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Math Help - Volume of the Cap of a Sphere

  1. #1
    Newbie Bahama.Llama's Avatar
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    Volume of the Cap of a Sphere

    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, Im 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
    Last edited by Bahama.Llama; April 8th 2006 at 11:48 PM.
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  2. #2
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    Quote 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, Im 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 8th Post !!!
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  3. #3
    Newbie Bahama.Llama's Avatar
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    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
    <br />
\pi \int^R_b \left( \sqrt{r^2-x^2} \right)^2 dx<br /> <br />
\pi ( \left r^2x - \frac{x^3}{3} \right|^R_b)<br /> <br />
\pi \left( (R^3 - \frac{R^3}{3})-(R^3-R^2h- \frac{(R-h)^3}{3}) \right)<br />
    Last edited by Bahama.Llama; April 9th 2006 at 10:59 AM.
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  4. #4
    TD!
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    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.r or (r-y).pi. Integrating:

    <br />
\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)<br />
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  5. #5
    TD!
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    Apparently, some more steps leading to the simplified answer are necessary. Here goes.

    <br />
\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)<br />


    <br />
\begin{gathered}<br />
   = \pi \left( {\frac{{2r^3  - 3r^2 \left( {r - h} \right) + \left( {r - h} \right)^3 }}<br />
{3}} \right) \hfill \\ \\<br />
   = \pi \frac{{2r^3  - 3r^3  + 3r^2 h + r^3  - 3r^2 h + 3rh^2  - h^3 }}<br />
{3} \hfill \\ \\<br />
   = \pi \frac{{3rh^2  - h^3 }}<br />
{3} = \frac{{\pi h^2 }}<br />
{3}\left( {3r - h} \right) \hfill \\ <br />
\end{gathered} <br />
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  6. #6
    Newbie Bahama.Llama's Avatar
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    you certainly are a super member! thanks for all of your help
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  7. #7
    TD!
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    You're welcome
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  8. #8
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    Oh not! not another 1337
    I am the only one allowed here.
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