`y = x, y = (2x)/(1 + x^3)` (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

The shell has the radius `x - (-1)` , the cricumference is `2pi*(x+1)` and the height is `(2x)/(1+x^3) - x,` hence, the volume can be evaluated, using the method of cylindrical shells, such that:

`V = 2pi*int_(x_1)^(x_2) (x+1)*((2x)/(1+x^3) - x) dx`

You need to find the endpoints, using the equation `(2x)/(1+x^3) - x = 0 => 2x - x - x^4 = 0 => 3x - x^4 = 0 => x(3 - x^3) = 0 => x = 0 ` and `x = root(3) 3`

`V = 2pi*int_0^(root(3) 3) (x+1)*((3x - x^4)/(1+x^3)) dx`

`V = 2pi*int_0^(root(3) 3) (x+1)*(3x - x^4)/((x+1)(x^2-x+1)) dx`

Reducing the like terms yields:

`V = 2pi*int_0^(root(3) 3) (3x - x^4)/(x^2-x+1) dx`

`V = 2pi*(int_0^(root(3) 3) (3x)/(x^2-x+1) -int_0^(root(3) 3) (x^4)/(x^2-x+1) dx`

`V = 2pi*(int_0^(root(3) 3) (3x)/(x^2-x+1)dx - int_0^(root(3) 3) x^2 dx - int_0^(root(3) 3) x dx + int_0^(root(3) 3) (x)/(x^2-x+1)dx)`

`V = 2pi*(int_0^(root(3) 3) (4x)/(x^2-x+1)dx - int_0^(root(3) 3) x^2 dx - int_0^(root(3) 3) x dx)`

`V = 2pi*(2*int_0^(root(3) 3) (2x+1-1)/(x^2-x+1)dx - x^3/3|_0^(root(3) 3) - x^2/2|_0^(root(3) 3))`

`V = 2pi*(2*int_0^(root(3) 3) (2x-1)/(x^2-x+1)dx+ 2*int_0^(root(3) 3) (1)/(x^2-x+1)dx - 1 - (root(3) 9)/2)`

You need to solve `int_0^(root(3) 3) (2x-1)/(x^2-x+1)dx ` using substitution `x^2-x+1 = t => (2x-1)dx =dt.`

`V = 2pi*(2*ln(x^2-x+1)|_0^(root(3) 3)+ 2*int_0^(root(3) 3) (1)/((x-1/2)^2 + ((sqrt3)/2)^2)dx - 1 - (root(3) 9)/2)`

`V = 2pi*(2*ln((root(3) 9)-(root(3) 3)+1)+ (4/sqrt3)*arctan (2x-1)/sqrt3|_0^(root(3) 3) - 1 - (root(3) 9)/2)`

`V = 2pi*(2*ln((root(3) 9)-(root(3) 3)+1)+ (4/sqrt3)*arctan (2(root(3) 3)-1)/sqrt3 -(4/sqrt3)*(pi/6) - 1 - (root(3) 9)/2)`

Hence, evaluating the volume, using the method of cylindrical shells, yields `V = 2pi*(2*ln((root(3) 9)-(root(3) 3)+1)+ (4/sqrt3)*arctan (2(root(3) 3)-1)/sqrt3 -(4/sqrt3)*(pi/6) - 1 - (root(3) 9)/2)`