Labware - MA35 Multivariable Calculus - Three Variable Calculus



Stokes' Theorem for Function Graphs (Page: 1 | 2 )


Stokes' Theorem states that for some vector field F and oriented surface S with boundary curve s,

∫∫Scurl F ⋅ dS = ∫s+F ⋅ ds.

Proof of Stokes Theorem:

At the beginning of the proof of Green's theorem, we examined a problem relating the value of some function P(x, y) at the point (x0, c) to that value at the point (x0, d). A more general problem is to find how the value of a function P(x, y, z) changes from a point (x0, c, z(x0, c)) to a point (x0, d, z(x0, d)), if we confine ourselves to the function graph of z(x, y).

If we travel a distance dy in the y direction, two events will occur. Firstly, there will be some change in P(x, y, z) due to the change in y, given by Py(x, y, z)*dy. This occurrence is familiar to us from Green's theorem.

Secondly, since we are confining ourselves to the function graph of z(x, y), there must be some change dz in z as a result of the change in y, given by dz = zy(x, y) * dy. This change in z causes a change in P(x, y, z) by the amount P_z(x, y, z) * dz = P_z(x, y, z) * zy(x, y) * dy.

The net change in P(x, y, z) from y = c to y = d, then, is given by

P(x0, d, z(x0, d)) - P(x0, c, z(x0, c)) = ∫cd(Py(x0, y, z(x0, y)) + Pz(x0, y, z(x0, y))*zy(x0, y))dy

By integrating the result above with respect to x from x = a to x = b, with c and d expressed as functions of x, we get a comparison between integrals over two different curves given by a double integral:

abP(x, d(x), z(x, d(x)))dx - ∫abP(x, c(x), z(x, c(x)))dx = ∫abc(x)d(x)(Py(x, y, z(x, y)) + Pz(x, y, z(x, y))*zy(x, y))dydx

Now multiply by -1 and apply Fubini's theorem:

abP(x, c(x), z(x, c(x)))dx - ∫abP(x, d(x), z(x, d(x)))dx = ∫abc(x)d(x)(-Py(x, y, z(x, y)) + Pz(x, y, z(x, y))*-zy(x, y))dxdy

If we wanted to include curves that connect (x, c(x), z(x, c(x))) to (x, c(x), z(x, c(x))) at x = a and x = b so that the left side of the equation above became a closed curve, this would be equivalent to adding 0, since x would not be changing along these curves. This allows us to write the following equation:

s+P(x, y, z(x, y))dx = ∫∫S(-Py(x, y, z(x, y)) + Pz(x, y, z(x, y))*-zy(x, y))dxdy

for some surface S (part of a function graph) and boundary s ("+" indicates that the direction of travel, projected onto the x-y plane, is counterclockwise when viewed from above). Let us refer to this result, for now, as part 1 of Stokes' Theorem for function graphs.

We can follow a similar process, switching the roles of x and y and considering some function Q(x, y, z) instead of P(x, y, z) to get "part 2":

s+Q(x, y, z(x, y))dy = ∫∫S(Qx(x, y, z(x, y)) - Qz(x, y, z(x, y))*-zx(x, y))dxdy

"Part 3" is discussed on the next page.



  • 1. For each of these demos, describe what happens when you set z(x, y) equal to:
    • 0
    • Any constant
    • Any linear function of x and y (ax + by + c)
  • 2. Show that both demos work for graphs that project onto right triangular regions.