ditlhamowusp

2023-02-19

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

Brisa Fitzgerald

The paraboloid has equation $y=c\left({x}^{2}+{z}^{2}\right)$ and is a surface of revolution about the y axis of the curve $y=c{x}^{2}$. Although "paraboloid" can refer to more intricate shapes, the comparison to the circumscribed cylinder suggests that the circular form is what is intended.
first, determine the paraboloid's enclosed volume.
The volume enclosed by a surface of revolution of a positive curve f around an axis y is a known result:
$V=\pi {\int }_{a}^{b}{\left(f\left(y\right)\right)}^{2}dy$
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 problem challenges us to demonstrate the formula for any height of paraboloid with no upper limit. Therefore, we just let it float as the variable h (for "height").
We need to express the parabolic formula in terms of y rather than x, which is easily done: $y=c{x}^{2}$ becomes $x=\sqrt{\frac{y}{c}}$. We can now use the volume of revolution formula to determine the paraboloid's volume:
${V}_{par}=\pi {\int }_{0}^{h}{\left(\sqrt{\frac{y}{c}}\right)}^{2}dy=\pi {\int }_{0}^{h}\frac{y}{c}dy$
So
${V}_{par}=\frac{\pi }{c}{\left[\frac{1}{2}{y}^{2}\right]}_{0}^{h}=\frac{\pi {h}^{2}}{2c}$
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 $x=\sqrt{\frac{y}{c}}$ as $r=\sqrt{\frac{h}{c}}$.
Thus the cylinder volume is
${V}_{cyl}=h\cdot \pi {\left(\sqrt{\frac{h}{c}}\right)}^{2}=h\cdot \pi \frac{h}{c}=\frac{\pi {h}^{2}}{c}$
And so, no matter what height you cut the volume of the paraboloid off at, ${V}_{cyl}=2{V}_{par}$.

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