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Volume bounded by graphs

  1. Sep 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Set up the integral (but do not solve) for the volume of the object created by rotating the region bounded by y = arctan(x) and y = arcsin(x) in the first quadrant.

    2. Relevant equations

    I = ∏∫(f(x)^2 - g(x)^2) dx

    3. The attempt at a solution


    a.) rotate about he x axis

    I came up with

    I = ∏∫(arcsin(x))^2 - (arctan(x))^2 dx between [0,1]

    b.) rotate about the y -axis

    I came up with

    I = ∏∫(arctan(x))^2 - (arcsin(x))^2 dx between [0,∏/2]

    Did I do this correctly.
    Thank you.
     
  2. jcsd
  3. Sep 4, 2013 #2

    Zondrina

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    Lets start with part a).

    Re-write your functions as ##x(y)##, otherwise you wont be able to find the limits of integration ( The only solution to ##arcsin(x) = arctan(x)## is ##x=0##. ##x=1## is not a solution. ).

    The solid formed from this process will be sort of diamond shaped.
     
    Last edited: Sep 4, 2013
  4. Sep 4, 2013 #3

    LCKurtz

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    Those two curves do not enclose an area in the first quadrant.
     
  5. Sep 4, 2013 #4
    LCkurtz that is what the question said exactly straight to the teeth from my course pack.


    Once again word for word.

    Set up the integral (BUT DO NOT SOLVE) for the volume of the object created by rotating the region bounded by y = arctan(x)and y = arcsin(x) in the first quadrant.
    a.) rotate about the x axis
    b.) rotate about the y axis

    That is word for work LCkurtz. You see I chose my boundary for revolving around x to be 1 because the domain of arcsin(x) goes form -1 to 1. And I chose ∏/2 when revolving around the y axis because the range of arcsin(x) and arctan(x).
    Those are my thoughts.
     
  6. Sep 4, 2013 #5

    Zondrina

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    That's why I said, you have to change the functions into functions of ##y## and use a different method entirely. You wont be able to solve this with the usual method you've been using. The functions as is in the problem only have ONE intersection and therefore do not bound any particular area.
     
  7. Sep 4, 2013 #6

    LCKurtz

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    I don't care if it is "straight to the teeth", written in gold ink, and blessed by the Pope. Those two curves do not enclose a region in the first quadrant. A reasonable GUESS might be to add the line ##y = \frac \pi 2## to the area description. That would define a region that is infinite in extent requiring improper integrals. You have to properly define your region before you try setting up any integrals.
     
  8. Sep 4, 2013 #7
    Haha. The Pope doesn't belong in physics forum come on!
    But seriously though I said the upper bound for the integral was PI/2. But because the PI/2 is not included in the range of arctan(x) does this mean that it is infinite in extent? What about the the upper limit of integration I gave when it is around the x axis? I said I wanted it to go from 0 to 1 because that is the range of arcsin(x).
    I'm kind of lost with what to do because like I said this is straight from my course pack.
     
  9. Sep 4, 2013 #8

    Zondrina

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    Here's a graph to help you understand why you can't do this the way you're thinking :

    http://gyazo.com/f163b065b07aa3a19ab25798611de7c6

    The graphs will never intersect again as ##x → ∞##. Therefore there is no other point of intersection as to form a closed area with.

    Re-writing you get ##x = sin(y)## and ##x = tan(y)##.

    When does ##sin(y) = tan(y)##? One point is ##y=0##. Solving the rest will give you your limits for ##y##. Then you need to integrate with respect to ##y## and not ##x##.
     
  10. Sep 4, 2013 #9
    OK, but I have one method to use which is the disk method. If I use the disk method and revolve it around the x axis wouldn't I need to integrate with respect to x?
     
  11. Sep 4, 2013 #10

    LCKurtz

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    Here is a picture of the presumed region, assuming the upper boundary is ##y = \frac \pi 2##. The actual region continues on infinitely far to the right as the arctan nears its asymptote.

    shaded.jpg
     
    Last edited: Sep 4, 2013
  12. Sep 4, 2013 #11
    Thanks, I see what you are saying but how am I supposed to do this problem then? Revolving it around the x axis and using the disk method? Maybe I could do it with respect to x and go from 0 to 1 then go from 1 to infinity? I mean this makes no sense as a question considering the chapter of information I'm supposed to use.
     
  13. Sep 4, 2013 #12

    LCKurtz

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    Yes, if you want to use the disk method for revolving about the x axis, you would need two integrals just as you say from 0 to 1 and from 1 to infinity, because the upper radius changes.

    You can do the disk method around the y axis with one integral. Show us what you get for each.
     
  14. Sep 4, 2013 #13
    Well I'm just supposed to set up the integral not actually do it.

    This is what I set up though.


    For about x I have
    I_1 = ∏∫(arcsin(x))^2 - (arctan(x))^2 dx from 0 to 1.
    I_2 = ∏∫((∏/2)^2 - (arctan(x))^2 dx and from 1 to infinity

    Volume is then I_1 + I_2 = A

    About y

    I = ∏∫(tan(x))^2 - (sin(x))^2 dy from 0 to ∏/2.

    EDIT: I think it is OK now.
     
    Last edited: Sep 4, 2013
  15. Sep 4, 2013 #14

    LCKurtz

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    You mean volume, not area.

    Your one about x is OK, but the second one isn't. You need things in terms of y and dy integrals.
     
  16. Sep 4, 2013 #15
    OK look at the post again. I did mean volume and yes I forgot to put it in terms of y.
     
  17. Sep 4, 2013 #16

    LCKurtz

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    No, close but still not right. It should be of the form ##\int x_{right}^2 - x_{left}^2~dy## where the ##x## values are given ##x## in terms of ##y## from the curves.
     
  18. Sep 4, 2013 #17
    About y

    I = ∏∫(tan(y))^2 - (sin(y))^2 dy from 0 to ∏/2.

    ? Opps
     
  19. Sep 4, 2013 #18
    good?
     
  20. Sep 4, 2013 #19

    LCKurtz

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    That's correct now.
     
  21. Sep 4, 2013 #20
    Yeah I just totally didn't pay attention my brain said one thing hand did the other.
    Thanks
     
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