Flux Integral for a Surface Above a Disc with Downward Orientation

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QUESTION: Compute the flux of the vector field, F , through the surface, S.

F = xi+yj+zk

S is the part of the surface z = x^2 + y^2 above the disc x^2 + y^2 ≤ 4 , oriented downward.

I am just wondering why the integrand is from o to 4 while the radius is only 2.

bewjcz.jpg
 
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It isn't. That's incorrect. It should be 2.
 
My solution guide says otherwise, along with my online homework
 
HallsofIvy said:
It isn't. That's incorrect. It should be 2.

I just emailed my teacher and he said there was an error in the online homework. The correct answer is indeed used with radius 2. Sorry about that.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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