For the vector function
in the x-y plane:
The vector function
is obviously well-defined (and infinitely
differentiable) everywhere except the origin.
The distance of the origin (0, 0) from the point (2, 2) is
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The only singularity is outside
.
Green’s theorem is therefore valid everywhere on
.
Green’s theorem
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everywhere in the domain, including within C .
![]()
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[The direct evaluation of the line integral is very difficult and should be avoided!]
For the vector field
:

Therefore
is irrotational everywhere and a potential function
exists.
The potential function
is such that

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where f (y, z) is an
arbitrary function of integration
![]()
![]()
where g (z) is another
arbitrary function of integration
![[match partial dphi/dz]](t2/q2b5.gif)
The potential function is therefore
![]()
where C is an
arbitrary constant of integration.
However, we require that the potential be zero at the
origin:
![]()
The potential function is therefore
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A sheet is in the shape of that part of the circular
paraboloid
that lies between the circular cylinders
![]()
The sheet has a surface density of

Let r2 =
x2 + y2
then the equation of the circular paraboloid becomes
Use the parametric grid (r, q), such
that
![]()

The ranges of the parameters are
2 < r < 4 and
0 < q < 2p.
The tangent vectors along the coordinate grid lines are
![]()
A normal vector to the surface at every point is

![]()
![[evaluating integrals]](t2/q3a6.gif)
![]()
By symmetry,
![]()
Taking moments about the x-y plane,


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Therefore the centre of mass of the sheet is located at
![]()
[Note that the centroid is located exactly half way between the bottom and top of the sheet. As z increases, the decreasing surface area compensates exactly for the increasing surface density.]
An alternative solution, using the projection method, is available from this link.