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# Stokes' Theorem Problem watch

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1. Considering a semicircle-faced cylinder with radius a and height h, and the vector field V=(x+y+z)i+zj+yzk. Perform the closed loop integral around the cross-section at z=0 in the anticlockwise direction, and verify using the surface integral.

I keep getting two different answers for each part.

For the closed loop integral:

The semicircle edge:

Convert to plane polar form:
x=rcosθ, y=rsinθ, r=a
dx=-asinθdθ, dy=acosθdθ

V.dr=a^2dθ(-sinθ)(cosθ+sinθ)

Limits: 0<θ<π
-a^2∫(sinθ)(cosθ+sinθ)dθ=-(πa^2)/2

The flat edge:

dr=dxi
V.dr=(x+y+z)dx=xdx

Limits: -a<x<a

As the integrand is an odd function, this yields zero.

Thus, the closed loop integral is -(πa^2)/2.

For the surface integral:

The semicircle edge:

curl(V)=(z-1)i+j-k

Surface can be parameterised by x and z

r=xi+sqrt(a^2-x^2)j+zk
δr/δx=i-(x/sqrt(a^2-x^2))jδr/δz=k

dS=[-(x/sqrt(a^2-x^2))i-j]dxdz

curl(V).dS=[1+x(1-z)/sqrt(a^2-x^2)]dxdz

Limits: -a<x<a, 0<z<h

∫∫[1+x(1-z)/sqrt(a^2-x^2)]dxdz=2ah

The flat edge:

y=0

r=xi+zk
δr/δx=iδr/δz=k

dS=-dxdzj

curl(V).dS=-dxdx

Limits: -a<x<a, 0<z<h

-∫∫dxdz=-2ah

The surface integrals of the two edges add to give zero.

OK, what went wrong here?
2. (Original post by Nuclear Ghost)
x
Eurgh I seem to have made a sign error somewhere, so I've got two answers - differing by a factor of .
That said, you're missing one of the surfaces - the half-disk at .

Edit: Noticed my mistake, you're good apart from the additional surface.
Also it seems more natural to parameterise the cylinder with cylindrical polars, but I guess that's a matter of taste.
3. (Original post by joostan)
Eurgh I seem to have made a sign error somewhere, so I've got two answers - differing by a factor of .
That said, you're missing one of the surfaces - the half-disk at .

Edit: Noticed my mistake, you're good apart from the additional surface.
Also it seems more natural to parameterise the cylinder with cylindrical polars, but I guess that's a matter of taste.
I did use cylindrical coordinates, but it did get pretty messy. Either a careless mistake somewhere or just the nature of this question. Anyways, thanx.

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