Calculus: Early Transcendentals, Chapter 6, 6.3, Section 6.3, Problem 20

The shell has the radius -2 - y , the cricumference is 2pi*(-2 - y) and the height is 2 - x , hence, the volume can be evaluated, using the method of cylindrical shells, such that:
V = 2pi*int_(y_1)^(y_2) (-2 - y)*(2-x) dy
You need to evaluate the endpoints from equation y^2+1=2 => y^2 = 1 => y_(1,2) = +-1
V = 2pi*int_(-1)^1 (-2 - y)*(2 - y^2 - 1) dy
V = 2pi*int_(-1)^1 (-2 - y)*(1 - y^2) dy
V = 2pi*int_(-1)^1 (-2 + 2y^2 - y + y^3)dy
V = 2pi*(int_(-1)^1 (-2 dy) + int_(-1)^1 2y^2 dy - int_(-1)^1 ydy + int_(-1)^1 y^3 dy)
Using the formula int x^n dx = (x^(n+1))/(n+1) yields:
V = 2pi*(-2y + 2y^3/3 - y^2/2 +y^4/4)|_(-1)^1
V = 2pi*(-2 + 2/3 - 1/2 + 1/4 + 2 +2/3 + 1/2 - 1/4)|_(-1)^1
V = 2pi*(+ 4/3)
V = (8pi)/3
Hence, evaluating the volume, using the method of cylindrical shells, yields V = (8pi)/3.