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Integration problem

  1. Sep 28, 2005 #1
    Right, I was looking over some past exam papers from these science-maths scholarship exams I was looking to do. It's all revision for me but it's been a while. Anyway, skimming over the double integration questions. Now if I recall correctly it's a fairly simple process, integrate the inner bit, then the outer et voila right? Ok ya got a bit more with oddly shaped areas (type I and II etc) and whatnot but it's still pretty straightforward.

    I hit a bump, embarrassingly just with the mechanics of integrating itself, I just couldn't damn remember what to do when the question came up like this.

    To elaborate, I'd come accross this question:

    Now, if I'm doing this correctly I initially ignore the outer nested bits and concentrate on this:

    Wtf do I do with that? I put it through mathematica and it comes up with some "null" argument jive.
    Skimmed numerous times through all relevant sections of my calculus book (by Anton) to no avail..

    Looked at the previous year's question and sure enough there was one just like it:

    With no such example in my obviously inferior recommended text I'm left to ask for help.

  2. jcsd
  3. Sep 28, 2005 #2


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    Dearly Missed

    Change the description of your region so that "y" runs from 0 to 2, and x runs from 0 to 1/2y.

    I ended up with something like 13/18.
    Last edited: Sep 28, 2005
  4. Sep 28, 2005 #3
    I'm at a loss, why would I do that? The limits are 2 to 2x for y and 0 to 1 for x, where did you get those other numbers from?
    And I don't understand how changing the limits would help at all, it's before that. I mean how does one integrate sqrt(y^3 + 1) ?
  5. Sep 28, 2005 #4


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    DRAW the region in the xy-plane!!
    See that your region could equally well be described by my limits.
    Do the x-integration first, and you'll see that all your problems are solved.
    Last edited: Sep 28, 2005
  6. Sep 28, 2005 #5
    Ah, my apologies, I see where you get this now. Thank you. It has been over a year since I've studied mathematics.

    Okay I did this with the new limits but I still came back to the same problem. After doing the x-integration I get:
    1/4 *y^2 * sqrt(y^3 + 1). Which I can't integrate for y, I tried substitution and tabular integration. Looked over the previous integration and couldnt see any slip.

    Thanks for your patience so far by the way..
  7. Sep 28, 2005 #6
    Try u=y3+1.
  8. Sep 28, 2005 #7
    OH MY GOD.. I'm so slow today. It's obvious. Thanks

    This is what I get for not studying anything the last 5 months. Brainrot.. :redface:
  9. Sep 28, 2005 #8
    Was this the actual question? It seems like it should be a double integral...
  10. Sep 29, 2005 #9
    The actual question is to evaluate this:

    Which I think is a missprint (I'd heard there was one on this paper) as otherwise I found it impossible to do. I substituted y/2 for x/2 and it worked out like the previous one.
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