Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 362: 18

$$V = \frac{3}{{10}}\pi $$

$$\eqalign{ & {\text{We have }}y = {x^2},\,\,\,\,x = {y^2} \cr & y = {x^2} \Rightarrow x = \sqrt y , \cr & {\text{Let }}x = x \cr & {y^2} = \sqrt y \cr & {\text{The intersection points are }}\left( {0,0} \right){\text{ and }}\left( {1,1} \right).\,\,\,\sqrt y \geqslant {y^2}{\text{ on }}0 < y < 1 \cr & {\text{The volume of the solid can be calculated using}} \cr & V = \int_c^d {\pi \left( {{{\left[ {w\left( y \right)} \right]}^2} - {{\left[ {v\left( y \right)} \right]}^2}} \right)dy} \cr & {\text{Let }}w\left( y \right) = \sqrt y {\text{ and }}v\left( y \right) = {y^2} \cr & {\text{Thus}}{\text{,}} \cr & V = \pi \int_0^1 {\left( {{{\left[ {\sqrt y } \right]}^2} - {{\left[ {{y^2}} \right]}^2}} \right)dy} \cr & V = \pi \int_0^1 {\left( {y - {y^4}} \right)dy} \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{1}{2}{y^2} - \frac{1}{5}{y^5}} \right]_0^1 \cr & V = \pi \left[ {\frac{1}{2}{{\left( 1 \right)}^2} - \frac{1}{5}{{\left( 1 \right)}^5}} \right] - \pi \left[ {\frac{1}{2}{{\left( 0 \right)}^2} - \frac{1}{5}{{\left( 0 \right)}^5}} \right] \cr & V = \pi \left[ {\frac{3}{{10}}} \right] - \pi \left[ 0 \right] \cr & V = \frac{3}{{10}}\pi \cr} $$