Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x^2 and y^2=x about the x-axis.

y=x^2 and y^2=x

Let us first find the intersections of these two graphs.....substituting the first function into the second, we have

x^4 = x

x^4 - x = 0

x(x^3 - 1)  = 0

So, setting each factor to 0 ....the intersection points are 0  and 1

And let us re-write the second function as y = √x

So.....the cross-sectional area, A(x) =  pi [ (√x)^2  - (x^2)^2 ]  = pi [ x - x^4]

So....the volume is given by :

1

∫    A(x) dx   =

0

    1

pi  ∫   x - x^4   dx  =

   0

pi  [ (1)^2/ 2   -  (1)^5 / 5 ] =

pi [ 1/2   -  1/5]  =

(3/10)pi  units^3