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Suppose we have a cylinder of x^2 +y^2 = a^2 and we are interested in working out the flux of a vector field going through the curved surface, given that the surface element is defined as: Ds = Dr/du × Dr /dv

I tried suggesting that r = xi + yj +zk and then replacing z with zero and differentiating before calculating the cross product to give 1k dxdy

However this does not seem to be correct. Any help would be appreciated.

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The trouble with your parametrization of $(x,y,0)$ is that it does not define a cylinder. This defines the $xy$-plane. And, anyway, you can see it really can't be the cylinder, because setting $x=y=0$ gives the origin which is obviously not part of your cylinder (unless $a=0$, but you get the idea).

One way you can parametrize the cylinder is this way: $$ \vec S(\theta, z) = (a \cos \theta,\ a \sin \theta,\ z) $$ You can see it satisfies the equation $x^2 + y^2 = a^2$ and it's capable of describing any point on the cylinder because $a \cos \theta$, $a \sin \theta$ fully describes a circle, and $z$ says there is a circle at every height. So this will work.

Next you can compute the area element like so: \begin{align} d\vec S &= \frac{\partial\vec S}{d\theta} \times \frac{\partial\vec S}{\partial z}\, d\theta\, dz \\ &= (a \cos \theta,\ a \sin \theta,\ 0) \end{align} which is outward-oriented which is probably what you want (if not, you can always negate it). Therefore the flux is $$ \Phi = \int_{-\infty}^\infty \int_0^{2\pi} \vec F\Big(\vec S(\theta, z) \Big) \cdot \left(\frac{\partial\vec S}{d\theta} \times \frac{\partial\vec S}{\partial z} \right)\, d\theta\, dz $$ where $\vec F(x,y,z)$ is the vector field in question.

This assumes you want the flux over the entire infinite cylinder. If you just want part of it, you'll have to adjust the bounds accordingly (and possibly add in integrals for the top and bottom of the cylinder too, if that matters).

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