Question

Find the volume V of the described in the image below.
A cap of a sphere with radius r and height h.

Answer 1

image:

Answer 2

I can safely assume this:

\[chord_{length}=2\sqrt{r^2-h^2}\]
h = height

\[V_{sphere}=\int\limits_{-r}^{r}\pi y^2 dx = \int\limits_{-r}^{r}\pi (r^2-x^2) = \frac{4}{3}\pi r^3\]

Answer 3

This was my first thought, probably not the simplest.

My initial reasoning is to find a triple integral for the "ice cream cone" shape, and then to subtract out the area of the cone under it.

I concluded these bounds in spherical coordinates:
\[0\le \rho \le r\]\[0 \le \phi \le \arcsin((r-h)/r)\]\[0 \le \theta \le 2\pi\]

Constructing the integral:
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\arcsin((r-h)/r)}\int\limits_{0}^{r}\rho^2\sin \phi d \rho d \phi d \theta\]
Should give the volume of the ice cream cone shape, but now we must subtract out the volume cone.

\[V _{cone}=(1/3)\pi r^2 h\]
In this case,
\[V _{cone}=(1/3)\pi (r-(r-h)^2)(r-h)\]

Assuming that I didn't make a mistake (and that this is even calculatetable by hand), I conclude that this may be accurate:
\[V_{total}=\int\limits_{0}^{2\pi}\int\limits_{0}^{\arcsin((r-h)/r)}\int\limits_{0}^{r}\rho^2\sin \phi d \rho d \phi d \theta-(1/3)\pi (r-(r-h)^2)(r-h)\]

Answer 4

I found this easier using cylindrical coordinates.

Answer 5

abstracted, your bottom integral, in theory, should work, but it should be \[\arcsin (h/r)\]

Answer 6

for your upper bound

Answer 7

Thinking about it, with cylindrical you could probably skip the subtracting the cone part with a little extra work

Answer 8

Considering the way you did it, using the upper bound I provided should make "subtracting" the cone unnecessary, since the region provided goes between 0 and arcsin(h/r) on the azimuthal plane.

Answer 9

But even then, you would have to integrate with respect to the azimuth first, and then r to get the answer. to save time, though, cylindrical is the way to go.

Answer 10

Very interesting! I'll definitely take a closer look at this tomorrow when I have more time. I don't think the technique intended was triple integrals, a single integral method was recommended in the section it was from. So the solution is probably using geometric shortcuts of some kind. Thank you both for your time invested in helping out :-)