`pi(int_-1^1(1 - y^2)^2 dy)` Each integral represents the volume of a solid. Describe the solid.

The formula provided represents the volume of the solid obtained by rotating the region enclosed by the curves `x = y^2, x = 1` , about x = 1, using washer method:

`V = pi*int_a^b (f^2(x) - g^2(x))dx, f(x)>g(x)`

You need to find the endpoints by solving the equation:

`y^2 = 1 => y^2 - 1 = 0 => (y-1)(y+1) = 0 => y = 1, y = -1`

`V = pi*int_(-1)^1 ((1 - y^2)^2 - (1 - 1)^2)dy`

`V = pi*int_(-1)^1 ((1 - y^2)^)dy`

`V = pi*int_(-1)^1 (1 - 2y^2 + y^4)dy`

`V = pi*(int_(-1)^1 dy - 2int_(-1)^1y^2 dy + int_(-1)^1 y^4dy)`

`V = pi*(y - 2y^3/3 + y^5/5)|_(-1)^1`

`V = pi*(1 - 2/3 + 1/5 + 1 - 2/3 + 1/5)`

`V = pi*(2 - 4/3 + 2/5) => V = pi*(30 - 20 + 6)/15`

`V =(16pi)/15`

Hence, evaluating the volume of the solid obtained by rotating the region enclosed by the curves `x = y^2, x = 1` , about x = 1, using washer method, yields `V =(16pi)/15.`