# Evaluating a double integral over a possible unbounded region

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

-
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's a basic tutorial and quick reference. There's an "edit" link under the question. – joriki Oct 14 '12 at 9:07
The title talks about a double integral, but the question contains no integral. Perhaps you mean "integrate" instead of "evaluate"? – joriki Oct 14 '12 at 9:11
Yes, sorry, evaluate the integral of the function over the given region. – CAF Oct 14 '12 at 9:12
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. – joriki Oct 14 '12 at 9:14
You edited the question, but now it still doesn't mention any integral? – joriki Oct 14 '12 at 9:26

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.

-
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) – CAF Oct 14 '12 at 9:18
@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. – joriki Oct 14 '12 at 9:22
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? – CAF Oct 14 '12 at 9:29
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. – joriki Oct 14 '12 at 9:30
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? – CAF Oct 14 '12 at 9:33

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

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 $a=2x$.

-
Graphs help so much with problems of this sort! – amWhy Nov 27 '13 at 13:44
@amWhy: Thanks my friend. Thanks. :-) – S. Snape Nov 27 '13 at 13:46