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Double integrals over finite region

  1. Oct 13, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data

    Evaluate [tex] \int_{R} \int \frac{xy^2}{(4x^2 + y^2)^2} dA [/tex] where R is the finite region enclosed by [itex] y = x^2\,\,\text{and}\,\, y = 2x [/itex]

    3. The attempt at a solution

    I think the easiest way to integrate is to first do it wrt x and then wrt y, i.e [tex] \int_{0}^{4} \int_{\frac{y}{2}}^{\sqrt{y}} \frac{xy^2}{(4x^2 +y^2)^2} dx dy, [/tex] where [tex] R = [(x,y) : \frac{y}{2} ≤ x ≤ \sqrt{y} , 0 ≤ y ≤ 4 ] [/tex]

    Compute inner integral: Take [itex] y^2 [/itex] out of the integrand and let [itex] u = 4x^2 + y^2 [/itex]. Doing so and evaluating at [itex] x = \sqrt{y}, x = \frac{y}{2} [/itex] gives [tex] \frac{y^2}{8} [ -\frac{1}{4y + y^2} + \frac{1}{2y^2} ] [/tex]

    Can somebody confirm this is correct up to here? When I do the subsequent integration wrt y, I get a negative answer for volume.
    EDIT: It looks like this can be evaluated via polar coordinates which I will do once I have an answer using Cartesian.
     
    Last edited: Oct 13, 2012
  2. jcsd
  3. Oct 13, 2012 #2
    Did you notice that your integrand is undefined at the origin? In fact, it is unbounded near the origin, and the limit doesn't exist at the origin (you should be able to verify this). So the conditions of Fubini's Theorem (which allows one to compute double integrals by using iterated integrals) may not be satisfied.

    I can confirm your first antiderivative (w.r.t. x), but I got a positive answer, which one would expect given that the integrand is positive on the region of integration. So you must've made an error somewhere during the second computation.

    I would advise you to also try integrating w.r.t. y first. The resulting problem is much more difficult computationally; you'll need to use a trig sub or partial fraction decomposition. I got a different answer for the dydx version than I did for the dxdy version (though it's possible I made a mistake somewhere). If I didn't make a mistake and they really are different, then that means ...
     
  4. Oct 13, 2012 #3

    CAF123

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    Here is what I did for my second computation wrt y: [tex] \int_{0}^{4} \frac{y^2}{8} \frac{1}{4y + y^2} dy + \int_{0}^{4} \frac{y^2}{8} \frac{1}{2y^2} dy, [/tex] which when simplified gives [tex] - \int_{0}^{4} \frac{y}{32} + \frac{1}{8} dy + \int_{0}^{4} \frac{1}{16}dy [/tex] Evaluating this gives -1/2?

    If you got a different answer using dydx and dxdy then that must mean that Fubini's thm is not applicable? In general, how else would you compute?
     
  5. Oct 13, 2012 #4
    There's some scary algebra in that first simplified integral. Might want to check that.

    Well, Fubini says if the "real" double integral is finite, then the iterated integrals will be the same and equal to the "real" integral (actually Fubini requires the integral of the absolute value to be finite, but our integrand is nonnegative and so the absolute value doesn't change anything).

    So if our integral was finite, then the iterated integrals would be the same. But they're not (I think). So what does that say about the "real" integral?
     
  6. Oct 13, 2012 #5

    CAF123

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    Oops, I feel like an idiot.
    We have [tex] -\int_{0}^{4} \frac{1}{8} - \frac{1}{2(y+1)} dy + \int_{0}^{4} \frac{1}{16} dy, [/tex] which gives [itex] -\frac{1}{4} + \frac{1}{2}ln(2) [/itex]

    If the double integral is not finite then it is not defined?
    If the iterated integrals are not the same, then does this mean what I have done above is meaningless?
     
  7. Oct 13, 2012 #6
    Yes. That is what I got.

    Basically. You should check your text to see what your version of Fubini's theorem says. If it is a standard undergrad engineering type text (as opposed to an advanced text geared more towards pure math majors), then it most likely phrases things in terms of Riemann integrability instead of finiteness.

    I wouldn't say that. As annoying as the iterated integral computations are here, it is still probably the easiest way to show that the given integrand is not Riemann integrable on the given region. Also, you have a nice, "tangible" example that demonstrates when iterated integrals don't match up. And you're getting some practice with some techniques of integration that you're likely a bit rusty using.

    Basically, you're learning. Isn't that the point? :biggrin:
     
  8. Oct 13, 2012 #7

    CAF123

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    My text says: An integral can be expressed as an iterated integral (in either order) so long as f is bounded on R, f is discontinuous only on a finite number of smooth curves and the iterated integrals exist.

    I am not entirely sure I understand what you are saying - is the value I computed correct or is it the case that if I tried to compute the 'iterated' integral the other way, I would find a different answer so that suggests the function is not integrable in the given region? (I.e there is no answer)

    Just for interest, can I do this by polar coordinates? The denominator suggests I might be able to.
     
  9. Oct 13, 2012 #8
    Um ... so I redid my computations for the dydx integral, and it looks like I did make an error in my first attempt. Now I'm getting them to be equal. I'd still suggest that you verify that.

    Even if the iterated integrals are equal, that's still not enough to say that the double integral is "good". I'm curious as to how you've been taught to deal with these improper integrals.

    Yes, you should be able to use polar coordinates to deal with this problem, but the resulting computations aren't any nicer.
     
  10. Oct 13, 2012 #9

    CAF123

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    I have only been working on double integrals for a week. The only improper integral we touched on was that of e^(-x^2) from -infinity to infinity.
    Going back to Fubini's thm for a moment: if the function is discontinuous at (0,0), then iterated integrals still work because the theorem states that the function can be discontinuous at a few places and still hold? Sorry if this appears to be quite a silly question, my professor said the proof is quite complex and requires deeper study of analysis.
     
  11. Oct 13, 2012 #10
    It's the unboundedness of the integrand that makes this one "bad", not the fact that it's discontinuous at the origin.

    I wouldn't worry too much about this problem if I were you. It sounds like you haven't been given all of the necessary tools for justifying why the iterated integrals work here. I suppose it's meant as an example of how math problems in the "real world" can be much nastier than those that are tailored for undergrad courses.
     
  12. Oct 13, 2012 #11

    CAF123

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    This question is a hand in question. Is it not still discontinuous at (0,0) because the limit at x=0 is non existent?
    EDIT: I see you say that the function is still discontinuous at (0,0) on reading your last post again, so disregard my last question. But what exactly does bounded mean?
     
  13. Oct 14, 2012 #12

    CAF123

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    Someone today suggested splitting the region into 2 smaller regions and then doing the integral. I don't see how that would help since (0,0) is at the edge of the boundary.
     
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