From the graph we can see that the volume we seek is between the two functions. In order to find this, we must find the volume of revolution of #f(x)=sqrt(x)# and subtract the volume of revolution of #f(x)=x^2#. This is shown as the shaded area.

First we need to find the upper and lower bounds. We know the lower bound is #0# since #f(x)=sqrt(x)# is undefined for #x<0#. The upper bound is where the functions intersect:

#:.#

#x^2=sqrt(x)#

#x^2/x^(1/2)=1#

#x^(3/2)=1#

Squaring:

#x^3=1#

#x=root(3)(1)=1#

**Volume of** #bb(f(x)=sqrt(x))#:

#pi int_(0)^(1)(x^(1/2))=pi[2/3x^(3/2)]_(0)^(1)#

#=pi{[2/3x^(3/2)]^(1)-[2/3x^(3/2)]_(0)}#

**Plugging in upper and lower bounds:**

#=pi{[2/3(1)^(3/2)]^(1)-[2/3(0)^(3/2)]_(0)}=(2pi)/3# cubic units

**Volume of** #bb(f(x)=x^2)#

#pi int_(0)^(1)(x^2)=pi[1/3x^3]_(0)^(1)#

#=pi{[[1/3x^3]^(1)-[1/3x^3]_(0)}#

**Plugging in upper and lower bounds:**

#=pi{[[1/3(1)^3]^(1)-[1/3(0)^3]_(0)}=pi/3# cubic units.

**Required volume is:**

#(2pi)/3-pi/3=##color(blue)(pi/3 "cubic units.")#

**Volume of revolution:**