Question

Prove that the volume of any paraboloid is always half the volume of the circumscribed cylinder?

Prove that the volume of any paraboloid is always half the volume of the circumscribed cylinder.

Answer 1

Calculate volumes of the solids and compare. Use the surface of revolution technique for the paraboloid.

Explanation:

The paraboloid has equation `y=c(x^2+z^2)` (where `z` is the axis coming out of the page) and is a surface of revolution about the `y` axis of the curve `y=cx^2`. There are more complicated shapes called "paraboloid", but the circular form must be the one meant due to the comparison to the circumscribed cylinder.

First, calculate the volume enclosed by the paraboloid

The volume enclosed by a surface of revolution of a positive curve `f` around an axis `y` is a known result:
`V=pi int_a^b (f(y))^2 dy`

Regarding our limits of integration, note that they are in `y`. The lower limit is obvious - the extreme of the surface is at `y=0` when both `x` and `y` are 0. Negative `x` or `y` means that the `x^2` and/or `y^2` contributions become greater than 0. The upper limit is unspecified - the problem asks us to prove the formula for any height of paraboloid. So we simply let it float as variable `h` (for 'height').

We need to express the parabolic formula in terms of `y` rather than `x`, which is easily done: `y=cx^2` becomes `x=sqrt(y/c)`. Now we may apply the volume of revolution formula to find the volume of the paraboloid:
`V_{par}=pi int_0^h (sqrt(y/c))^2 dy=pi int_0^h y/c dy`
So
`V_{par}=pi/c [1/2 y^2]_0^h = (pi h^2)/(2c)`

Second, calculate the volume enclosed by the cylinder

The volume of a cylinder is its height multiplied by the area of its circular cross-section. The height we chose: `h`. The radius is given by the rearranged parabola formula `x=sqrt(y/c)` as `r=sqrt(h/c)`.

Thus the cylinder volume is
`V_{cyl}=h*pi (sqrt(h/c))^2=h*pi h/c=(pi h^2)/c`

And so, no matter what height you cut the volume of the paraboloid off at, `V_{cyl}=2V_{par}`.