求匀质心形线 $ r = 1 + \cos\theta $ 的上半部分弧（$ 0 \leq \theta \leq \pi $）的质心。 由于是匀质弧线，质心坐标公式为：

$$ \bar{x} = \frac{\displaystyle\int_{L} x \, ds}{\displaystyle\int_{L} ds}, \quad \bar{y} = \frac{\displaystyle\int_{L} y \, ds}{\displaystyle\int_{L} ds} $$

其中弧长微元为：

$$ ds = \sqrt{r^2 + \left( \frac{dr}{d\theta} \right)^2} \, d\theta $$

首先计算 $ r = 1 + \cos\theta $，则

$$ \frac{dr}{d\theta} = -\sin\theta $$

于是

$$ r^2 + \left( \frac{dr}{d\theta} \right)^2 = (1+\cos\theta)^2 + \sin^2\theta = 1 + 2\cos\theta + \cos^2\theta + \sin^2\theta = 2 + 2\cos\theta = 2(1+\cos\theta) $$

因此

$$ ds = \sqrt{2(1+\cos\theta)} \, d\theta $$

利用半角公式 $ 1+\cos\theta = 2\cos^2\frac{\theta}{2} $，得

$$ ds = \sqrt{2 \cdot 2\cos^2\frac{\theta}{2}} \, d\theta = 2\left|\cos\frac{\theta}{2}\right| d\theta $$

在 $ 0 \leq \theta \leq \pi $ 时，$\frac{\theta}{2} \in [0, \frac{\pi}{2}]$，$\cos\frac{\theta}{2} \geq 0$，所以

$$ ds = 2\cos\frac{\theta}{2} \, d\theta $$

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**第一步：求弧长**

$$ L = \int_{L} ds = \int_{0}^{\pi} 2\cos\frac{\theta}{2} \, d\theta = 2 \cdot \left[ 2\sin\frac{\theta}{2} \right]_{0}^{\pi} = 4 \left( \sin\frac{\pi}{2} - \sin 0 \right) = 4 $$

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**第二步：求 $\bar{x}$**

极坐标下 $ x = r\cos\theta = (1+\cos\theta)\cos\theta $，所以

$$ \int_{L} x \, ds = \int_{0}^{\pi} (1+\cos\theta)\cos\theta \cdot 2\cos\frac{\theta}{2} \, d\theta $$

利用 $ 1+\cos\theta = 2\cos^2\frac{\theta}{2} $，得

$$ = \int_{0}^{\pi} 2\cos^2\frac{\theta}{2} \cdot \cos\theta \cdot 2\cos\frac{\theta}{2} \, d\theta = 4 \int_{0}^{\pi} \cos^3\frac{\theta}{2} \cos\theta \, d\theta $$

再用 $\cos\theta = 2\cos^2\frac{\theta}{2} - 1$，得

$$ = 4 \int_{0}^{\pi} \cos^3\frac{\theta}{2} \left( 2\cos^2\frac{\theta}{2} - 1 \right) d\theta = 4 \int_{0}^{\pi} \left( 2\cos^5\frac{\theta}{2} - \cos^3\frac{\theta}{2} \right) d\theta $$

令 $ u = \frac{\theta}{2} $，则 $ d\theta = 2 du $，当 $\theta=0$ 时 $u=0$，$\theta=\pi$ 时 $u=\frac{\pi}{2}$，得

$$ = 4 \int_{0}^{\pi/2} \left( 2\cos^5 u - \cos^3 u \right) \cdot 2 du = 8 \int_{0}^{\pi/2} \left( 2\cos^5 u - \cos^3 u \right) du $$

利用公式 $\displaystyle\int_{0}^{\pi/2} \cos^{2n+1} u \, du = \frac{(2n)!!}{(2n+1)!!}$：

$$ \int_{0}^{\pi/2} \cos^3 u \, du = \frac{2}{3}, \quad \int_{0}^{\pi/2} \cos^5 u \, du = \frac{4 \cdot 2}{5 \cdot 3} = \frac{8}{15} $$

因此

$$ \int_{L} x \, ds = 8 \left( 2 \cdot \frac{8}{15} - \frac{2}{3} \right) = 8 \left( \frac{16}{15} - \frac{10}{15} \right) = 8 \cdot \frac{6}{15} = \frac{48}{15} = \frac{16}{5} $$

所以

$$ \bar{x} = \frac{16/5}{4} = \frac{4}{5} $$

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**第三步：求 $\bar{y}$**

极坐标下 $ y = r\sin\theta = (1+\cos\theta)\sin\theta $，所以

$$ \int_{L} y \, ds = \int_{0}^{\pi} (1+\cos\theta)\sin\theta \cdot 2\cos\frac{\theta}{2} \, d\theta $$

同样用 $ 1+\cos\theta = 2\cos^2\frac{\theta}{2} $，得

$$ = \int_{0}^{\pi} 2\cos^2\frac{\theta}{2} \cdot \sin\theta \cdot 2\cos\frac{\theta}{2} \, d\theta = 4 \int_{0}^{\pi} \cos^3\frac{\theta}{2} \sin\theta \, d\theta $$

利用 $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$，得

$$ = 4 \int_{0}^{\pi} \cos^3\frac{\theta}{2} \cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \, d\theta = 8 \int_{0}^{\pi} \cos^4\frac{\theta}{2} \sin\frac{\theta}{2} \, d\theta $$

令 $ u = \cos\frac{\theta}{2} $，则 $ du = -\frac{1}{2}\sin\frac{\theta}{2} d\theta $，即 $\sin\frac{\theta}{2} d\theta = -2 du$。 当 $\theta=0$ 时 $u=1$，$\theta=\pi$ 时 $u=0$，因此

$$ \int_{L} y \, ds = 8 \int_{1}^{0} u^4 \cdot (-2) du = 16 \int_{0}^{1} u^4 du = 16 \cdot \frac{1}{5} = \frac{16}{5} $$

所以

$$ \bar{y} = \frac{16/5}{4} = \frac{4}{5} $$

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**结论**：质心坐标为

$$ \boxed{\left( \frac{4}{5},\ \frac{4}{5} \right)} $$

难度：★★☆☆☆