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Consider the part of the surface $z = xy$, which lies within the cylinder $x^2+y^2 = 9$ and call it $S$. Compute the upward flux of $F = (y,x,3)$ through $S$.

Clearly, the normal to the given surface is: $\mathbb{n} = <y,x,-1>$, so we have the integral: $$ \int \int_D x^2+y^2 -3$$ to calculate, or at least that's what I got.

However, I'm a bit confused on the domain of integration here. Clearly, it cannot be the entire disc. Can someone give me pointers on how to find the domain of integration in such cases?

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Yes, you can do it like that. But you want the upward flux. So $$ \iint_D(3-x^2-y^2)\,dA=\int_0^{2\pi}\int_0^3(3-r^2)r\,dr\,d\theta=-\frac{27\pi}2 $$

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  • $\begingroup$ Feel like such an idiot... I forgot the $r$ when converting the coordinates. As a follow up, does the flux not depend on the z-coordinate at all? What if we were finding the flux through the cylinder between z=xy and z = 10? Since it's completely independent of z, would we expect the flux to remain the same? $\endgroup$ – Gummy bears Apr 20 at 6:40
  • $\begingroup$ That makes no sense. The flux is across a surface. If you use a different surface, you'll have a different normal. $\endgroup$ – Martin Argerami Apr 20 at 7:03

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