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Math Help - volume and cube

  1. #1
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    Question volume and cube

    Given a fixed surface area for a cube and a sphere.

    A= 1m^2

    Determine the volume of the cube and the sphere. Which one has greater volume?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by caligyrl4lyfe View Post
    Given a fixed surface area for a cube and a sphere.

    A= 1m^2

    Determine the volume of the cube and the sphere. Which one has greater volume?

    Well...the surface area of a square is SA=6s^2, where s is the side of the cube, and SA=4\pi r, where r is the radius of the sphere.

    First find s and r, given that SA=1, then find the volume, noting that V_{\text{cube}}=s^3, and V_{\text{sphere}}=\tfrac{4}{3}\pi r^3.

    Do these processes make sense?

    --Chris
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  3. #3
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    I'm still a little confused.

    So, would I solve SA= 6s^2 for s.


    And would I find r by solving <br /> <br />
SA=4\pi r<br />
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by caligyrl4lyfe View Post
    I'm still a little confused.

    So, would I solve SA= 6s^2 for s.


    And would I find r by solving <br /> <br />
SA=4\pi r^{\color{red}2}<br />
    Yes, where SA=1.

    So you would be solving 1=6s^2 for s, and 1=4\pi r^{\color{red}2} for r.

    Then you would plug these values into the corresponding volume equations I gave you. Then see which one is larger.

    --Chris

    EDIT: I caught a tiny mistake on my part. The surface area of a sphere is given by SA=4\pi r^{\color{red}2}, not SA=4\pi r.

    My mistake!
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  5. #5
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    So, is

    s= 2.24

    and

    r=3.40
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Yes, where SA=1.

    So you would be solving 1=6s^2 for s, and 1=4\pi r^{\color{red}2} for r.

    Then you would plug these values into the corresponding volume equations I gave you. Then see which one is larger.

    --Chris

    EDIT: I caught a tiny mistake on my part. The surface area of a sphere is given by SA=4\pi r^{\color{red}2}, not SA=4\pi r.

    My mistake!
    Quote Originally Posted by caligyrl4lyfe View Post
    So, is

    s= 2.24

    and

    r=3.40
    How did you get those values?

    1=6s^2\implies \tfrac{1}{6}=s^2\implies s=\sqrt{\tfrac{1}{6}}\approx \color{red}0.41

    1=4\pi r^2\implies r^2=\frac{1}{4\pi}\implies r=\sqrt{\frac{1}{4\pi}}\approx \color{red}0.28.

    Now that we know what s and r are, find V_{\text{cube}} and V_{\text{sphere}}.

    --Chris
    Last edited by Chris L T521; September 30th 2008 at 09:49 PM. Reason: darn typos!
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  7. #7
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    WHEW!

    I finally got it!

    The sphere has a larger volume...

    Thanks Chris~
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