Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.3 Volumes By Cylindrical Shells - Exercises Set 5.3 - Page 370: 3

$$V = \frac{1}{3}\pi $$

$$\eqalign{ & {\text{From the graph we have }}x = 2y - 2{y^2},\,\,\,x = 0,\,\,\,0 \leqslant y \leqslant 1 \cr & \cr & {\text{The volume of the solid can be calculated using cylindrical shells}} \cr & V = \int_c^d {2\pi y\left[ {f\left( y \right) - g\left( y \right)} \right]dy} \cr & {\text{Let }}f\left( y \right) = 2y - 2{y^2}{\text{ and }}g\left( y \right) = 0 \cr & V = 2\pi \int_0^1 {y\left[ {2y - 2{y^2} - 0} \right]dy} \cr & V = 4\pi \int_0^1 {\left( {{y^2} - {y^3}} \right)} dy \cr & {\text{Integrating}} \cr & V = 4\pi \left[ {\frac{1}{3}{y^3} - \frac{1}{4}{y^4}} \right]_0^1 \cr & V = 4\pi \left[ {\frac{1}{3}{{\left( 1 \right)}^3} - \frac{1}{4}{{\left( 1 \right)}^4}} \right]_0^1 - 4\pi \left[ 0 \right]_0^1 \cr & V = 4\pi \left( {\frac{1}{{12}}} \right) \cr & V = \frac{1}{3}\pi \cr} $$