Find the volume of the solid bounded by the curves y = 4 - x^2 and y = x revolved about the x axis.

Answer 1

intersect them first
x = 4-x^2
x^2 + x - 4 = 0
x = (-1 ± √32)/2

I am skeptical about your typing of the problem
The region you are rotating has part above and part below the x-axis.
This is not your usual kind of question of this type.
Just try to visualize the kind of solid you could get, I can't.

Answer 2

Yeah that's why I'm confused about this one, I saw that the area went above and below. So I'm guessing the equation was supposed to be y = (4-x)^2.?

Answer 3

The region R is bounded by the curves x=y2+2, y=x-4, and y=0

Write a single integral that gives the volume of the solid generated when R is revolved about the x-axis, as well as an intergral revolved around the y-axis. Do not evaluate the intergrals.