区域 $R$ 由直线 $y=0$、$y=\dfrac{5}{2}\sqrt{3}$ 和曲线 $x^2 = 25 - y^2$ 围成。将 $R$ 绕 y 轴旋转 $\pi$ 弧度。利用换元法 $y = 5\sin\theta$，求所得旋转体的准确体积。

1. Set up the volume integral for a rotation of $\pi$ radians.

The solid is generated by rotating the region $R$ through $\pi$ radians (half a revolution) about the $y$-axis. The volume generated is half of the standard volume of revolution for a full $2\pi$ rotation.
$V = \frac{1}{2} \int_{y_1}^{y_2} \pi x^2 \, dy$
$V = \frac{\pi}{2} \int_0^{\frac{5}{2}\sqrt{3}} (25 - y^2) \, dy$

2. Differentiate the substitution and determine the new limits of integration.

Using the given substitution $y = 5\sin\theta$:
$\frac{dy}{d\theta} = 5\cos\theta \implies dy = 5\cos\theta \, d\theta$
Finding the new limits:
When $y = 0$, $5\sin\theta = 0 \implies \theta = 0$
When $y = \frac{5}{2}\sqrt{3}$, $5\sin\theta = \frac{5}{2}\sqrt{3} \implies \sin\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}$

3. Apply the substitution to the volume integral and simplify.

Substitute into the volume integral:
$V = \frac{\pi}{2} \int_0^{\frac{\pi}{3}} \left[25 - (5\sin\theta)^2\right] (5\cos\theta) \, d\theta$
$V = \frac{\pi}{2} \int_0^{\frac{\pi}{3}} 25(1 - \sin^2\theta) (5\cos\theta) \, d\theta$
$V = \frac{125\pi}{2} \int_0^{\frac{\pi}{3}} \cos^2\theta \cdot \cos\theta \, d\theta$
$V = \frac{125\pi}{2} \int_0^{\frac{\pi}{3}} \cos^3\theta \, d\theta$

4. Evaluate the trigonometric integral to find the exact volume.

Evaluate the integral by splitting $\cos^3\theta = \cos\theta(1 - \sin^2\theta)$:
$V = \frac{125\pi}{2} \int_0^{\frac{\pi}{3}} (\cos\theta - \sin^2\theta\cos\theta) \, d\theta$
$V = \frac{125\pi}{2} \left[ \sin\theta - \frac{1}{3}\sin^3\theta \right]_0^{\frac{\pi}{3}}$
$V = \frac{125\pi}{2} \left[ \left(\sin\frac{\pi}{3} - \frac{1}{3}\sin^3\frac{\pi}{3}\right) - (0) \right]$
$V = \frac{125\pi}{2} \left[ \frac{\sqrt{3}}{2} - \frac{1}{3}\left(\frac{\sqrt{3}}{2}\right)^3 \right]$
$V = \frac{125\pi}{2} \left( \frac{\sqrt{3}}{2} - \frac{3\sqrt{3}}{24} \right)$
$V = \frac{125\pi}{2} \left( \frac{3\sqrt{3}}{8} \right) = \frac{375\sqrt{3}}{16}\pi$