0
$\begingroup$

Attempt no 2. Well, it is my first question here, so you may expect some errors, even in grammar, because english is not my mother language. So, I was asked to 'verify' the Stokes Theorem in these questions, and I would like to use differential forms, because it is the content that we are discussing now (and by verify I mean solve both sides of Stokes equation and verify if they are equal):

a) $\omega$ =$(x+3y)dx+(2x-y)dy$ and $\Sigma=\{(x,y)|x^2+2y^2\le2\}$

b) $\omega=xdx+zdy$ and $\Sigma=\{(x,y,z)|x^2+y^2=z^2, 0\le z\le1\}$

c) $\omega=zdxdy$ and $\Sigma=\{(x,y,z)|x^2+y^2+z^2\le 1, z\ge0\}$

I tried to solve the a), watched some video courses, but I simply can't understand how to do it. The only part that I did was $d\omega$ from a), and I got by result $dydx$.

Thanks since now.

$\endgroup$
0
$\begingroup$

Stokes' theorem says $\int_\Sigma d\omega=\int_{\partial\Sigma}\omega.$

For part a), on the left-hand side integrand $d\omega=3dy\wedge dx+2dx\wedge dy=-dx\wedge dy.$ So our integral is

$$ \int_\Sigma d\omega=-\iint_\Sigma dx\,dy=-\int_{-1}^{+1}\int_{-\sqrt{2-2y^2}}^{+\sqrt{2-2y^2}}dx\,dy=-2\int_{-1}^{+1}\sqrt{2-2y^2}\,dy=-\pi\sqrt{2} $$

(Draw a picture to help write these bounds of integration for the region $\Sigma=\{x^2+2y^2\leq 2\}$).

Then to compute the right-hand side $\int_{\partial\Sigma}\omega$ we need a parametrization of the boundary ellipse $\partial\Sigma=\{x^2+2y^2=2\}$, which I will take to be $(\sqrt{2}\cos t,\sin t)$

We have

$$ \int_{\partial\Sigma}\omega = \int(x+3y)dx+(2x-y)dy \\= \int_0^{2\pi} -(\sqrt{2}\cos t+3\sin t)\sqrt{2}\sin t\,dt+(2\sqrt{2}\cos t -\sin t)\cos t\,dt \\=\int_0^{2\pi}(-3\sqrt{2}\sin^2t - 3\sin t\cos t+2\sqrt{2}\cos^2t)\,dt\\=-\pi\sqrt{2}. $$

For that last equality, you can use the identities that $\int_0^{2\pi}\sin t\cos t\,dt=0$ and $\int_0^{2\pi}\cos^2t\,dt=\int_0^{2\pi}\sin^2t\,dt=\pi.$ (If you have trouble remembering these identities, just imagine integrating the Pythagorean identity over one period.)

So Stokes' theorem is verified.

$\endgroup$
  • $\begingroup$ Thank you very much for the answer, but why did you used (2–√cost,sint) for parametrization? There is a specific reason, right? What should I use in the other ones? $\endgroup$ – Lucas Dec 29 '17 at 22:51
  • $\begingroup$ @Lucas parametrizations for conic sections and quadric surfaces are standard topics for a multivariable course, but let me add a few words to my answer about how to remember them, as well as what to do if you don't remember them $\endgroup$ – ziggurism Dec 30 '17 at 5:29
  • $\begingroup$ @Lucas I started to write up a lengthy explanation of how to find bounds for integrals, and how to find parametrizations for curves and surfaces. But this is pretty far removed from the actual question you asked. If you have questions about evaluating basic multivariable integrals, maybe you should post a new question (or search for existing questions)? $\endgroup$ – ziggurism Dec 30 '17 at 6:18
0
$\begingroup$

a) By Greens / Stokes theorem:

$\oint (x+3y)\ dx + (2x-y)\ dy = \iint \frac {\partial}{\partial x} (2x-y) - \frac {\partial}{\partial y}(x+3y) \ dA$

Left side first:

$x = \sqrt 2 \cos t\\ dx = -\sqrt 2\sin t\\ y = \sin t\\ dy = \cos t$

$\int_0^{2\pi} -2 \cos t\sin t - 3\sqrt{2}\sin^2 t + 2\sqrt 2\cos^2 t - \sin t\cos t\ dt$

$-\sqrt 2 \pi$ (I am leaving it to you to actually crank through the integration)

Right side:

Taking those partial derivatives and simplifying.

$\iint - 1\ dA = -A = -\sqrt 2 \pi$

$\endgroup$

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.