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Math Help - Volume

  1. #1
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    Volume

    The base of a solid is the region bounded by the parabola y=\frac {1}{2}x^2 and the line y = 2. Each plane section of the solid perpendicular to the y-axis is an equilateral triangle. Find the volume of the solid.

    I did not understand the line "Each plane ............ equilateral triangle"
    What shape does the solid have?
    I have drawn the graph, but volume of which solid is it asking?
    Attached Thumbnails Attached Thumbnails Volume-graph34.jpg  
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  2. #2
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    Quote Originally Posted by Shyam View Post
    The base of a solid is the region bounded by the parabola y=\frac {1}{2}x^2 and the line y = 2. Each plane section of the solid perpendicular to the y-axis is an equilateral triangle. Find the volume of the solid.

    I did not understand the line "Each plane ............ equilateral triangle"
    What shape does the solid have?
    I have drawn the graph, but volume of which solid is it asking?
    a volume whose cross sections perpendicular to the y-axis are equilateral triangles ... it doesn't have a special name.

    V = \int_c^d A(y) \, dy

    A = \frac{\sqrt{3}}{4} s^2 , where s = 2x = 2\sqrt{2y}

    V = \sqrt{3} \int_0^2 2y \, dy
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  3. #3
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    different base, but check out the link for a good visualization ...

    http://mathdemos.gcsu.edu/mathdemos/...ross75slab.gif
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  4. #4
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    Quote Originally Posted by skeeter View Post
    different base, but check out the link for a good visualization ...

    http://mathdemos.gcsu.edu/mathdemos/...ross75slab.gif
    The answer says

    Area \;A = \frac{1}{2}(2x)(\sqrt 3 x)=x^2 \sqrt3

    dV = A dy

    V = \int\limits_0^2 {x^2 \sqrt 3 {\text{ }}dy}  = \int\limits_0^2 {2y\sqrt 3 {\text{ }}dy}  = 4\sqrt 3

    I got it now. Thanks a lot.
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