# How do you find the volume of y = 11sinx , y=0, [ 0,pi ] revolving about the x axis?

Jun 10, 2015

$V = \frac{121}{2} {\pi}^{2} \approx 597.111$.

#### Explanation:

This is the formula for calculating the volume of a solid obtained from a revolution around the x-axis:
$V = \pi \cdot {\int}_{a}^{b} {f}^{2} \left(x\right) \mathrm{dx}$
Our function $f \left(x\right) = 11 \sin x \mathmr{and} a = 0 , b = \pi$.
So the volume is:
$V = \pi \cdot {\int}_{0}^{\pi} 121 {\sin}^{2} \left(x\right) \mathrm{dx}$
$V = 121 \pi \cdot {\int}_{0}^{\pi} {\sin}^{2} \left(x\right) \mathrm{dx}$
$V = 121 \pi \cdot {\left[\frac{1}{2} \left(x - \sin \left(x\right) \cos \left(x\right)\right)\right]}_{0}^{\pi} = \frac{121}{2} \pi {\left[x - \sin \left(x\right) \cos \left(x\right)\right]}_{0}^{\pi}$
$V = \frac{121}{2} \pi \left(\pi - 0 - 0 + 0\right) = \frac{121}{2} {\pi}^{2} \approx 597.111$.