# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=x^(1/2)#, #y=0#, and #x=4# rotated about the x axis?

##### 1 Answer

This solid to-be-revolved looks like:

graph{(y - sqrtx)(y)sqrt(4-(x-0))/sqrt(4-(x-0)) <= 0 [-5, 5, 0, 5]}

If you want to do it with the shell method, convert your functions to their inverses.

You get:

Now your domain is your range and your interval is

graph{(y - x^2)sqrt(4-(y-0))/sqrt(4-(y-0)) >= 0 [0, 8, -1, 5]}

*Note: if you evaluated this using #f(x) = x^2# instead, you would coincidentally get the same answer, but for the wrong work, so if you did that on a test, you wouldn't get full credit.*