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Finding volume using cross sections

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    find the volume of the solid whose base is the region bounded by y^2=9x and x=1 in the xy-plane. each cross section perpendicular to the x-axis is an isosceles right triangle with its hypotenuse in the xy-plane.


    2. Relevant equations
    [tex]V=\int_{a}^{b}A(x)dx[/tex]


    3. The attempt at a solution
    well first i solved for y and got y=3sqrt(x) and y=-3sqrt(x), then i figured out the length of the hypotenuse to be 3sqrt(x)-(-3sqrt(x))=6sqrt(x)
    but the formula for the area of an isosceles right triangle is 1/2a^2 where a is one of the triangles legs. how do i figure out the length of one of the legs?
     
  2. jcsd
  3. Dec 10, 2011 #2

    Mark44

    Staff: Mentor

    I hope you have drawn sketches of everything, especially the triangles. The base of each triangle is 2y = 6√x. Each triangle is made up of two isosceles right triangles. The base of each is y = 3√x. Is that enough to get you started?
     
  4. Dec 10, 2011 #3
    yeah got it, thanks!
     
  5. Dec 10, 2011 #4
    just a quick question to check if i got the right answer, the area will then be 1/2(3sqrt(2x))^2=1/2*9*2x=9x
    and integrating i get a volume of 9/2 since my limits will be from 0 to 1?
     
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