The solution packet made by the UBC Math Club has some errors in it. I will post the errors here (I will update this post if there are more reported to me). After the exam, I will report these errors to the math club so that they can improve next year’s packet.

Some errors are quite serious, and some are excellent learning opportunities. Please report any new errors you find to me.

April 2007, Problem 7 (d)

This is a TERRIBLE mistake. It highlights something I’ve been trying to explain all semester! (And it looks like they tried to correct it from a previous also wrong answer!) The answer is that is NOT conservative, because if it were, then all line integrals on closed loops would be zero, and we found a counterexample in part (c). The fact that would usually tell us that is zero, but that test doesn’t work here, because the domain of is not simply connected.

December 2009, Problem 5

The solutions give an incorrect domain to the parametrisation. They give and , which is a piece of the surface lying over the 1st quadrant of the xy-plane. This surface is a parabolic bowl, facing downward, and shifted up, and we are asked to use the part of the bowl above the xy plane. This piece looks like an upside-down bowl. However, the solutions give a domain that would give just the noon-to-3-o’clock portion of the bowl. If you have any doubts, please graph the two parametrised surfaces. (In particular, if we dipped the bowl down into a sink of water, it would hold air. But the piece of surface over the domain the solutions suggest wouldn’t hold any air!) This error is an excellent learning opportunity — please do investigate what I’ve described in this paragraph.

December 2009, Problem 3 (c)

The curl is computed incorrectly (the first component) but it does not seem to affect the final answer.

April 2009, Problem 9, True-False #4

The answer is TRUE. (The solutions claim it is false). Both expressions are well-defined. The right-hand side is zero because the curl of the gradient is always zero. The left hand side is zero because whatever is, it is parallel to itself, and parallel vectors have zero cross-product.

April 2009, Problem 4

An algebra error in the integral. Your classmate explains it well:

“The solution applies Green’s theorem, which I can follow, but the integration from the 3rd to 4th lines of the computation doesn’t seem to be right. When I did it myself before looking at the solutions, I got 54 as my answer.. and I think the difference in in the integration. Simplifying the expression in the integral gives -8x^3 +24x^2 then they antidifferentiated it to be -2x^4+12x^3. I think it should be -2x^4+8x^3, though.. which changes the final answer to 54.”

April 2007, #4

The integral is set up completely wrong. And they should do it directly anyway; it’s much better to use the Divergence Theorem. This is the same question as Stewart, Section 17.9 #11, and the answer is .

April 2007, #5

There’s an algebra error in the integral.

April 2007, #8

There’s a missing r in the double integral.

April 2009, #8

There’s an algebra error in the integral. The answer should be .

April 2005, #4

There’s an error in the algebra of u’s and v’s somewhere.

April 2009, #4

The algebra seems highly suspect (integral calculation).

December 2009, #3 (c) (ii)

Typo: it should read .

December 2009, #6 (c)

The solution makes vague reference to the “winding number” but it seems to claim the integral is for any field with a hole in its domain. That’s not true: the integral must depend on the vector field. Not all vector fields with a hole at the origin have integrals for curves circling the origin twice. (Imagine multiplying the field by 7, for example; the value of the integral would change.) So the solution isn’t a solution, it’s just a guess.

MORE IN THE COMMENTS BELOW!

Comments on:"Mistakes in Math Club Solution Packet" (1)Serenasaid:in the exam packet, question 4 april 2005:

we find that the z component of ru x rv =

v^2u^2 – 4(u^2v^2)

which they calculate as -2(u^2v^2) instead of -3(u^2v^2)