Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question was a question in an exam years ago.

Find the values of the following integrals.
(i) $$\int_\Gamma\dfrac{xdy-ydx}{x^2+y^2},$$ where $\Gamma$ is the curve $x=t\cos t$, $y=t^2\sin t$, for $t\in [2\pi,6\pi]$.
(ii) $$\int_A\dfrac{dx}{(1+\|x\|_2^4)^{\frac{1}{4}}},$$ where $A=\{x=(x_1,x_2,x_3)\in \mathbb R^3| x_2\gt0, \|x\|_2\le3\}$.

For $(i)$, I tried substituting the parametrisation of $\Gamma$ into the integral, but got nothing. I thought that this integral might be exact, but found no exact anti-derivatives...
For $(ii)$, I wrote it as $$\frac{4\pi}{2}\int_0^3\dfrac{r^2dr}{(1+r^4)^{\frac{1}{4}}}.$$ The I made the change of variables $s=(1+r^4)$ to re-write it as $$\frac{\pi}{2}\int_1^{82}\dfrac{ds}{(s(s-1))^{1/4}}.$$ Then this becomes an improper integral! Since that integrand is bounded by $(s-1)^{-1/4}$ and $\int (s-1)^{-1/4}ds$ converges in that interval, I can also prove the convergence of this improper integral. But how should I obtain the value? It seems that partial fraction decomposition works not so well here...
Thanks in advance for any hint or help, and edits.

share|cite|improve this question
$$\int \dfrac{xdy-ydx}{x^2+y^2}=\int\dfrac{d\frac yx}{1+\left(\frac yx\right)^2}$$.Put $\frac yx=\tan\theta$ – lab bhattacharjee Jul 25 '13 at 15:58
The first one is not exact. – Integral Jul 25 '13 at 16:01
Are you sure that $y=t^2\sin(t)$? – Mhenni Benghorbal Jul 25 '13 at 16:22
Yes, at least that is the version I saw. Is there any inconsistence? – awllower Jul 25 '13 at 16:23
The first integral is the winding number of your curve around $0$, so its value can be found without any computation. For the second one, try to use spherical coordinates. – Etienne Jul 25 '13 at 16:37

HINT: $ 1)$Note that $$\frac{xdy-ydx}{x^2+y^2}=d\left(\tan^{-1}\left(\frac{y}{x}\right)\right)=d\left(\tan^{-1}\left(t\tan t\right)\right)$$ on the curve $\Gamma$.

share|cite|improve this answer
This is not valid in the entire domain, because you got $x=0$ for some $t$. – Integral Jul 25 '13 at 16:03
But there are a finite number of discontinuities at $t=5\pi/2,7\pi/2,9\pi/2,11\pi/2,$ you can write the integral as the sum of the integrals in the intervals in between, right? – Samrat Mukhopadhyay Jul 25 '13 at 16:09
These are discontinuities of second kind, and I dont know you can think that way in this case. – Integral Jul 25 '13 at 16:14
Might I ask do you have any suggestion on the second question? It is funny that I surmised that the second should be solved very quickly, while the first might take a long time... :D – awllower Jul 25 '13 at 16:16
Aha, I see, the integral is not quite straightforward as it has this $\|x\|^4$ in it. – Samrat Mukhopadhyay Jul 25 '13 at 16:44

You can write the integral as

$$ \int_\Gamma\dfrac{xdy-ydx}{x^2+y^2}= \int_\Gamma\dfrac{xdy}{x^2+y^2}+\int_\Gamma\dfrac{ydx}{x^2+y^2}=I_1+I_2 .$$

Using the given parametrization, we have

$$ I_1= \int_{2\pi}^{6\pi}\dfrac{t\cos(t) (2t\sin(t)+t^2\cos(t)) }{t^2\cos^2(t)+t^4\sin^2(t)}dt = \dots.$$

You can do the $I_2$ the same way.


$$ x=t\cos(t)\implies dx -t\sin(t)+\cos(t) dt$$ $$ y=t^2\sin(t) \implies dy= 2t\sin(t)+t^2\cos(t) dt. $$

share|cite|improve this answer
And I do not think that the transformed integrals are obvious to solve?Also, you missed some $d$ in the end. – awllower Jul 25 '13 at 16:33
@awllower: This why I asked you about $y$. Note that, if $y=t\sin(t)$ then the integral will be easy to integrate. You can use Maple or Mathematica to evaluate these integrals. Anyhow, this is the technique. – Mhenni Benghorbal Jul 25 '13 at 16:34
If $y=t\sin t$, then the denominator would be $t^2$. And I would not post here. haha In any case, thanks for the attention. – awllower Jul 25 '13 at 16:38
@awllower: Sorry, it won't make a difference. I think you need to use Maple or Mathematica to evaluate these integrals. – Mhenni Benghorbal Jul 25 '13 at 16:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.