Stokes Theorem Problem: Surface Integral on Ellipse with Curl and Normal Vector

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Homework Statement


F = xi + x3y2j + zk; C the boundary of the semi-ellispoid z = (4 - 4x2 - y2)1/2 in the plane z = 0

Homework Equations



(don't know how to write integrals on here, sorry)

double integral (curl F) . n ds

The Attempt at a Solution



curl F = 3y2x2k
n = k

curl F . n = 3y2x2

So I have a surface integral, which I think I can change to dA since the differential of the surface area is just 1dA...

Now this is where I'm stuck. How do i do the double integral with an ellipse? I tried it in rectangular coordinates but got some function I don't know how to integrate. Help!:confused:
 
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Can you do this by integrating F dot dr over the boundary curve instead?

r(theta)=cos(theta) I+2sin(theta) J+0 K
 
What function do you get that you can't integrate? It looks like it's just a trig substitution to me.
 
what you can do is use Stoke's theorem to convert to the line integral of the vector filed and then if you use Green's theorem to convert that to a double integral, use a trig sub and then integrate. (integration takes a little bit of work) I think the answer is pi. Anybody agree?
 
Yes, it's pi. Both ways.
 
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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