# 2015年考研数二第16题解析：定积分、旋转体的体积

## 解析

$$V_{1} = \pi \int_{0}^{\frac{\pi}{2}} y^{2}(x) dx = \pi \int_{0}^{\frac{\pi}{2}} (A \sin x)^{2} dx;$$

$$V_{2} = 2 \pi \int_{0}^{\frac{\pi}{2}} x |y(x)| dx = 2 \pi \int_{0}^{\frac{\pi}{2}} A x \sin x dx. ①$$

[1]. 在 $(0,\frac{\pi}{2})$ 区间内，始终有 $A \sin x > 0$ 存在，因此 $①$ 式不需要写成如下形式：

$V_{2} = 2 \pi \int_{0}^{\frac{\pi}{2}} x \cdot | A \sin x| dx.$

$$V_{1} =$$

$$A^{2} \pi \int_{0}^{\frac{\pi}{2}} \sin^{2}x dx =$$

$$\frac{1}{2} A^{2} \pi \int_{0}^{\frac{\pi}{2}} (1-\cos 2x) dx =$$

$$\frac{1}{2} A^{2} \pi [\int_{0}^{\frac{\pi}{2}} 1 dx – \int_{0}^{\frac{\pi}{2}} \cos 2x dx ] =$$

$$\frac{1}{2} A^{2} \pi [ \frac{\pi}{2} – \frac{1}{2} \sin 2x |_{0}^{\frac{\pi}{2}} ] =$$

$$\frac{1}{2} A^{2} \pi [ \frac{\pi}{2} – \frac{1}{2} (0-0)] =$$

$$\frac{1}{4} A^{2} \pi^{2}.$$

$$V_{2} =$$

$$2 A \pi \int_{0}^{\frac{\pi}{2}} x \sin x dx =$$

$$-2 A \pi \int_{0}^{\frac{\pi}{2}} x d(\cos x) =$$

$$-2 A \pi [ x \cos x |_{0}^{\frac{\pi}{2}} – \int_{0}^{\frac{\pi}{2}} \cos x dx] =$$

$$-2 A \pi [ 0 – \int_{0}^{\frac{\pi}{2}} \cos x dx] =$$

$$-2 A \pi [ 0 – \sin x|_{0}^{\frac{\pi}{2}}] =$$

$$-2A \pi \cdot [-1 \cdot (1-0)] =$$

$$-2A \pi \cdot (-1) = 2 A \pi.$$

$$V_{1} = V_{2}.$$

$$\frac{1}{4} A^{2} \pi^{2} = 2 A \pi \Rightarrow$$

$$\frac{1}{4} A \pi = 2 \Rightarrow$$

$$A \cdot \frac{\pi}{4} = 2 \Rightarrow$$

$$A = 2 \cdot \frac{4}{\pi} \Rightarrow$$

$$A = \frac{8}{\pi}.$$