Volume of Sold with a Known Crosssection

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Main Question or Discussion Point

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 ?
 
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Answers and Replies

  • #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.
 
  • #3
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hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
(1/4)y^2. Right?
 
  • #4
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hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
(1/4)y^2. Right?
yes, you are right.
 

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