Answer to Question #123569 in Calculus for Mr S Moodley

Question

What is the volume of the solid obtained by rotating the region under the curve y=cosx between x=0 and x=π/12 about the x-axis.

Answer 1

The volume of solid obtained by rotation of the curve $f(x)$ between $x=a$ and $x=b$ about the x-axis can be calculated as

$V = \pi \int\limits_a^bf^2(x)\,dx$ . In our case

$V = \pi \int\limits_0^{\pi/12}\cos^2 x\,dx = \pi \int\limits_{0}^{\pi/12}\dfrac{1+\cos(2x)}{2}\,dx = \pi \int\limits_{0}^{\pi/12}\dfrac12\,dx + \pi\int\limits_{0}^{\pi/12} \dfrac{\cos(2x)}{2}\,dx = \dfrac{\pi^2}{24} + \pi \int\limits_{0}^{\pi/6}\dfrac{\cos \tau}{4}\,d\tau = \dfrac{\pi^2}{24} + \dfrac{\pi}{4} \sin\tau\Big|_0^{\pi/6} = \dfrac{\pi^2}{24} +\dfrac{\pi}{8}.$