# Verify Stokes’ theorem for a hemisphere S .Verify Stokes’ theorem for a hemisphere S defined as x^2+y^2+z^2=9 (z>=0) where a vector field F=z^2i + 2xj-yk exists over the surface and...

Verify Stokes’ theorem for a hemisphere *S* *.*

Verify Stokes’ theorem for a hemisphere *S* defined as x^2+y^2+z^2=9 (z>=0) where a vector field F=z^2i + 2xj-yk exists over the surface and around its boundary *c*.

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This has been asked before, and I answered it already: http://www.enotes.com/math/q-and-a/verify-stokers-theorem-hemisphere-s-define-x-2-y-2-186509

I've reposted the solution here (only difference is the nomenclature, boundary was C in this case, and not B)

Prove ∫∫(S)(∇xF)dS = ∫BFdr

Let u be bounded by [0,pi/2]

Let v be bounded by [0, 2pi]

Let t be bounded by [0,2pi]

r(u,v) = (3sin(u)cos(v), 3sin(u)sin(v), 3cos(v))

(dr/du)(dr/dv) = (9sin^2(u)cos(v), 9sin^2(u)sin(v),9cos(u)sin(u))

r(t) = (3cos(t), 3sin(t), 0)

∇xF = (dFz/dy - dFy/dz)i + (dFx/dz - dFz/dx)j +(dFy/dx - dFx/dy)k

= (-1 - 0)i + (2z-0)j + (2-0)k

= -i + (2z)j + 2k

∫∫(S)(∇xF)dS

= ∫∫(-1,6cosv,2)(9sin^2(u)cos(v),9sin^2(u)sin(v),9cos(u)sin(u))dA

= ∫∫(-9sin^2(u)cos(u) + 54sin^2(u)sin(v)cos(v) + 18cos(u)sin(u))dv[0,2pi]du[0,pi/2]

= ∫36pi*cos(u)sin*u)du[0,pi/2]

= 18pi

∫BFdr

=∫(0,6cos(t),-3sin(t))(-3sin(t),3cos(t),0)dt[0,2pi]

=∫(18cos^2(t)dt[0,2pi]

=∫(9 + 9cos(2t))dt[0,2pi]

=18pi

Since 18pi = 18pi, the theorem is proven ;)