1
$\begingroup$

Let $C$ be parametrization $\mathbf{r}=\sin(t) \mathbf{i}+\sin(2t) \mathbf{j}$, $t \in [0, 2\pi]$. Sketch picture and investigate how the surface orientates. Calculate line integral $\oint_C \mathbf{F}\bullet d \mathbf{r}$ using Green's theorem, where $\mathbf{F}$ is vector field $F(x,y)=ye^{x^2} \mathbf{i}+x^3e^y \mathbf{j}$.

EDIT: Well using Green's $$\oint_C ye^{x^2} dx +x^3e^y dy = \iint_R \left(\frac{\partial x^3e^y}{\partial x}- \frac{\partial ye^{x^2}}{\partial y}\right)\; dA$$ So how can find boundaries to $R$. I guess it's trivial if you know how to sketch the picture? Any hints/tips?

Maybe: $\mathbf{r}=\langle \sin(t), \sin(2t)\rangle$ and $\mathbf{r}'=\langle \cos(t), 2\cos(2t)\rangle$ $$\begin{align}\oint_C ye^{x^2} dx +x^3e^y dy&=\oint_0^{2\pi} \left(F_1(t)\frac{\partial}{\partial t}x(t)+F_2(t)\frac{\partial}{\partial t}y(t)\right) dt \\ &=\oint_0^{2\pi} \underbrace{\sin(2t)e^{\sin^2 (t)}\cdot\cos(t)}_{\text{odd?=0 $\in [0,2\pi]$}}+\sin^3(t)e^{\sin(2t)}\cdot2\cos (2t) dt \end{align}$$ If I evaluated the integral right this seems a bit complex (maybe use even-ness to evaluate the int?) . It's zero?

$\endgroup$
  • $\begingroup$ isn't this stokes's theorem? $\endgroup$ – Ellya Apr 1 '14 at 14:47
  • $\begingroup$ This would be easier to calculate using a standard line integral as you already know the boundary $C$. However using Green's Theorem you have to use the surface. $\endgroup$ – George1811 Apr 1 '14 at 14:49
  • $\begingroup$ @ellya No this is Green's theorem notice the $\textbf{k}$ rather than the normal. It is a certain case of Stoke's Theorem $\endgroup$ – George1811 Apr 1 '14 at 14:51
  • $\begingroup$ I see your edit, makes sense now. $\endgroup$ – Ellya Apr 1 '14 at 14:55
  • $\begingroup$ Hmmm. Now i'm confused myself. I'm not sure which one to use. $\endgroup$ – ELEC Apr 1 '14 at 14:56
1
+50
$\begingroup$

Now this is a bit messy:

$x=sin t$

$y=2sin 2t$

enter image description here

We are looking only at first quarter of the double lattice on the picture.

One can see that for $t \in [0,\pi/2]$ we have $x>=0$ and $y>=0$ what enables us to do:

$y=2sin 2t = 2 sin t cos t= 2 x \sqrt{1-x^2}$

So the limits of integration for first quadrant would be:

$x=0 \rightarrow x=1$

and for $y$:

$y=0 \rightarrow y=2 x \sqrt{1-x^2}$

Now similar procedure can be arranged for other three quadrants with respect to signs of $x$ and $y$ but it seems to me that this would end up in evenly complicated integration...

$\endgroup$
2
$\begingroup$

I know this is a late answer, but the solution is

$$\oint_C \mathbf{F} \bullet d\mathbf{r}=0$$

The Green theorem states that the contour integral is equal to the curl of the function in the area (is a particular case of Stokes', in 2D). But this depends of the orientation of the area: in other words, it depends on where the normal vector is pointing, with respect to the parametrization.

As shown in the graph by Markisa, the region is simmetric with respect to the Y-axis. But, the parametrization defines the contour as: enter image description here

Note that the area in the right side of the Y-axis is positive-oriented, and the area on the left is negative-oriented (because the path is defining the area on the other side).

As both areas are symmetric, and have different orientation, their integrals will be equal in magnitude but different in sign. Then, the full integral is 0, because it is equal to the sum of both areas.

I know this is an "intuitive" answer, but it's all that you need to solve the problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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