Evaluate , where C is the closed curve parameterized and oriented by the path .

Hint: Show that the path lies on the surface .

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- May 3rd 2010, 10:16 PMdavesface[SOLVED] Stokes' Theorem
Evaluate , where C is the closed curve parameterized and oriented by the path .

Hint: Show that the path lies on the surface . - May 4th 2010, 04:02 AMHallsofIvy
"Parameterized", yes, but this doesn't give an orientation. Are we to assume that t is going from 0 to ? That would give an orientation.

[/quote]Hint: Show that the path lies on the surface .[/QUOTE]

Have you done the "hint"? Do you know any identity for "sin(2t)"?

Stokes' theorem, since you title this thread "Stokes' Theorem", in its simplest form says that is a vector valued function then .

Here, and is the differential tangent vector, .

Since you know that the path lies entirely in the plane z= 2xy, you can write that as the position vector . The two derivative vectors, and are in the tangent plane to the surface (which here is the plane itself) and their cross product, is normal to the plane and its length gives the "differential of surface area". Assuming that the path integral was to be from 0 to (rather than from 0 to ), the normal is to be oriented by positive z so .

You still need to determine the bounds on x and y. - May 4th 2010, 05:04 PMdavesface
I don't know what to say about the orientation. I've posted all the information given about the problem. I'm guessing it's the negative orientation, though, since the given answer is .

I have no clue how they got that, as my work has boiled down to by taking the cross product and dot product where indicated and then substituting in 2xy for the z value that showed up from the dot product.

What am I missing here? - May 4th 2010, 05:17 PMdavesface
Wow, you know what? It's way easier than I made it seem. All you have to do is convert it to polar and reverse the sign on the integral. r goes from 0 to 1 and theta goes from 0 to 2pi. Thanks for the help.