Computing a Line Integral: Stokes' Thm

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Homework Statement



Compute the line integral of v = 6i + yz^2j + (3y + z)k along the path (0,0,0) -> (0,1,0) -> (0,0,2) -> (0,0,0). Check your answer using Stokes' Thm

Homework Equations





The Attempt at a Solution



I've tried breaking into three pieces. The first with dx = dz = 0, second dx = 0 and third dx = dy = 0. The solution is given as 8/3 but I can't seem to come up with that. Do I have to parametrize the curve or what?
 
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Well I was able to do it by paramatrizing the 3 paths individually. A little tedious, but given that you can't use Stokes' theorem except to check your answer, it's the best you can do. The paths are all straight lines. Perhaps you could use green's theorem here, since the path lies on a plane and is closed, but the function you're taking the path integral of has 6i in it.
 
Ok, I managed to get it. Thanks :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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