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Volume of Sold with a Known Crosssection

  1. Apr 7, 2007 #1
    the base is bounded by x^2 + y^2 = 4
    the solid is formed by isosceles right triangles perpendicular to x axis. the hypotenuse of each triangle is sitting on the base of the figure

    i just need to know what area formula i would start out with, is it just bh/2 ?
     
    Last edited: Apr 7, 2007
  2. jcsd
  3. Apr 7, 2007 #2

    HallsofIvy

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    An "isosceles right triangle perpendicular to the x axis" has hypotenuse y. Since it is isosceles, a2+ a2= 2a2= y2. That is, each leg has length [itex]a= y\sqrt{2}[/itex] and so its area is (1/2)bh= (1/2)(a*a)= (1/2)(2y)= y.
     
  4. Apr 9, 2007 #3
    hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
    (1/4)y^2. Right?
     
  5. Apr 10, 2007 #4
    yes, you are right.
     
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