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I'm trying to solve the following exercise:

Let $$\omega = \frac{xdy-ydx}{x^2+y^2}$$ on $\mathbb{R}^2 \setminus (0,0)$ (which is the standard example of a closed but not exact form). Let $g\colon[0,2\pi] \to \mathbb{R}^2 \setminus (0,0)$ be defined by $$g(t)=(e^t \sin(17t),e^{t^2} \cos(17t))$$ Calculate the integral of $g^*(\omega)$ over $[0, 2\pi]$

Just by looking at this question, I think it's not good idea to actually calculate (a tried... :) ) this. Which means this thing has to be $0$ or $ 2 \pi$. I'm tending towards $2 \pi$ since although $g$ is not a closed curve, we're still somehow doing a full rotation around (0,0).

But I just have no idea where to start and or what I have to show. Could someone give me hint on how to solve such a problem?

Thanks for any help.

Edit, here's g:

enter image description here

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The picture kind of obscures the (important) fact, that the curve winds around the origin several times. – user20266 Aug 21 '12 at 17:41

Ok, since this is homework I'll tell you what you have to do but won't do it for you ;-)

  • extend the curve to, say, $[0,4\pi]$ by adding a curve segment $h(t)$ such that $g$ and $h$ together form a closed, piecewise smooth, curve $f$.
  • calculate the integral of the closed curve $f$ using a theorem from your course notes. It is not $0$ and not $2\pi$ if you took the simplest and obvious choice for $h$. Have a close look at the parametrization of your $g$. This curve runs several times around the origin.
  • Calculate the integral of $\omega$ along $h$ -- make sure to choose $h$ in such a way that this is easy to do.
  • Take the difference of the results from the second and third step.


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Got it. Thanks! – snom Aug 21 '12 at 17:23
Sorry, im still confused: I added a picture of g in the question (only have excel at work :) ). The obvious way to make it a closed curve would be joining start- and endpoint by a straight line, is that correct? But then it goes around the origin only 1 time. wouldn't then The integral of f be $2 \pi$? (Sorry, but i'm really having a hard time with this sort of questions...). Tanks a lot for your help. – snom Aug 21 '12 at 17:41
@snom $\sin(0)=\sin(17*2*\pi)=0, \cos(0)=\cos(2*17*\pi)=1$. How does your extension pass through $0$? If it does, it is not a valid extension anyway, since $0$ is not allowed. (Did you ignore the $2$ from the interval?) – user20266 Aug 21 '12 at 17:44
@snom I see. Maybe you should draw $(\sin(17t), (1+t)\cos(17t)$ to see what really happens. $\exp$ grows so fast you see it winding around $0$ only once. – user20266 Aug 21 '12 at 17:50
Ok, somethings wrong with my curve. Ill have to have a look at this again when i'm at home. I guess excel is not the right tool to draw this sort of things :) – snom Aug 21 '12 at 17:51

HINT: Let $g(t) = (x(t),y(t))$. Then use: $$ g^\ast(\omega)(t) = \frac{x(t) y'(t)-y'(t) x(t)}{x^2(t) + y^2(t)} \mathrm{d} t = \mathrm{d} \left(\arctan(x(t), y(t))\right) $$ The $\arctan(x,y)$ is the arc-tangent of two variables.

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Hint: Convert to polar coordinates: $$ \begin{array}{ccc} \mbox{$\begin{align} x&=r\cos(\theta)\\ \mathrm{d}x&=\cos(\theta)\mathrm{d}r-r\sin(\theta)\mathrm{d}\theta \end{align}$} &\quad& \mbox{$\begin{align} y&=r\sin(\theta)\\ \mathrm{d}y&=\sin(\theta)\mathrm{d}r+r\cos(\theta)\mathrm{d}\theta \end{align}$} \end{array} $$ and you get $\omega=\mathrm{d}\theta$.

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