Figure 8.67
旋转面积.

旋转区域是图所示的阴影区域。 8.67.

(a) 旋转区域的面积由下式给出

$$\begin{aligned} A &= \int_2^5 y , dx = \int_2^5 (x - 2)^{1/2} , dx \ &= \left[ \frac{2}{3}(x - 2)^{3/2} \right]_2^5 = 2\sqrt{3} \text{ square units} \end{aligned}$$

如果我们把该面积的形心坐标记为 $(\bar{x}, \bar{y})$ then, from (8.37),

$$\begin{aligned} \bar{x} &= \frac{1}{A} \int_2^5 xy , dx = \frac{1}{A} \int_2^5 x(x - 2)^{1/2} , dx = \frac{1}{A} \int_2^5 [(x - 2)^{3/2} + 2(x - 2)^{1/2}] , dx \ &= \frac{1}{A} \left[ \frac{2}{5}(x - 2)^{5/2} + \frac{4}{3}(x - 2)^{3/2} \right]_2^5 = \frac{1}{A} \left[ \frac{2}{5}(3)^{5/2} + \frac{4}{3}(3)^{3/2} \right] = \frac{1}{A} \frac{38}{5} \sqrt{3} \end{aligned}$$

代入该值 $A = 2\sqrt{3}$ 前面得到的给出 $\bar{x} = \frac{19}{5}$ .

Likewise, from (8.40),

$$\begin{aligned} \bar{y} &= \frac{1}{A} \int_2^5 \frac{1}{2} y^2 , dx = \frac{1}{A} \int_2^5 \frac{1}{2} (x - 2) , dx \ &= \frac{1}{A} [\frac{1}{4}(x - 2)^2]_2^5 = \frac{9}{4A} \end{aligned}$$

Inserting $A = 2\sqrt{3}$ 则给出 $\bar{y} = \frac{3}{8}\sqrt{3}$ 从而质心的坐标为 $(\frac{19}{5}, \frac{3}{8}\sqrt{3})$ .

(b) From (8.36) 体积 $V$ 所形成的旋转体的体积为

$$\begin{aligned} V &= \pi \int_2^5 y^2 , dx = \pi \int_2^5 (x - 2) , dx \ &= \pi [\frac{1}{2}x^2 - 2x]_2^5 = \frac{9}{2}\pi \text{ cubic units} \end{aligned}$$

如果我们用

（原文此处不完整，仅提供了句子开头，无法继续翻译后续内容。） $(\bar{X}, \bar{Y})$ then, from (8.41) and (8.42),

$$\bar{Y} = 0$$

and

$$\begin{aligned} \bar{X} &= \frac{\pi}{V} \int_2^5 xy^2 , dx = \frac{\pi}{V} \int_2^5 x(x - 2) , dx \ &= \frac{\pi}{V} [\frac{1}{3}x^3 - x^2]_2^5 = \frac{\pi}{V} [(\frac{125}{3} - 25) - (\frac{8}{3} - 4)] \ &= \frac{18\pi}{V} \end{aligned}$$

代入该值 $V = \frac{9}{2}\pi$ 前面得到的给出 $\bar{X} = 4$ 使得重心的坐标为 $(4, 0)$ .