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Really ugly graph-volume problem

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid generated by revolving the region about the x-axis.
    Region = the area separated by given function and lines:

    x = sqrt(2y)/(y^2 +1)
    y=0.
    3. The attempt at a solution

    i just need to solve the integral of ∫x dy... How? Is there any simple way to solve it?
     
  2. jcsd
  3. Oct 4, 2012 #2

    LCKurtz

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    Don't you want ##\pi\int x^2\, dy##? Write out your integral, complete with limits so we know exactly what your problem is.
     
  4. Oct 5, 2012 #3
    that would be the case if I wanted to revolve it around the Y-axis.

    Here's a graph showing the problem:http://desmond.imageshack.us/Himg14/scaled.php?server=14&filename=63679970.jpg&res=landing [Broken]

    i need to find the area between the y=1 and the graph (which I require the integral of x to find). that way I can find the average radii of the solid, remake it into a cylinder with the same area and revolve it to find a volume identical to the 1 I'm supposed to find.

    Duno if this will work, tho.
     
    Last edited by a moderator: May 6, 2017
  5. Oct 5, 2012 #4

    LCKurtz

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    Yes, I was thinking you were revolving about the Y axis. I see you have added the line y=1 to the problem. To revolve about the x axis given you have x in terms of y, you want to use a dy integral and the shell method with an integral of this form:$$
    V =2\pi\int_a^b yf(y)dy= 2\pi\int_0^1\frac{\sqrt 2 y^\frac 3 2}{1+y^2}\, dy$$which isn't quite the form in your original post. Whether that is an easy integral to work isn't obvious to me at the moment.
     
    Last edited by a moderator: May 6, 2017
  6. Oct 6, 2012 #5
    Yes, your method sounds much better than mine. I just learned the shell method yesterday so I don't have any problems with that.

    But the integral of x needs to be solved anyway (and it's without thinking about 2*pi outside, imo).
     
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