Stroke's Theorem

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Stoke's Theorem

1. The problem statement, all variables and given/known data

F(x,y,z,) = (3x+y)i - zj + y²k;
S = the union of the cylinder {(x,y,z) : x² + y² = 4, 2 < z < 4} and the hemisphere {(x,y,z) : x² + y² +(z-2)² = 4, z < 2}, oriented by P |--> n(P) with n(0,0,0) = -k
Use Stoke's Theorem to evaluate the double derivate of curl(f) [dot] ndS for the vector field F and the oriented surface.

2. Relevant equations

3. The attempt at a solution

I really don't know how to use stoke's theorem when considering a union, normally I would solve the second piece for when z = 2, thus x² + y² = 4 which gives you a parametrization of <cos(t),sin(t),2> but I don't know how to consider the cylinder. Would I just say <cos(t),sin(t),4> instead to account for the opening and then note that there is a negative orientation?
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I my personal sense, stokes theorem may apply in the surface when it is union.You just need to let z=4,and evaluate like your former attempt.

I am not strongly convinced,cause I understand the world with field theory.
And don't sleep so late.
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Science Advisor
enricfemi, you "understand the world with field theory" and can't do a problem like this? This is the mathematical core of field theory!

Stokes' theorem, in this situation, says that the integral [itex]\nabla x \vec{f}\cdot d\vec{S}[/itex] is equal to the integral of [itex]\vec{f}[/itex] around the boundary of the surface.

In other words, just integrate (3x+y)i - zj + y²k around the boundary of the figure. Normally a finite cylinder has two boundaries: the circle around the top and bottom. Here, the bottom is "capped" by the (inverted) hemisphere but the only thing that matters is that the only boundary is now the circle at the top: {(x,y,z) : x² + y² = 4, z= 4}. That should be easy.
Did I say someting wrong?
Physics world do have something different form maths ,while HallsofIvy are a professor of mathematics .Just like I cann't sure the discontiguous surface whether fit to stoke's theory out of Maxwell's equations.
:!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!)
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