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Math Help - another solid revolution / intergration T_T

  1. #1
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    another solid revolution / intergration T_T

    a hemispherical bowl of radius a units is filled with water to a depth of a/2 units.Findby intergration, the volume of the water.
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  2. #2
    Member SengNee's Avatar
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    Quote Originally Posted by chibiusagi View Post
    a hemispherical bowl of radius a units is filled with water to a depth of a/2 units.Findby intergration, the volume of the water.
    When r=a, the d=a, as the bowl is hemisphere, d=r.
    Therefore, when d=0.5a, r=0.5a.

    Volume of a hemisphere
    =\frac{2}{3}\pi r^3
    =\frac{\pi a^3}{12}


    I only know this method.
    I don't know the method by integration, sorry.
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  3. #3
    Senior Member JaneBennet's Avatar
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    You should get V=\pi\int_{\frac{a}{2}}^a{(a^2-x^2)}\,\mathrm{d}x.
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  4. #4
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    When r=a, the d=a, as the bowl is hemisphere, d=r.
    Therefore, when d=0.5a, r=0.5a.

    Volume of a hemisphere
    =\frac{2}{3}\pi r^3
    =\frac{\pi a^3}{12}
    If you cut a hemisphere in half the two parts are not hemispheres so this method doesn't work.

    I would use the equation for a circle y = \sqrt {a^2-x^2} and rotate the region between a/2 and a around the x-axis.
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