Use cylindrical shells to find the vol of the solid that results when the region enclosed by y=x^3-5x^2+6x over [0,2] is revolved about the y axis

To find the volume of the solid that is formed when the region enclosed by the curve y = x^3 - 5x^2 + 6x over the interval [0, 2] is revolved about the y-axis, we can use the method of cylindrical shells.

1. First, let's denote the function y = x^3 - 5x^2 + 6x as f(x). We need to rewrite this equation in terms of x = g(y), so we can integrate with respect to y. Solving for x in terms of y, we have:

x = g(y) = (y + 5y^2 - y^3)^(1/3)

2. We now need to determine the height of the cylindrical shell, which is the vertical distance between the curve and the y-axis at a given y-value. The height is given by h(y) = g(y) - 0 = g(y).

3. The radius of the cylindrical shell is equal to the x-coordinate of the curve y = f(x). In this case, the radius is equal to x.

4. Now, we can express the volume of each cylindrical shell as:

dV = 2πrh(y)dy

where r is the radius and h(y) is the height.

5. To find the total volume, we need to integrate the expression for dV from y = 0 to y = f(2):

V = ∫[0, f(2)] 2πrh(y)dy

V = ∫[0, f(2)] 2πx(y)x' dy

V = 2π∫[0, f(2)] x(y)dy

6. Lastly, we substitute x(y) = g(y) into the integral and evaluate it:

V = 2π∫[0, f(2)] [(y + 5y^2 - y^3)^(1/3)] dy

Integrating this expression will give you the volume of the solid formed when the region enclosed by y = x^3 - 5x^2 + 6x over the interval [0, 2] is revolved about the y-axis.