Stoke's Theorem: Calculating Int_C vecF dot dr on Ellipse & Cylinder

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Homework Statement


Use Stoke's Theorem to calculate \int_C\vec{F}\cdot\, dr, where
\vec{F}=<x^2z\, ,xy^2\,,z^2> and C is the curve of the intersection of the plane x+y+z=1 and the cylinder
x^2+^2=9.
(C is oriented clockwise when viewed from above.)
Answer: \frac{81}{2}\pi.



Homework Equations

Stoke's Theorem



The Attempt at a Solution



Okay, let me try to explain where I am getting lost. Firstly, I know that the premise of Stoke's Theorem is that is relates a line integral to a Surface integral.

When I graph this, I get a cylinder that is symmetrical about the z-axis and it is intersected by a plane which results in an ellipse. (see terrible drawing below)

Now, I need to parametrize (how do you spell that anyway?) S. Now S is the surface that is bounded by C right? If not, please stop me here.

Photo1.jpg
 
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Saladsamurai said:
Now, I need to parametrize (how do you spell that anyway?) S. Now S is the surface that is bounded by C right? If not, please stop me here.
Yes. S is any surface that is bounded by C. Usually the easiest surface to work with is the plane surface bounded by C if C can be visualised as lying on a plane.
 
Defennder said:
Yes. S is any surface that is bounded by C. Usually the easiest surface to work with is the plane surface bounded by C if C can be visualised as lying on a plane.

I know form doing this in class that we parametrized S as r=<x, y, 1-x-y>

but I am not so sure why. Oh wait... is that the surface bounded by the ellipse?

So now all I need to do is find dr/dx x dr/dy and compute the dot product of the resultant with F(r(x,y) and integrate?
 
You are told that x + y + z = 1 and now you parametrize that by <x,y,f(x,y)> and z = f(x,y) = 1 - x - y by algebra (is that part of your question?)
 
I guess my question is what is S? Is it the the surface of the ellipse? I believe it is, so I guess my question has been answered.
: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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