All of the points mentioned below are covered in your book and in the online class notes – with examples. Of course everything was covered with examples in class as well. You have plenty of homework assigned so I will not suggest more here. for examples look at the other online notes and in the text.

1. Line integrals

• Definition of path, curve, simple path/curve, closed path/curve, Jordan path/curve, smooth curve, and piecewise smooth curve.
• You need to know what and mean and how to calculate these.
• You should know how to use line integrals to calculate mass, average value of along , center of mass, moments of inertia, etc.
• You need to understand the notion of orientation of a curve.
• You need to understand and know how to calculate, flux through a curve, flow along a curve, and work along a curve.
• You need to know what circulation is and what the notation means. You need to know what the standard (outward) orientation of a Jordan curve is.

2. Del

• You should know what is and what (divergence of ), (curl of ) are.
• You should know some trivial facts: (since this is just ) so in particular the curl of the gradient is always . – recall is the triple product and is the same as , so since the determinant of a matrix with two identical rows is . So the divergence of the curl is always .

3. Conservative fields

• Definition of conservative field as well as restrictions on the domain so that the definition makes sense, i.e., open and path connected.
• Path independence, equivalence of path independence with conservative.
• Gradient fields/potential functions, equivalence with conservative fields under certain assumptions.
• Fundamental theorem for line integrals, where is a conservative field defined on an open path connected region , is a piecewise smooth curve from to lying in , and for some potential function .
• Test to check when a field is conservative, including conditions on when the test works. (Open, path connected, simply connected region; has continuous second partials.)
• Be able to find a potential function for a conservative field and use this to evaluate a line integral.

4. Surface integrals

• You need to know what a smooth surface is and how they fit together to form piecewise smooth surfaces.
• You need to know and understand what an orientation of a smooth/piecewise smooth surface is and how to find an orientation if it exists.
• You need to know what surface integrals are and how to compute them. In particular you need to know what means, how to interpret – in the notes there are several ways of interpreting mentioned.
• You should know how to apply surface integrals to find surface area, mass of a thin surface, center of mass of a thin surface, moment of inertia of a thin surface, average value of on , etc.
• You should know the definition of flux of through and how to compute this.

5. The fundamental theorems of calculus – Green’s, Stoke’s, Divergence

• You should know and be able to use the normal and tangential forms of Green’s theorem in the plane. In particular you must know the hypotheses of Green’s theorem and when it applies. You must know what the induced orientation of the boundary of is and how to compute it.
• You should be able to use Green’s to simplify the calculation of certain line integrals and conversely to compute area of planar regions.
• You should have some idea of how Green’s gives the test for path independence and why simple connectedness is important.
• You should understand the way Green’s theorem can be used in the case that or . For example given a field with divergence on a region and a Jordan curve in , then where might be taken to be a much simpler curve. (This might be the point of the problem, it is up to you to pick a simple and evaluate the line integral there.)
• Stoke’s theorem is essentially just the tangential form of Green’s theorem for arbitrary surfaces satisfying certain conditions – that you must know.
• You should be able to use Stoke’s theorem to calculate the flux of through some surface via computing a simple line integral on the boundary.
• You should know why you can replace a given surface by a much simpler one having the same boundary and use this in computations of . For example if I ask you to compute the flux of through the parabolic surface from to . Then you know where is just the unit disk centered on the -axis sitting at . (This is also a consequence of the divergence theorem.) Conversely, Stoke’s can be use to calculate a line integral via a corresponding surface integral where .
• The divergence theorem is just the three dimensional version of the normal form of Green’s theorem.
• One important case (just as with Green’s and Stoke’s) is that when (-divergence), then the flux integral through a surface is “independent of the surface” in some sense.
• You should know the relevance of fields satisfying (radial fields that are inversely proportional to the power of the distance from the origin and in particular that the divergence for these fields is when ).

Not to under emphasise this: Know the main theorems and definitions, know the conditions under which the theorems apply – i.e. the hypotheses of the theorems.

]]> http://ketcherscourses.wordpress.com/2009/05/10/math-275-may-10-final-notes/feed/ 0 ketchers http://ketcherscourses.wordpress.com/2009/05/03/math-275-may-5/ http://ketcherscourses.wordpress.com/2009/05/03/math-275-may-5/#comments Mon, 04 May 2009 05:03:03 +0000 ketchers http://ketcherscourses.wordpress.com/?p=220 ]]> Here is a printable version of this post.

1. Divergence Theorem

The divergence theorem is also known as Green’s theorem (for three dimensions) and as Gauss’ theorem.

Theorem 1 Let be a closed bounded region in with being the union of finitely many disjoint closed smooth surfaces. The boundary of is oriented with orientation so that the points into . Let be a vector field defined on an open region including and which is continuously differentiable in . Then

Notice that this really is Green’s theorem pushed up to 3 dimensions and in fact there is a general theorem that could just be called the “Fundamental Theorem of Calculus” that would cover all these (the usual FTC is just Green’s theorem in one dimension!) In the following take to get FTC from Calc I, gives Green’s Theorem (and hence Stoke’s Theorem), gives the Divergence Theorem.

Theorem 2 Let be a closed bounded region in with being the union of finitely many disjoint closed smooth -dimensional regions. The boundary of is oriented with orientation so that the points into . Let be a vector field defined on an open region including and which is continuously differentiable in . Then

The proof will be very much like that of Green’s theorem. First we will prove the theorem for particularly simple regions and then argue geometrically to extend the result to more general regions. The simple regions are called projectable.

Let be a solid and be the projection (shadow) of on the -plane. If there are are smooth so that for all such that can be viewed as the solid bounded by

then is called -projectable.

Similarly -projectable and -projectable are defined and is projectable if it is ,, and -projectable.

Proof: (Proof sketch for the divergence theorem) Let be projectable and let the outward normal , we want to show

Expanding this it becomes

so it suffices to show

Suppose is -projectable and we show

First calculate using Fubini.

so we must compute for . For , so . For , define by .

For , the norm must point outward, this is reverse of the standard orientation:

and and so

For ,

and (this is like the calculation for )

So we have

and hence

which is what we wanted. The other cases are proved in a similar fashion.

That every region as in the hypothesis of the divergence theorem can be sufficiently approximated by projectable regions is left to the imagination of the reader.

The divergence theorem itself provides a reason for calling the divergence. Let be a sphere centered at of radius . The mean value theorem for multiple integrals says that for some , where is the volume of a sphere of radius . So we have

so

That is

this is the flux density, i.e., flux/unit of volume at a point, and that is an interpretation of divergence.

Example 1. Compute the flux of outward through thesurface of the solid cone for first just directly using surface integrals and then using the divergence theorem.

I will call the cone the boundary of consists of two surfaces is the surface of the cone and is the top. I will use polar coordinates for the parameterization, so , (this is not the only way.) In each case and

A little thought gives

and little calculation gives

So

Since this becomes

So

Now using the divergence theorem

Using cylindrical coordinates this is

Example 2. One useful consequence of the Divergence theorem is that if on an open region of and are solids with , then

For example the electric field produced by a point charge at the origin is

See \href{http://en.wikipedia.org/wiki/Coulomb for more on this.

A little calculation shows for . The strength of the field passing through the surface of a solid containing the origin is

For complicated this might be difficult to compute, but we can find so that the ball of radius centered at the origin, i.e., , is completely contained within the solid bounded by . The boundary of , , where is the sphere of radius centered at the origin. The divergence theorem gives

This is not hard to calculate!

Parameterize by

A little calculation gives that points outwards and

So for any surface of a solid containing the origin

Example 3. The Divergence Theorem can be used to find volume just as Green’s Theorem could be used to find volume. Given a solid we can take any field with and get

Consider the solid, , formed by rotating the curve , for about the -axis. Take so . The boundary of , (a surface of rotation) is parameterized by by . A moments thought reveals the outward normal to be given by

so

hence

Problems:
14.8: 7, 13, 15, 17, 21, 25, 31

]]> http://ketcherscourses.wordpress.com/2009/05/03/math-275-may-5/feed/ 0 ketchers http://ketcherscourses.wordpress.com/2009/05/03/math-275-may-4/ http://ketcherscourses.wordpress.com/2009/05/03/math-275-may-4/#comments Sun, 03 May 2009 07:43:16 +0000 ketchers http://ketcherscourses.wordpress.com/?p=218 ]]> Here is a printable version of this post.

1. Stoke’s Theorem

Green’s theorem generalizes from regions in and to regions in the -plane viewed as a surface in and in this form it essentially does not change

Now if we replace with an oriented surface in with orientation with boundary consisting of finitely many non-intersecting Jordan curves oriented counterclockwise with respect to then this becomes

and this is Stoke’s Theorem.

Theorem 1 (Stoke’s Theorem) Let be a piecewise smooth oriented surface with orientation and with a boundary consisting of finitely many disjoint Jordan curves, then

where the orientation of the boundary is positive with respect to .

Written slightly differently Stoke’s theorem looks like

Proof: Let parameterize where is now a region in the -plane satisfying the hypotheses of Green’s Theorem. The proof essentially just parameterizes via a planar set and applies Green’s theorem in . The details get a bit involved!

Recall

and similarly for other combinations like , etc. Let

In this notation we have

So

and

Using the chain rule

So

So we aim to show

This splits into three parts:

By Green’s Theorem (applied in the -plane to the vector field ) gives:

Now from the product rule

and

Since we have

Now the chain rule gives:

A little computation gives

So the above becomes

So

This is what we wanted for , the cases for and are similar.

Just as we used Green’s theorem before we can use Stoke’s theorem to prove

Theorem 2 If is continuously differentiable in an open connected and simply connected region of and , then is conservative.

Proof: (Hint.) Take and in and consider two paths and from to . Follow from to and back. This will result in some shared edges in which the fact that the orientation is reversed will cancel out and some polygons to which we can apply Stoke’s and get that the integral along the boundary is . thus we get .

Example 1. Find where is the curve formed by the intersection of and where the orientation is the outward normal.

Consider the planar surface with boundary and normal where . We have

Viewing the plane as the level surface we have

So our integral is

Where is the projection of in the -plane. To find we compute the intersection of the plane and , this gives

rearanging gives

This is an ellipse centered at with

The area of this ellipse is So the integral is

Problems:
14.7: 3, 5, 7, 9, 13, 19, 21, 25

]]> http://ketcherscourses.wordpress.com/2009/05/03/math-275-may-4/feed/ 0 ketchers http://ketcherscourses.wordpress.com/2009/05/02/math-170-may-1/ http://ketcherscourses.wordpress.com/2009/05/02/math-170-may-1/#comments Sun, 03 May 2009 04:45:37 +0000 ketchers http://ketcherscourses.wordpress.com/?p=215 ]]> We will skip 5.7 and spend the last week on application of integrals.

As far as problems go you need to work every integral in 5.5 and 5.6 – NO KIDDING – this will be a huge part of the final and if you can’t use the substitution rule you won’t do well. If a problem is too easy skip it, but please work the integrals that aren’t easy for you.

1. Surface Integrals

Just as for curves to develop a theory of integration on surfaces we need to parametrize the surface. A parametrization of a surface is just a map that is continuous, 1-1, and onto . Often we write rather than .

The level curves of are those curves in that you get by fixing one of or and letting the other vary.

Example 1 – Torus:\label{torus} A parametrization of the torus centered at the origin with major radius and minor radius , assume , is given by: For set

So is given by

Fixing the -level curve (so varies) is a circle parallel to the -axis and in particular for this circle lies in the -plane and has radius . Fixing the -level curve is a circle of radius centered on the the -level curve. In this way you can see the surface is a torus. Here is a picture with level curves drawn for and at , etc.

Example 2 – Mobius Strip: A parametrization of the Mobius strip centered at the origin with major radius and width , assume , is given by: For and set

This parametrization describes taking a strip of paper of length and width and twisting it so as to bring the short ends together with one twist. Here is a picture with level curves drawn for at and at .

A surface is smooth there is parametrization so that

• has smooth boundary.
• has continuous first partials.
• the vectors and are non-zero and not colinear.

The first condition implies that has a smooth boundary. The third condition is equivalent to . A surface is piecewise smooth if it is composed of a finite number of smooth pieces that intersect at most on their boundaries. The tangent plane to at is given parametrically as

which, of course, is the same as the plane through with normal .

The transformation transforms the rectangle to a patch of the surface that has surface area . is approximated by the parallelogram given by the vectors and . So

This gives the differential

Often one variable is given in terms of the other two, for example for where is the projection of on the -plane. In this case the parameterization is

In this case

So

and

This might be remembered as follows, recall

So

and

This is really a bunch of formalism, but it works out nicely.

Given a real valued function we define the surface integral of on by fixing a parametrization then partitioning into little rectangles which get transformed into little patches of the surface whose surface area is . Thus we have

In particular the surface area of is

If is a density function (for a thin sheet ), then the mass is , similarly for first moments, moments of inertia, etc.

Example 3. Find the surface area of the \hyperlink{torus}{torus}. We have

so

So

and

This can be seen to be correct geometrically by cutting it and stretching it out into a cylinder of height with radius .

If a surface is given as a level curve with continuous first partials and , then at each point on we can find an open neighborhood of so that one variable is given in terms of the other two – hence a local parametrization. This follows from the implicit function theorem which in this case can be stated as

Theorem 1 Assume has continuous partials in a nbhd of and , then there is a nbhd of so that for some that is differentiable on . (Similarly we can solve for or if the appropriate partial is non-zero.)

Since is differentiable we can apply the chain rule using :

for in . So

So

This means

and

So

and hence

where is the “shadow” of in the -plane. (This is assuming , the same sort of expression works for projections in the other coordinate planes assuming the appropriate partial is non-zero.)

Example 4. Suppose the density of a hemispherical dome of radius is , find the mass. The surface is given by . The projection of on the -plane is the disk of radius , . Here so and since so

2. Flux through a surface

A surface is orientable if it is possible to continuously (in the variables ) assign a normal unit vector to each point on the surface.

If is smooth as witnessed by and is orientable, then

This need not be the case if is non-orientable as can be seen by the Mobius strip, which in non-orientable. The point is that is continuous as a function on (in the -plane), however, the induced assignment of normal vectors on the Mobius strip in as a function of is not continuous as is seen below.

If is given as the level surface to a continuously differentiable with , then

If is orientable with orientation and is a continuous vector field defined on a region including , then

so the flux through is

If on , then

Similarly for on and on .

If is given as a level surface and in , then

Similarly with or replacing . So

Example 5. Find the flux of the field through the hemisphere of radius lying above the -plane. The surface is given by . Here the normal . so and so

Here is the disk of radius on the -plane centered at the origin, . So

Example 6. Find flux of the gradient field of through the surface given by over the triangular region in the -plane with corners , , .

Here the parametrization is so

I’ll leave it to you to actually compute this. Problems:
14.5: 11, 13, 15, 21, 25, 29, 33, 34, 35, 43
14.6: 11, 13, 15, 21, 25, 29, 35, 41

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