# Find the volume of the solid generated by revolving the region bounded by x=Siny,x=1,and x-axis about y-axis?

May 22, 2018

$\textcolor{p u r p \le}{V = \pi {\int}_{0}^{\frac{\pi}{2}} {\left(1\right)}^{2} - \left({\sin}^{2} y\right) \mathrm{dy} = {\left(\pi\right)}^{2} / 4}$

#### Explanation:

show that the shaded region in the sketch below revolving about y-axis.

when x=1

$x = \sin y \Rightarrow \sin y = 1 \Rightarrow y = \frac{\pi}{2}$

the interval of the integral become $y \in \left[0 , \frac{\pi}{2}\right]$

now set up the integral

$\textcolor{b l u e}{V = \pi {\int}_{0}^{\frac{\pi}{2}} {\left(1\right)}^{2} - {\left(\sin y\right)}^{2} \cdot \mathrm{dy} = \pi {\int}_{0}^{\frac{\pi}{2}} 1 - {\sin}^{2} y \cdot \mathrm{dy}}$

$\textcolor{b l u e}{= \pi {\left[y - \frac{y - \sin \frac{2 \cdot y}{2}}{2}\right]}_{0}^{\frac{\pi}{2}} = \pi {\left[\frac{\sin \left(2 \cdot y\right) + 2 \cdot y}{4}\right]}_{0}^{\frac{\pi}{2}} = {\left(\pi\right)}^{2} / 4}$