`x = y^2 + 1, x = 2` Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves...

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.`