Volume of a rotated graph?

  1. Mar 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid found by rotating the area bound between
    the curves x = 1, y = x^2, y = 0, around x-axis.


    2. Relevant equations
    I know you have to solve this using the shell method, but I just get extremely confused while trying to do it. Maybe someone could explain it to me?


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 19, 2008 #2
    This is the formula for the cylindrical shell method:
    [tex]\int_a^b2\pi xf(x)dx[/tex]
    you can also use [tex]\int_a^bA(x)dx[/tex]
    The second one would be easier.
     
    Last edited: Mar 19, 2008
  4. Mar 19, 2008 #3
    Sorry! I was mistaken when I said the area was unbounded. The first thing you need to do is find where these curves intersect and determine the leftmost and rightmost boundaries of X.
     
  5. Mar 19, 2008 #4
    Normally to find points of intersection you would set the two curves you wish to find the x values that satisythis equation. However they are much more easily found in this particular problem by drawing a quick sketch. It comes out at x=0 to x=1

    Also, this does not require shell method at all. Shell method is for when you cannot create an infinite number of lines between between your boundaries that is perpundicular to your axis of revolution such that one of those lines would hit the same curve twice. In this case all you need to is:
    [tex]\pi\int_0^1\ (x^2)dx[/tex]
     
  6. Mar 19, 2008 #5
    Should be (f(x))^2.
     
  7. Mar 19, 2008 #6
    Would that end up being...

    [tex]\int^{1}_{0}(\pi r^2)dx[/tex]

    where r = "Outer radius" (1) minus "inner radius" ([tex]x^2[/tex])

    ?

    (I hope so, I have a test tomorrow)
     
  8. Mar 19, 2008 #7
    No, because x = 1 is the vertical boundary in your integration sign.
     
  9. Mar 21, 2008 #8
    Ok...I think I get it.

    [tex]\int^{1}_{0}\pi (x^2)^2 dx[/tex]

    = [tex]\pi*\int^{1}_{0}x^4 dx[/tex]

    = [tex]\pi*(1^5/5-0)[/tex]

    = [tex]\pi/5[/tex]

    Is that right?
     
  10. Mar 21, 2008 #9
    Looks right.

    I should've said x = 1 is the vertical line that limits how far horizontally you should integrate.
     
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