Volume of the Cap of a Sphere

**Bahama.Llama** · April 8th 2006, 10:22 PM

Well I figured out the volume of a Torus alright (deleted?), but now I'm strugglin' with this next problem

Basically I have to use integration techniques for find the formula for the volume of the cap of a sphere with a radius R and the cap having a height h.

Here's what I have (i feel like MacGyver saying that) so far:

I don't think I had an problems setting up the integral correctly... however once I solve for it, pi simply cancels out (but is needed in the final formula). I know R is a constant as well as h, however, I’m a little hesitant to drag out the R^2 to the front leaving the -x^2 to be integrated. Any help would be greatly appreciated.

I tried using math tags, however I couldn't get the integrals set up right with limits, so figured a scan of my work would be alright. Hope its readable

If I'm doing this the long why, by all means let me know

**ThePerfectHacker** · April 9th 2006, 07:13 AM

Originally Posted by Bahama.Llama

Well I figured out the volume of a Torus alright (deleted?), but now I'm strugglin' with this next problem

Basically I have to use integration techniques for find the formula for the volume of the cap of a sphere with a radius R and the cap having a height h.

Here's what I have (i feel like MacGyver saying that) so far:

I don't think I had an problems setting up the integral correctly... however once I solve for it, pi simply cancels out (but is needed in the final formula). I know R is a constant as well as h, however, I’m a little hesitant to drag out the R^2 to the front leaving the -x^2 to be integrated. Any help would be greatly appreciated.

I tried using math tags, however I couldn't get the integrals set up right with limits, so figured a scan of my work would be alright. Hope its readable

If I'm doing this the long why, by all means let me know

Consider a circle $y=\sqrt{r^2-x^2}$ having radius $r$ on interval $[0,r]$ . Draw a vertical line $x=h$ so that is passes through this quater-circle. You need to find the volume of the cap, meaning the volume of this circle segment rotated about the x-axis. Thus, you have,
$\pi \int^r_h \left( \sqrt{r^2-x^2 \right)^2 dx$
Thus,
$\pi \int^r_b r^2-x^2 dx$
$\pi \left( \left r^2x-\frac{1}{3}x^3 \right|^r_h \right)$
$\pi \left( r^3-\frac{1}{3}r^3-r^2h+\frac{1}{3}h^3 \right)$
$\pi \left(\frac{2}{3}r^3-r^2h+\frac{1}{3}h^3 \right)$

This is my 8

th Post !!!

**Bahama.Llama** · April 9th 2006, 09:29 AM

I'm not sure thats right... the cap has a height h... so wouldn't the integral have to be from R-h to R?

I also noticed that in your second integral... you have it bounded from b to R, is that a typo or is that a substitution that I'm missing

Grats on post #800!

b is R-h because i can't get the math deal to look right

$ \pi \int^R_b \left( \sqrt{r^2-x^2} \right)^2 dx \pi ( \left r^2x - \frac{x^3}{3} \right|^R_b) \pi \left( (R^3 - \frac{R^3}{3})-(R^3-R^2h- \frac{(R-h)^3}{3}) \right) $

**TD!** · April 9th 2006, 01:04 PM

I think you're on the right track but let me rewrite and continue it a bit.

We want to sum all the circle area's starting from heigth R-h until R to have the cap of heigth h. I choose y as the positive vertical axis, then a circle with radius r has area pi.r² or (r²-y²).pi. Integrating:

$ \pi \int\limits_{r - h}^r {r^2 - y^2 } dy = \pi \left[ {r^2 y - \frac{{y^3 }}{3}} \right]_{r - h}^r = \frac{{\pi h^2 }}{3}\left( {3R - h} \right) $

**TD!** · April 9th 2006, 01:25 PM

Apparently, some more steps leading to the simplified answer are necessary. Here goes.

$ \pi \left[ {r^2 y - \frac{{y^3 }}{3}} \right]_{r - h}^r = \pi \left( {\left( {r^3 - \frac{{r^3 }}{3}} \right) - \left( {r^2 \left( {r - h} \right) - \frac{{\left( {r - h} \right)^3 }}{3}} \right)} \right) $

$ \begin{gathered} = \pi \left( {\frac{{2r^3 - 3r^2 \left( {r - h} \right) + \left( {r - h} \right)^3 }} {3}} \right) \hfill \\ \\ = \pi \frac{{2r^3 - 3r^3 + 3r^2 h + r^3 - 3r^2 h + 3rh^2 - h^3 }} {3} \hfill \\ \\ = \pi \frac{{3rh^2 - h^3 }} {3} = \frac{{\pi h^2 }} {3}\left( {3r - h} \right) \hfill \\ \end{gathered} $

**Bahama.Llama** · April 9th 2006, 01:56 PM

you certainly are a super member! thanks for all of your help

**TD!** · April 9th 2006, 01:59 PM

You're welcome

**ThePerfectHacker** · April 9th 2006, 04:10 PM

Oh not! not another 1337

I am the only one allowed here.

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