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## Homework Statement

Use stokes theorem to find double integral curlF.dS where S is the part of the sphere x

^{2}+y

^{2}+z

^{2}=5 that lies above plane z=1.

F(x,y,z)=x

^{2}yzi+yz

^{2}j+z

^{3}e

^{xy}k

## Homework Equations

stokes theorem says double integral of curlF.dS = [itex]\int[/itex]

_{C}F.dr

## The Attempt at a Solution

boundary curve C is circle given by x

^{2}+y

^{2}=5, z=1.

Vector equation of c then is r(t)=[itex]\sqrt{5}[/itex]costi+[itex]\sqrt{5}[/itex]sintj+1k where 0<t<2pi

then r'(t)= -[itex]\sqrt{5}[/itex]sinti+[itex]\sqrt{5}[/itex]costj

[itex]\int[/itex]

_{C}F.dr = [itex]\int[/itex]F(r(t))dotr'(t).dt

F(r(t))=([itex]\sqrt{5}[/itex]cost)

^{2}([itex]\sqrt{5}[/itex]sint)i+([itex]\sqrt{5}[/itex]sint)j+e

^{[itex]\sqrt{5}[/itex]cost[itex]\sqrt{5}[/itex]sint}

then F(r(t))dotr'(t) = [([itex]\sqrt{5}[/itex]cost)

^{2}([itex]\sqrt{5}[/itex]sint)(-[itex]\sqrt{5}[/itex]sint)]i + [([itex]\sqrt{5}[/itex]cost)([itex]\sqrt{5}[/itex]sint)]j

simplifying down and i got to

5[itex]\int[/itex]sintcost-5cos

^{2}tsin

^{2}t.dt

how do i integrate that?

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