Yep, you can also do it with shells.
Consider the region in the plane bounded by the curves y = x^2 and y = x
Set up and evaluate the integral that finds the volume generated by rotating the region about the y-axis. Here is what I did:
x= y^(1/2) and x= y I put y=x^2 and y=x in the calculator and looked at the graph. I also got the limits by solving X^2 = x and got x=0 and x=1
then I did this: pi*integral from o to 1 {(sqrt(Y))^2 - (y)^2}dy =
pi *integral from 0 to 1 (y-y^2)dy= Then I took the intergral and got: pi(.5y^2-(1/3)y^3) from limit 0 to 1 and then putting in the limits i came up with: pi/6 is this correct?
Thank You,
Keith