Using the shell method to calculate the volume of a solid. Calculus help

I have the following cal problem. Should I solve for x first or it it ok to go ahead and set up the integral using y? How would I solve it?

Use the shell method to find the volume of the solid bounded by
y=8-x^2, y =x^2, x=0 about the y-axis. What is the first steps to set it up?

It's going to be way easier to integrate with respect to x, since the axis of rotation is vertical, the shell method requires slicing vertically, so we're good. Note that the curves intersect at x = 0 and x = 2.

V = 2π ∫ rh(x) dx from 0 to 2

First determine the radius. When x is 0 , the radius is 0 and when x is 2 the radius is 2, so the radius is just x. Now determine the height. Since 8 - x^2 is above x^2, subtract x^2 from 8 - x^2.

h(x) = 8 - x^2 - x^2 = 8 - 2x^2. Now sub all of that in:

V = 2π ∫ [x][8 - 2x^2] dx from 0 to 2
V = 2π ∫ [8x - 2x^3] dx from 0 to 2
V = 2π[4x^2 - (1/2)x^4] from 0 to 2
V = 2π[4(2)^2 - (1/2)(2)^4] <-- note that the lower limit of zero will be zero
V = 16π

Done!

If you use shells with a vertical axis of revolution, you have to integrate with respect to x. So you don't really have a choice (except to not use shells.)

The curves y = 8 - x² and y = x² intersect at (2, 4)---set the two y's equal and solve for x, it's very easy.

For each x between 0 and 2, you'd take a thin vertical strip (say with thickness Δx) and revolve this about the y-axis to form a thin cylindrical shell. If you draw is, it becomes clearer. You need the radius and height.

radius: distance from strip to axis of revolution----in this case it's just x

height: distance from top curve to bottom curve---in this case 8 - x² - x².

So one shell would have volume

V_{shell} = 2π r h Δx = 2π x(8 - 2x²) Δx.

Add these together from x = 0 to x = 2 in the limit as Δx -> 0 (becomes dx.)