Solution:

y = -2

(-2)^2 = x

x = 4

y^2 = +- 2

the point of intersection should be (0,0) and (4,2)

now im using the cylindrical method in obtaining the volume

$\displaystyle V = 2\pi\int_{0}^{4} R dA$

where dA = (y2-y1)dx

for

y2 = sqrt(x)

y1 = x

and

radius be r = x+2 since the revolved line is at the 2nd quadrant and x in the 1st quadrant and it needs to be added

$\displaystyle V = 2\pi\int_{0}^{4} (x+2) (y_2 - y_1) dx$

$\displaystyle V = 2\pi\int_{0}^{4} (x+2) (\sqrt{x} - x) dx$

V = -(416/15)pi ????