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Volume of cross sections using isosceles right triangle

  1. May 28, 2012 #1
    1. The problem statement, all variables and given/known data
    The area is bounded by the equation y=x^2, x axis and the line x=3 and is perpendicular to the x axis


    2. Relevant equations
    A=1/2bh
    Height=Base


    3. The attempt at a solution
    A=1/2B^2
    V=1/2B^2 from 0 to 3
    V=1/2 B^3/3 from 0 to 3
    V=B^3/6 from 0 to 3
    V=27/6

    please tell me if im doing something wrong, on the right track or i'm correct
     
  2. jcsd
  3. May 28, 2012 #2

    HallsofIvy

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    Are we to assume that the cross sections perpendicular to the x-axis are "isosceles right triangles"- with one leg perpendicular to the x-axis or the hypotenuse?

    I think what you are trying to do is correct but your notation is very strange. You are given information about x and y- use them, not new variables! Assuming it is the leg that lies perpendicular to the x-axis then the other leg is just y= [itex]x^2[/itex]. The area of such a triangle is [itex](1/2)(x^2)(x^2)= x^4/2[/itex]. The "thickness" of each such cross-section is [itex]\Delta x[/itex] so the volume is [itex](1/2)x^4\Delta x[/itex]. In the limit that becomes an integral.
     
  4. May 28, 2012 #3
    Its perpendicular to the x axis not the hypotenuse. Also I combined the variables as one because the height is equal to the base. Also, we weren't taught thickness so that information is not needed. However , in my problem i realize i forgot to plug in X^2

    So redoing my work it would be :
    V=x^5/10 from 0 to 3 which is equal to 243/10 , therefore your answer should agree with mine after you integrate your work. I'm just not sure where the thickness part come in since i have not learned that yet
     
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