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Volume of a solid using disks/washers

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid generated by rotating the region enclosed by [itex]y=\frac{1}{1+x^2}[/itex] , x=-1,x=1 and y=0 about the line y=2



    2. Relevant equations

    pi(outer radius)^2-pi(inner radius)^2

    3. The attempt at a solution

    Since i am rotating around a horizontal line i figured disks/washers would be a better method than cylindrical shells for this problem. First i sketched the graph. It is sort of an upside down parabola enclosed by the x axis and x=-1 and x=1. For the outer radius I got 2 and for the inner radius I got [itex]2-\frac{1}{1+x^2}[/itex]. Since the limits of integration are -1 and 1 and its a parabola I figured I can use symmetry to make it from 0 to 1 and multiply the whole thing by 2.
    Step 1:
    [itex]2\pi \int_0^1(2)^2-(2-\frac{1}{1+x^2})^2\,dx[/itex]

    Step 2:
    [itex]2\pi \int_0^1(4)-(4-\frac{4}{1+x^2}+\frac{1}{(1+x^2)^2})dx[/itex]

    Step 3:
    [itex]2\pi \int_0^1\frac{4}{1+x^2}-\frac{1}{(1+x^2)^2}dx[/itex]

    Step 4:
    [itex]2\pi \int_0^1\frac{4(1+x^2)-1}{(1+x^2)^2}dx[/itex]

    Step 5:
    [itex]2\pi \int_0^1\frac{4x^2}{(1+x^2)^2}dx[/itex]

    Here I hit a wall because I am not entirely sure how to integrate this. In fact I am not even sure if I am correct up to this point. I would appreciate it greatly if you guys could point me in the right direction.
     
  2. jcsd
  3. Apr 2, 2014 #2

    SammyS

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    Hello toothpaste666. Welcome to PF !

    It's not a parabola, but it is symmetric w.r.t the y-axis.

    4(1 + x2) - 1 = 4x2 - 3 ≠ 4x2 .
     
  4. Apr 2, 2014 #3
    Thank you! Wow I cant believe I made that mistake thank you for catching me on that. Wouldnt it be 4X^2+3? The rest of it is right so far though? I was a little unsure about my inner and outer radius and limits of integration. If that is the case I have been having trouble finding a method to integrate that. I played around with a couple of substitutions and partial fractions but I didn't make much progress.
     
    Last edited: Apr 2, 2014
  5. Apr 2, 2014 #4

    haruspex

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    I wouldn't do step 4. After step 3, the left hand integral is well-known. For the other, try the same trig substitution that solves the left hand one.
     
  6. Apr 3, 2014 #5
    Oh i see i should have split the integral. I ended up just making it more complicated. So i should let x = tan(theta) on the integral on the right?
     
  7. Apr 3, 2014 #6

    haruspex

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    That worked for me.
     
  8. Apr 4, 2014 #7
    I got it. Thank you so much!
     
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