Problem Statement
The region bounded by the curve $y = \sqrt{x} \sin x$, the $x$-axis, and the lines $x = 0$ and $x = \pi$ is rotated through $2\pi$ radians about the $x$-axis. Find the exact volume of the solid generated.
Verified Solution & Marking Scheme
Set Up Volume of Revolution Formula
$V = \pi \int_{0}^{\pi} y^2 \, dx = \pi \int_{0}^{\pi} (\sqrt{x} \sin x)^2 \, dx = \pi \int_{0}^{\pi} x \sin^2 x \, dx$
Use Half-Angle Identity sin² x = (1 - cos 2x)/2
$V = \frac{\pi}{2} \int_{0}^{\pi} x(1 - \cos 2x) \, dx = \frac{\pi}{2} \left[ \int_{0}^{\pi} x \, dx - \int_{0}^{\pi} x \cos 2x \, dx \right]$
$\int_{0}^{\pi} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\pi} = \frac{\pi^2}{2}$
Integrate x cos 2x by Parts
$\int_{0}^{\pi} x \cos 2x \, dx = \left[ x \frac{\sin 2x}{2} \right]_{0}^{\pi} - \int_{0}^{\pi} \frac{\sin 2x}{2} \, dx = 0 + \left[ \frac{\cos 2x}{4} \right]_{0}^{\pi} = \frac{\cos 2\pi - \cos 0}{4} = \frac{1 - 1}{4} = 0$
Thus:
$V = \frac{\pi}{2} \left( \frac{\pi^2}{2} - 0 \right) = \frac{\pi^3}{4} \approx 7.7515$