Question

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?

Answer 1

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:
`y = x^2`
`x = 0`
`y = 4`

Now your domain is your range and your interval is `[0,2]`.

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

`V = 2piintxf(x)dx`

`f(x) = x^2 in [0,2]`, bounded from above by `y = 4`, thus `f(x) = 4-x^2`

`V = 2piint_0^2 x*(4-x^2)dx`

`= 2pi[2x^2 - x^4/4]|_(0)^(2)`

`= 2pi[(2(2)^2 - (2)^4/4) - cancel((2(0)^2 - (0)^4/4))]`

`= color(blue)(8pi "u") ~~ 25.1327 "u"`

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.