Question

How do you find the volume of the solid generated by revolving the region bounded by `y=2x^2`, `y=0`, `x=2`, revolving on `y=8`?

Answer 1

It looks like this:

graph{(y-2x^2)(y-8)(x-2)(sqrt(x)/sqrtx) <= 0 [-0.5, 5, -0.35, 8.0]}

I'm assuming it's also in the domain of `x >= 0`, which was not stated. (This will make the volume finite.)

Let's make it easier to work with though; due to the symmetry of the revolved solid, if you just take the portion of the revolved solid that would be above the line `y = 8` and shift it down 16 units to touch the bottom edge of the portion below `y = 8`, the solid will change from a circular-base bowl to a curved "cone".

Different solid---same volume.

And no need to do any subtractions to specify a range for the graph---we can just use `y = 2x^2` as-is since now we're revolving around `y = 0`. The volume is written like this:

`V = sum_a^b pi(r(x))^2Deltax`

...since you are essentially stacking "circles" with radii that vary according to a function to generate a solid. These "circles" are perpendicular to the revolution axis. The "circle" is defined by `r(x) = y = 2x^2` bounded below by the x-axis. Your horizontal bounds are of course said to be `[0,2]`. Therefore, you just have:

`V = int_0^2 pi(2x^2)^2dx`

`= piint_0^2 4x^4dx`

`= pi[4/5x^5]_(0)^(2)`

`= pi[(4/5 (2)^5) - cancel(4/5 (0)^5)]`

`color(blue)(= (128pi)/5 "u"^3) ~~ 26.808 "u"^3`