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Math Help - Volume of water in a hemispheical bowl

  1. #1
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    Volume of water in a hemispheical bowl

    Need to prove that the formula for the volume of water V in a hemispherical bowl of radius r when the depth of water is h is given by:

    V = ((pi*h^2)/3) * (3r-h)

    I assume using integration. Solids of revolution maybe?

    Any help appreciated.

    Thanks
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by anonymous_maths View Post
    Need to prove that the formula for the volume of water V in a hemispherical bowl of radius r when the depth of water is h is given by:

    V = ((pi*h^2)/3) * (3r-h)

    I assume using integration. Solids of revolution maybe?

    Any help appreciated.

    Thanks
    Draw a diagram. Let x be the variable depth and R be the variable radius.

    (r-x)^2+R^2=r^2

    r^2+x^2-2rx+R^2=r^2

    R^2=2rx-x^2

    dV=\pi R^2dx

    = \pi(2rx-x^2)dx

    V=\int_{0}^{h}\pi(2rx-x^2)dx

    = \pi(rx^2-\frac{x^3}{3})\big|_0^{h}

    = \pi(\frac{3rx^2-x^3}{3})\big|_0^{h}

    = \pi(\frac{3rh^2-h^3}{3})

    = \frac{\pi h^2}{3}(3r-h)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by anonymous_maths View Post
    Need to prove that the formula for the volume of water V in a hemispherical bowl of radius r when the depth of water is h is given by:

    V = ((pi*h^2)/3) * (3r-h)

    I assume using integration. Solids of revolution maybe?

    Any help appreciated.

    Thanks
    Draw a diagram, and the method of disks looks like the easiest method.

    CB
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