# The region R is bounded by the curves y=x² (x»0) , y=9x² (x»0) and the line y=1. Find the volume of the solid obtained when the region R is rotated about the y-axis?

Dec 15, 2017

$V = \setminus {\int}_{y = 0}^{y = 1} \setminus {\int}_{x = \setminus \frac{\sqrt{y}}{3}}^{x = \setminus \sqrt{y}} \left(2 \setminus \pi y\right) \mathrm{dy} \mathrm{dx} = \frac{8 \setminus \pi}{15.}$

#### Explanation:

$V = \setminus {\int}_{y = 0}^{y = 1} \setminus {\int}_{x = \setminus \frac{\sqrt{y}}{3}}^{x = \setminus \sqrt{y}} \left(2 \setminus \pi y\right) \mathrm{dy} \mathrm{dx}$

$V = 2 \setminus \pi . \setminus {\int}_{y = 0}^{y = 1} y \left[\setminus {\int}_{x = \setminus \frac{\sqrt{y}}{3}}^{x = \setminus \sqrt{y}} \mathrm{dx}\right] \mathrm{dy}$

$V = 2 \setminus \pi . \setminus {\int}_{y = 0}^{y = 1} y \left[\setminus \sqrt{y} - \setminus \frac{\sqrt{y}}{3}\right] \mathrm{dy} = \frac{4 \setminus \pi}{3.} \setminus {\int}_{0}^{1} y \setminus \sqrt{y} \mathrm{dy}$

$V = \frac{4 \setminus \pi}{3.} {\left[\frac{2}{5} {y}^{\frac{5}{2}}\right]}_{0}^{1} = \frac{8 \setminus \pi}{15.}$