sproblems9

# sproblems9 - Dr H Khanal Spring 2008 MA 441 – Solved...

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Unformatted text preview: Dr. H. Khanal Spring 2008 MA 441 – Solved Problems #10. Stokes’s Theorem 1. Integrate the surface integral directly. Then check the result by integrating the corresponding line integral in the Stokes’s theorem. Z Z S ∇ × F · n dA = I C F · r ( s ) ds Sol. Here F = [ y 2 ,- x 2 , 0] and the surface S is given by x 2 + y 2 ≤ 4 , y ≥ , z = 0. We can express the surface parametrically as r ( u,v ) = [ u cos v, u sin v, 0] , ≤ u ≤ 2 , ≤ v ≤ π. We calculate the normal vector using the formula N = r u × r v . r u = [cos v, sin v, 0] r v = [- u sin v, u cos v, 0] N = r u × r v = fl fl fl fl fl fl fl fl i j k cos v sin v- u sin v u cos v fl fl fl fl fl fl fl fl = [0 , , u ] . ∇ × F = fl fl fl fl fl fl fl fl i j k ∂ ∂x ∂ ∂y ∂ ∂z y 2- x 2 fl fl fl fl fl fl fl fl = [0 , ,- 2 x- 2 y ] . = ⇒ ∇ × F ( u,v ) = [0 , ,- 2 u (cos v + sin v )] , ∇ × F · N = [0 , ,- 2 u (cos v + sin v )] · [0 , , u ] =- 2 u 2 (cos v + sin v ) and the surface integral is Z Z...
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sproblems9 - Dr H Khanal Spring 2008 MA 441 – Solved...

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