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Math Help - sphere cone problem

  1. #1
    Member GAdams's Avatar
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    sphere cone problem

    If a cone with perpendicular height of 6h and radius 2h has the same volume as a sphere of radius r, show that r= cuberoot 6h.

    I got:

    1/3 pi r squared x h = 4/3 pi x r cubed substitute

    1/3 pi (6h) squared = 4/3 pi x r cubed divide by 3

    pi (2h) squared x 6h = 4 pi r cubed divide by pi

    (2h) squared x 6h = 4 r cubed

    24h cubed = 4r cubed

    ...where am i going wrong?
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  2. #2
    Junior Member
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    Quote Originally Posted by GAdams View Post
    If a cone with perpendicular height of 6h and radius 2h has the same volume as a sphere of radius r, show that r= cuberoot 6h.

    I got:

    1/3 pi r squared x h = 4/3 pi x r cubed substitute

    1/3 pi (6h) squared = 4/3 pi x r cubed divide by 3

    pi (2h) squared x 6h = 4 pi r cubed divide by pi

    (2h) squared x 6h = 4 r cubed

    24h cubed = 4r cubed

    ...where am i going wrong?
    Things seem to be going wrong once you've substituted. Why are you squaring the height (6*h)?

    Let me show you how to do it:
    \frac{1}{3}\pi (R_1)^2 H_1=\frac{4}{3}\pi (R_2)^3
    R_1=2h
    H_1=6h
    R_2=r
    \frac{1}{3}\pi [2h]^2 [6h]=\frac{4}{3}\pi [r]^3
    Where brackets signify substitution.

    First, I'll multiply both sides by \frac{4}{3} and divide both sides by \pi:
    \frac{1}{4} (2h)^2 (6h)=r^3
    Next, I'll distribute the squared out:
    \frac{1}{4} 4h^2 (6h)=r^3
    The four from what I just did cancels with the \frac{1}{4}. Also, I will combine the h^2 and the h.
    6h^3=r^3
    Take the third root of both sides, and you're done!
    r=h\sqrt[3]{6}
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