Stroke's Theorem

  • Thread starter Tonyt88
  • Start date
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?
 
Last edited:
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.
 
Last edited:

HallsofIvy

Science Advisor
41,621
821
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.
 
LOL
Did I say someting wrong?
:cry:
 
complementarity:
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.
:!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!) :!!)
greets
 
Last edited:

Want to reply to this thread?

"Stroke's Theorem" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top