Let's see what this looks like:
graph{(y+2-x)(sqrt(2-(x-2)))/(sqrt(2-(x-2))) <= 0 [-5, 6, -0.3, 6]}
Revolving around #x = -1# means doing the shell method makes this easier (you're revolving to make a shell, and you don't have to convert to #f(y)# functions).
#V = 2piint_a^b xf(x)dx#
where:
#x# is the radius of the shell
#f(x)# is the thickness of the shell
#2pi# indicates that the shell is fully revolved (#2pi# radians = 1 full revolution)
Now it's just a matter of figuring out what your equations are.
#f(x) = x-2#
(the thickness is from the inside surface of the shell to the outside surface, which is what #f(x)# from 2 to 4 revolves to become)
Interval: #[2,4]#
(implied by #x = 2# and #x = 4# in the question)
The radius of the shell is determined by the distance from the axis of #x = -1# to the current value of #x# in the interval #[2,4]#. Therefore, the radius is #x - (-1) = x+1#.
Overall, we get:
#V = 2piint_2^4 (x+1)(x-2)dx#
#= 2piint_2^4 x^2 - x - 2dx#
#= 2pi[x^3/3 - x^2/2 - 2x]|_(2)^(4)#
#= 2pi[(4^3/3 - 4^2/2 - 2(4))-(2^3/3 - 2^2/2 - 2(2))]#
#= 2pi[(64/3 - 16/2 - 8)-(8/3 - 4/2 - 4)]#
#= 2pi[(64/3 - 48/3)-(8/3 - 18/3)]#
#= 2pi[64/3 - 48/3 - 8/3 + 18/3]#
#= 2pi[26/3]#
#color(blue)(= (52pi)/3 "u"^3)#
