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Math Help - Volume of a sphere cap

  1. #1
    Super Member angel.white's Avatar
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    Volume of a sphere cap

    Can someone help me out on this problem?

    We are supposed to find the volume using integrals. Meaning as x or y changes, the cross-sectional area will change, so we find the equation for the cross sectional area, and then take the integral of it across the height.

    In this one, I figured the cross-sectional area = \pi x^2
    And if we set the bottom of the blue section along the x-axis, then it would be integrated from 0 to h, so

    volume = \pi \int_0^h x^2~dx

    volume = \pi [\frac 13 x^3]_0^h

    volume = \frac 13 \pi~h^3

    But somehow I'm supposed to incorporate the radius.



    Back of the book says the answer is \pi h^2(r-\frac 13 h)

    edit: well, part of my problem, I think is that I should be puting the cross-sectional area in terms of y. And I think I should be using the equation of a circle: x^2=r^2-y^2. That still doesn't get me the right answer, but it's much closer to theirs, as this would become \pi \int_0^h (r^2-y^2)~dy = \pi ( r^2h-h^3/3) so getting closer :/ But this method seems like it would integrate just the top half of the circle. So somehow I need to remove that gap between the circle's center, and the bottom of h.
    Attached Thumbnails Attached Thumbnails Volume of a sphere cap-volume001.jpg  
    Last edited by angel.white; March 9th 2008 at 05:05 AM.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by angel.white View Post
    Can someone help me out on this problem?

    We are supposed to find the volume using integrals. Meaning as x or y changes, the cross-sectional area will change, so we find the equation for the cross sectional area, and then take the integral of it across the height.

    In this one, I figured the cross-sectional area = \pi x^2
    And if we set the bottom of the blue section along the x-axis, then it would be integrated from 0 to h, so

    volume = \pi \int_0^h x^2~dx

    volume = \pi [\frac 13 x^3]_0^h

    volume = \frac 13 \pi~h^3

    But somehow I'm supposed to incorporate the radius.



    Back of the book says the answer is \pi h^2(r-\frac 13 h)

    edit: well, part of my problem, I think is that I should be puting the cross-sectional area in terms of y. And I think I should be using the equation of a circle: x^2=r^2-y^2. That still doesn't get me the right answer, but it's much closer to theirs, as this would become \pi \int_0^h (r^2-y^2)~dy = \pi ( r^2h-h^3/3) so getting closer :/ But this method seems like it would integrate just the top half of the circle. So somehow I need to remove that gap between the circle's center, and the bottom of h.
    Just rotate the part of the circle x^2 + y^2 = r^2 between y = r - h and y = r around the y-axis:

    V = \pi \int_{y = a}^{y = b} x^2 \, dy = \pi \int_{r - h}^{r} (r^2 - y^2) \, dy = ....
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Just rotate the part of the circle x^2 + y^2 = r^2 between y = r - h and y = r around the y-axis:

    V = \pi \int_{y = a}^{y = b} x^2 \, dy = \pi \int_{r - h}^{r} (r^2 - y^2) \, dy = ....
    Oh that's brilliant!

    Thank you, it would have taken me forever to see that.
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