# What is the volume of the solid produced by revolving f(x)=cosx, x in [0,pi] around the x-axis?

Nov 26, 2016

$\frac{1}{2} {\pi}^{2} = 4.93$ (3sf.)

#### Explanation:

General formula for volumes of revolution:

${V}_{x} = {\int}_{b}^{a} \pi {y}^{2} \mathrm{dx}$
${V}_{y} = {\int}_{b}^{a} \pi {x}^{2} \mathrm{dy}$

(the pi can be taken out of the integral in either case to simplify the calculation)

$y = \cos x$
${y}^{2} = {\cos}^{2} x$

The question requires solving:

$\pi {\int}_{0}^{\pi} {\cos}^{2} x \mathrm{dx}$

This is made easier using double angle formulae:

$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x = 2 {\cos}^{2} x - 1$
${\cos}^{2} x = \frac{1}{2} \left(\cos 2 x + 1\right)$

By taking the $\frac{1}{2}$ out of the integral it becomes:

$\frac{1}{2} \pi {\int}_{0}^{\pi} \cos 2 x + 1 \mathrm{dx}$

=$\frac{1}{2} \pi {\left[\frac{1}{2} \sin 2 x + x\right]}_{0}^{\pi}$

=$\frac{1}{2} \pi \left\{\left(\frac{1}{2} \sin 2 \pi + \pi\right) - \left(\frac{1}{2} \sin 0 + 0\right)\right\}$

=$\frac{1}{2} \pi \left(\pi\right)$

=$\frac{1}{2} {\pi}^{2}$