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Math Help - Finding maximum volume of an ellipse (Optimisation under differentiation)

  1. #1
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    Finding maximum volume of an ellipse (Optimisation under differentiation)

    OK
    So, I have half an ellipse with the equation y = sqrt(1296-x^2). Here is the graph:
    Image Upload - ti
    It is actually a curved 3d structure with the height 36m, so I found out the above equation by using the equation of the ellipse.
    The question is:
    To find the maximum volume of this structure (or curve)?

    I can find the area by using integration but that wouldnt be the same as volume?
    Any ideas?

    PLS PLS HELP

    PS. THE BUILDING LOOKS LIKE THIS:
    Image Upload - 123
    Is it fair to say that it is an ellipse (half)?


    [IMG]file:///C:/Users/Ameya/AppData/Local/Temp/moz-screenshot-7.png[/IMG]

    THANK YOU SO MUCH!!!!
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  2. #2
    Junior Member BariMutation's Avatar
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    You'll need to use a triple integral (at least, that's the easiest way imo). I would do it in the order of dydxdz, but that's up to you. Your limits on y would be 0 to sqrt(1296-x^2), your limits on x would be found be using 0 = sqrt(1296 - x^2) and solving for x, and your limits on z would be -36 to 36. I believe this would be the correct answer, someone correct me if I'm wrong.
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  3. #3
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    It would make more sense to call that a semi-circle.
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  4. #4
    Junior Member BariMutation's Avatar
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    Actually, I just thought of this - isn't that building more of 1/4 of an ellipse? It's cut in half by the ground, but it also seems to be cut in half again just by design on the front.
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  5. #5
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    Well, the building is not the design! And, again, it is not an ellipse, it is a circle: If y= \sqrt{1296- x^2} then y^2= 1296- x^2 so x^2+ y^2= 1296= 36^2. That is a circle with center at (0, 0) and radius 36. But we are talking about volume and for that we need three dimensions. What happens behind the semi-circle that we see? Is that a volume of rotation (hemisphere) or does it go straight back- a semi-circular cylinder?

    The volume of a hemisphere is \frac{2}{3}\pi r^3, about 73287 for r= 36. The volume of a semi-circular cylinder is the area of the semi-circle, \frac{1}{2}\pi r^2, about 2035 for r= 36, multiplied by the depth of the cylinder that we do not know.

    Actually, the shape in the photo looks more like part of a hyperbola to me.
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