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Evaluate the (double) integral of $\frac{xy^2}{(4x^2 + y^2)^2}$ over the finite region enclosed by $y= x^2$ and $y = 2x$. My question is: I have tried this by the method of iterated integrals but then I noticed that at $(0,0)$ the function is undefined.

How would you go about solving this? Also is the region here unbounded?

Many thanks

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    You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". [Here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)'s a basic tutorial and quick reference. There's an "edit" link under the question.2012-10-14
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    The title talks about a double integral, but the question contains no integral. Perhaps you mean "integrate" instead of "evaluate"?2012-10-14
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    Yes, sorry, evaluate the integral of the function over the given region.2012-10-14
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    The function isn't discontinuous at the origin; it's undefined there. Please edit any clarifications into the question itself; people shouldn't have to delve into the comments to understand the question. There's an edit button underneath the question.2012-10-14
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    You edited the question, but now it still doesn't mention any integral?2012-10-14

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As @joriki noted in his second post, we have a bounded region (see below) :

enter image description here

Now try to intersect the functions $x^2=y=2x$ to find out that $0\le x\le 2$ and so you get, as plot tells us, that: $$\int_0^{x=2}\int_{x^2}^{y=2x}\frac{xy^2}{(4x^2+y^2)^2}dydx$$ It is not hard exploiting the substitution $y=a\tan(\theta)$ for $$\int\frac{y^2}{(a^2+y^2)^2}dy$$ where $a=2x$.

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    Graphs help so much with problems of this sort!2013-11-27
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    @amWhy: Thanks my friend. Thanks. :-)2013-11-27
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On the first question: The integrand grows like $1/r$ at the origin, but the width of your region also decreases as $r$, so you should be OK. I'd integrate over $x$ first, since the numerator contains the inner derivative of the denominator.

On the second question: The problem explicitly says to integrate over the finite region enclosed by the curves. "Bounded" is just the more formal term for what they call "finite", so no, the region is not unbounded.

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    Exactly what I did. I integrated wrt x and then wrt y and got a positive answer which is as I expected since the function is positive over region. So the discontinuity at x=o is ok, for the reasons you described? (I also recall from Fubini's theorem that if there is a small number of discontinuities then iterated integrals is still valid)2012-10-14
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    @CAF: As I wrote under the question, this is not a discontinuity. You can interpret this integral as the limit as you extend the region ever closer to the origin. The integrals over all the finite regions are all unproblematic, so the question is just whether the results converge as the region approaches the origin, which they do.2012-10-14
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    Can I ask what you mean by 'the results converge as the region approaches the origin'. I think the limit is non existent at x=0, does this not imply a discontinuity?2012-10-14
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    The limit of what is not existent at $x=0$? Imply a discontinuity in what? Please take more care to use language precisely; it makes communication a lot more efficient.2012-10-14
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    Apologies. The limit of the function as x tends to 0 does not exist and so would this not imply that the graph of the function has a discontinuity there?2012-10-14
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    @CAF: I don't know what it means for the graph of a function to have a discontinuity. Formally, the graph of a function is defined as the set of pairs of its inputs and outputs, and there is no notion of continuity for sets (that I know of). If you mean that the function itself is discontinuous at the origin, the answer is no, since the function isn't defined there. Asking whether a function is continuous only makes sense on the function's domain.2012-10-14
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    Ah I think the limit as x tends to 0 of the function is equal to 0 but at x=0, it is undefined. Just to check I understand correctly, my method of iterated integrals is valid?2012-10-14
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    One more question: in your earlier post, you said all 'results converge as the region approaches the origin'. What does this mean?2012-10-14
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    @CAF: It means that if you define an increasing sequence of regions on which the function is defined such that the union of the regions is the entire region of interest minus the origin (for instance, the $n$-th region could be the region of interest with a circle of radius $1/n$ around the origin cut out), then the sequence of their integrals converges, and it's its limit that we're trying to calculate. There are also other ways of defining the integral if the function isn't defined on the whole boundary, but all this is a bit needlessly formal; the point is that the integral doesn't blow up.2012-10-14
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    Or the region could be the part on or to the right of the line $x=t$ where $0. Then as $t$ goes to $0$ from above, the region increases toward the full domain of definiton (part between graphs minus origin). This approach is set up to work with one of the ways of setting up as iterated integral.2012-10-14
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    @coffeemath: Yes, that's slightly simpler and more practical :-)2012-10-14
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    Since the integrand is positive in the region and the objective is to evaluate the integral, questions of convergence can be ignored at first - just set up the integral and "do it". If you can do the integrals, you will get a positive number or infinity.2012-10-17
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    @CAF : I put the @ symbol here to make sure you read my comment above.2012-10-17