In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

We now generalise Green’s theorem to vector fields in The following result is by Sir George Gabriel Stokes.

Stokes’ theorem

Let be a simple closed positively oriented curve in the plane parameterised by a continuously differentiable function and let be the region enclosed by Let be a twice continuously differentiable and one-to-one parameterisation of a surface . The boundary of is parameterised by the positively oriented curveThen

Notice that the orientation of the surface and boundary curve is specified in the theorem above. This orientation of the surface and curve can also be described in the following way: If you walk in the positive direction along the boundary curve with your head pointing in the direction of the normal vector , then the surface will always be on your left.

**Exercise**

Use Stokes’ theorem to evaluate the integral where and is the part of the sphere that lies inside the cylinder and above the -plane.

**Exercise**

Prove that if everywhere on , then is conservative.

**Exercise**

We have previously seen that Green’s theorem does not apply to the case where the curve is the circle centered at with radius and where since is not defined at

Consider now the vector field

Notice that for the first two components of are the same as the first two components of Hence where is the circle centered at with radius in the -plane. Let the surface be the upper hemisphere, i.e. and Verify Stokes’ theorem for this case.

** Exercise**

Let the curve be the boundary curve of the Möbius strip parameterised by

where Calculate the line integral

Note that the integrand is conservative. Further recall that the Möbius strip is not orientable and hence the surface integral over the Möbius strip is not defined.

**Proof of Stokes’ theorem**

Let the vector field Let the parameterisation of the surface be given by

and let

Then the boundary is parameterised by

We can now write the line integral as

Using the chain rule we have

To shorten the expressions we assume in the following that and that and similarly for Thus

where the line integral is over the boundary curve of in positive orientation.

Since the last 3 integrals are over a region in and the boundary curve of is positively oriented, we can use Green’s theorem to obtain

We now apply the chain rule to the partial derivatives appearing above to obtain

By simplifying also the other expressions in the same manner we obtain

One of the sections listed above has the title example, but it seems more like a question? (J. Changed it to Exercise.)

I’m not quit understand the equation of stocks’ theorem above. Is that(Tu*Tv) a unit normal ? why it is dS not dA? thanks (J. Updated it now, thanks.)

Plz help me how to work out this problem?

Q.

Let W = { (x, y, z) є R3 : 1 ≤ x2 + y2 + z2 ≤ 4 } and F : W → R3 be defined by

F(x, y, z) = (x, y, z) / [x2 + y2 + z2 ]3/2 for (x, y, z) є W .if ∂W denotes the boundary of W oriented by the outward normal n to W ,then ∫∫ F. n d S is equal to

a) 0 b) 4pi c) 8pi d) 12 pi