Let $\mathcal C$ be the curve that traces the unit circle once (counterclockwise) in $\mathbb R^2$. The starting- and endpoint is (1,0). I need to figure out a parameterization for $\mathcal C$ and calculate the following integral.

$\oint_{\mathcal C}\begin{pmatrix}x_2^2 \cos x_1 \\ 2x_2(1+\sin x_1)\end{pmatrix} dx$


$t\rightarrow(\cos t, \sin t), t\in[0,2\pi]$

$\vec{x}(t)=\begin{pmatrix}\cos t \\ \sin t\end{pmatrix}$; $\space \space \frac{\partial \vec{x}(t)}{\partial t} =\begin{pmatrix}-\sin t \\ \cos t\end{pmatrix}$ $\iff d \vec{x}=\begin{pmatrix}-\sin t \\ \cos t\end{pmatrix} d t$

However, this gives me a really complicated term that I can't simplify properly.

$\oint_{0}^{2\pi} \begin{pmatrix}\sin^2 t\cdot \cos(\cos t) \\ 2 \sin t (1+\sin(\cos t)) t\end{pmatrix}\begin{pmatrix}-\sin t \\ \cos t\end{pmatrix} dt$

Any idea how to solve this?


Since the vector field $\langle x_2^2 \cos(x_1), 2x_2(1 + \sin(x_1))\rangle$ is the gradient of the scalar field $\phi(x_1,x_2) = x_2^2(1 + \sin(x_1))$, by the fundamental theorem of line integrals, your integral is zero.

  • $\begingroup$ Since the OP doesn't know about Green's theorem, I guess he doesn't know about the theorem that you invoke either. Apparently he really wants to do things "by hand". $\endgroup$ – Alex M. Apr 26 '15 at 15:00
  • $\begingroup$ I don't think that's the case, since this theorem is usually covered well before Green's. I'll just have to wait and see how the OP will respond. $\endgroup$ – kobe Apr 26 '15 at 15:05
  • $\begingroup$ @kobe I think we briefly touched on this in our last lecture but I am not too sure what you did there. Can you maybe elaborate a bit. How do I know that vector field is the gradient of the scalar field? $\endgroup$ – qmd Apr 26 '15 at 15:25
  • $\begingroup$ If $C$ is a path from $\mathbf{a}$ to $\mathbf{b}$, then $\int_C \nabla \phi\cdot d\mathbf{x} = \phi(\mathbf{b}) - \phi(\mathbf{a})$. In your case, $C$ starts and ends at $(1,0)$, so your integral evaluates to $\phi(1,0) - \phi(1,0) = 0$. $\endgroup$ – kobe Apr 26 '15 at 16:30
  • 1
    $\begingroup$ @Rzeta Let $\mathbf{F} = \langle x_2^2\cos(x_1), 2x_2(1 + \sin(x_1))\rangle$. Then $$\nabla \phi = \langle \phi_{x_1}, \phi_{x_2}\rangle = \langle x_2^2\cos(x_1),2x_2^2(1 + \sin(x_1))\rangle = \mathbf{F}$$ Your integral is $\int_C \mathbf{F}\cdot d\mathbf{x}$, which is $\int_C \nabla \phi\cdot d\mathbf{x}$. Since $C$ starts and ends at $(1,0)$, by the comment I made above, this integral evaluates to $\phi(1,0) - \phi(1,0)$, which is $0$. $\endgroup$ – kobe Apr 26 '15 at 18:19

Stop torturing yourself with parametrizations and use Green's theorem: you'll get $0$ in just a few seconds.

  • $\begingroup$ I don't know green's theorem. I should be able to solve this in another way. $\endgroup$ – qmd Apr 26 '15 at 14:21

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