How do you find the volume bounded by #y^2=x^3# and #y=x^2# revolved about the y-axis?

1 Answer
Jun 10, 2018

#pi/20# units ^3

Explanation:

We need to find where these two curves intersect to find the bounds of integration.

#y^2=x^3 and y=x^2#, squaring the second expression, #y^2=x^4# Solving for #y^2#,.......... [#x^4=x^3#] i.e, #x^3[x-1]=0#.

So #x=1, x=0# are the points of intersection.

From the graphs of these expressions in can be seen that #y=sqrt[x^3]# has a greater area than #y= x^2# so we must find the area under #y =x^2# and subtract it from the area under #y=sqrt[x^3]# and then revolve this area about the #x# axis between the bounds #x=1, x= 0#

Volume of revolution is given by #piint_a^by^2dx# So the volume of revolution = #[piint_0^1x^3dx# - # piint_0^1x^4dx]#.= #pi[x^4/4- x^5/5]# [ after integration by the general power rule] and evaluated for #x=1, x=0# will result in #pi[ 1/4-1/5]# = #pi/20#.