`y = x^2, y = 6x - 2x^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 `x` , the cricumference is `2pi*x` and the height is `6x - 2x^2 - x^2` , hence, the volume can be evaluated, using the method of cylindrical shells, such that:

`V = 2pi*int_(x_1)^(x_2) x*(6x - 3x^2) dx`

You need to evaluate the endpoints `x_1` and `x_2` , such that:

`x^2 = 6x - 2x^2 =>3x^2 - 6x = 0 => 3x(x - 2) = 0 => 3x = 0 => x = 0 and x - 2 = 0 => x = 2`

`V = 2pi*int_0^2 x*(6x - 3x^2) dx`

`V = 2pi*(int_0^2 6x dx - int_0^2 3x^2 dx)`

Using the formula `int x^n dx = (x^(n+1))/(n+1)` yields:

`V = 2pi*(6x^2/2 - 3x^3/3)|_0^2`

`V = 2pi*(3x^2 - x^3)|_0^2`

`V = 2pi*(3*2^2 - 2^3)`

`V = 2pi*4`

`V = 8pi`

Hence, evaluating the volume, using the method of cylindrical shells, yields `V = 8pi.`