You are Here: Home >< Maths

# Stokes' Theorem Problem

Announcements Posted on
Would YOU be put off a uni with a high crime rate? First 50 to have their say get a £5 Amazon voucher! 27-10-2016
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.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 4, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

Find out here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams