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Math Help - Confusing Volume Problem

  1. #1
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    Exclamation Confusing Volume Problem

    I'm completely stumped as far as this problem goes. I've got to find the volume of this solid and I can't figure out the area of the cross section. It starts out like this:

    The solid lies between planes perpendicular to the x-axis at x=0 and x=4. The cross sections perpendicular to the axis on the x interval [0,4] are squares whose diagonals run from y=-√x to y=√x.
    After talking to my instructor, I realized that the cross section I can see on the graph is half of the square cross section. Basically I figure that I find the volume of what I can see from the graph and double it.

    So I find that the base of the triangle is 2√x. Unfortunately I can't seem to go any farther, because I have neither a height nor any of the two equal sides. I'm wondering if there's a way I haven't thought of to find the area of this triangle.

    I realize, by the way, that this is more of a geometry problem than a calculus one, but I posted it here anyway in case there was something in the initial concept I missed.
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  2. #2
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    Quote Originally Posted by SethP View Post
    I'm completely stumped as far as this problem goes. I've got to find the volume of this solid and I can't figure out the area of the cross section. It starts out like this:



    After talking to my instructor, I realized that the cross section I can see on the graph is half of the square cross section. Basically I figure that I find the volume of what I can see from the graph and double it.

    So I find that the base of the triangle is 2√x. Unfortunately I can't seem to go any farther, because I have neither a height nor any of the two equal sides. I'm wondering if there's a way I haven't thought of to find the area of this triangle.

    I realize, by the way, that this is more of a geometry problem than a calculus one, but I posted it here anyway in case there was something in the initial concept I missed.
    The triangles are isosceles right triangles so the height is \sqrt{x}
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  3. #3
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    Dude, thanks! I totally forgot geometry class.
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