A container with height [tex]4.5[/tex] is created by rotating the curve [tex]y = 0.5x^2[/tex] [tex]0 \leq x \leq 3 [/tex] around [tex] x = -3 [/tex] and putting a plane bottom in the box. Find the volume [tex]V[/tex] of the box.
The Attempt at a Solution
I want to solve this by using the shell method. I have put up the following integral, which will be the volume of the revolution. It is however not correct, and I haven't really used the information about the height given in the text.. could anyone help me?
[tex]2\pi\int_0^3 (x+3)(0.5x^2) \, dx[/tex]