Stokes Theorem

thetguy

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I am trying to solve [MATH]\oint_C F \cdot dr[/MATH], where F=y2i, and C is the intersection of the cone [MATH]z = \sqrt{x^2+y^2}[/MATH] and z = 5, clockwise as seen from the origin. I am using Stokes theorem, so I know the curl is -2yk. From here, how would I parameterize the surface? I know I can use the parameterization to find the partial derivatives, cross those partial derivatives to find a vector normal to the surface, dot that with the curl, and integrate. I just need help understanding how to find the parameterization of the surface.
 
I am trying to solve [MATH]\oint_C F \cdot dr[/MATH], where F=y2i, and C is the intersection of the cone [MATH]z = \sqrt{x^2+y^2}[/MATH] and z = 5, clockwise as seen from the origin. I am using Stokes theorem, so I know the curl is -2yk. From here, how would I parameterize the surface? I know I can use the parameterization to find the partial derivatives, cross those partial derivatives to find a vector normal to the surface, dot that with the curl, and integrate. I just need help understanding how to find the parameterization of the surface.
Z = 5 is a plane - parallel to the X-Y plane. Do you see that the intersection of the said plane with the given cone will be a circle? What will be the equation of the circle?
 
Ah, it is a circle! I did not think about that. So would a parameterization of the circle be <r*cos(u),r*sin(u),5>, since we are told z=5, and x=r*cos(u) and y=r*sin(u) in cylindrical coordinates? then 0<r<5 and 0<u<2pi for my bounds in the integral when it is time to calculate?
 
Ah, it is a circle! I did not think about that. So would a parameterization of the circle be <r*cos(u),r*sin(u),5>, since we are told z=5, and x=r*cos(u) and y=r*sin(u) in cylindrical coordinates? then 0<r<5 and 0<u<2pi for my bounds in the integral when it is time to calculate?

That's just the top though.

note that [MATH]z=r[/MATH]
so you have [MATH] ( z \cos(u), z \sin(u), z),~0\leq u < 2\pi, ~0 \leq z \leq 5[/MATH] as the surface of the rest of the cone.
 
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