题目
设 $A>0$, $D$ 是由曲线段 $y=$ $A \sin x$ $(0 \leqslant x \leqslant \frac{\pi}{2})$ 及直线 $y=0$, $x = \frac{\pi}{2}$ 所围成的平面区域，$V_{1}$, $V_{2}$ 分别表示 $D$ 绕 $x$ 轴与绕 $y$ 轴旋转所成旋转体的体积，若 $V_{1}=V_{2}$, 求 $A$ 的值.
解析
由题可得：
$$
V_{1} = \pi \int_{0}^{\frac{\pi}{2}} y^{2}(x) dx = \pi \int_{0}^{\frac{\pi}{2}} (A \sin x)^{2} dx;
$$
$$
V_{2} = 2 \pi \int_{0}^{\frac{\pi}{2}} x |y(x)| dx = 2 \pi \int_{0}^{\frac{\pi}{2}} A x \sin x dx. ①
$$
注：
[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：
$V_{2} = 2 \pi \int_{0}^{\frac{\pi}{2}} x \cdot | A \sin x| dx.$
于是：
$$
V_{1} =
$$
$$
A^{2} \pi \int_{0}^{\frac{\pi}{2}} \sin^{2}x dx =
$$
$$
\frac{1}{2} A^{2} \pi \int_{0}^{\frac{\pi}{2}} (1-\cos 2x) dx =
$$
$$
\frac{1}{2} A^{2} \pi [\int_{0}^{\frac{\pi}{2}} 1 dx – \int_{0}^{\frac{\pi}{2}} \cos 2x dx ] =
$$
$$
\frac{1}{2} A^{2} \pi [ \frac{\pi}{2} – \frac{1}{2} \sin 2x |_{0}^{\frac{\pi}{2}} ] =
$$
$$
\frac{1}{2} A^{2} \pi [ \frac{\pi}{2} – \frac{1}{2} (0-0)] =
$$
$$
\frac{1}{4} A^{2} \pi^{2}.
$$
$$
V_{2} =
$$
$$
2 A \pi \int_{0}^{\frac{\pi}{2}} x \sin x dx =
$$
$$
-2 A \pi \int_{0}^{\frac{\pi}{2}} x d(\cos x) =
$$
$$
-2 A \pi [ x \cos x |_{0}^{\frac{\pi}{2}} – \int_{0}^{\frac{\pi}{2}} \cos x dx] =
$$
$$
-2 A \pi [ 0 – \int_{0}^{\frac{\pi}{2}} \cos x dx] =
$$
$$
-2 A \pi [ 0 – \sin x|_{0}^{\frac{\pi}{2}}] =
$$
$$
-2A \pi \cdot [-1 \cdot (1-0)] =
$$
$$
-2A \pi \cdot (-1) = 2 A \pi.
$$
又：
$$
V_{1} = V_{2}.
$$
于是：
$$
\frac{1}{4} A^{2} \pi^{2} = 2 A \pi \Rightarrow
$$
$$
\frac{1}{4} A \pi = 2 \Rightarrow
$$
$$
A \cdot \frac{\pi}{4} = 2 \Rightarrow
$$
$$
A = 2 \cdot \frac{4}{\pi} \Rightarrow
$$
$$
A = \frac{8}{\pi}.
$$