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August 4

Why can I not verify a straightforward divergence theorem? ? ??? ???? ??

The vector field $(x^3,y^3,0)$ over the cylinder of radius 2 and height 5.

$\int \int \underset{V}{\int }div(x^3,y^3,0)dV$

$=\int \int \underset{V}{\int }3x^2+3y^{2}dV$

$=5\int \underset{Disc}{\int }3x^2+3y^{2}d(Disc)$

$=5\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{2}{\int }}(3(r\cos \theta {)}^2+3(r\sin \theta {)}^2)rdrd\theta$

$=15\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{2}{\int }}{r}^{3}drd\theta$

$=60\underset{0}{\overset{2\pi }{\int }}d\theta$

$=120\pi$

$\int \underset{S}{\int }(x^3,y^3,0)\cdot \overrightarrow{n}dS$

$=\frac{1}{2}\int \underset{S}{\int }{x}^4+y^{4}dS$

$=\frac{5}{2}\underset{C}{\int }{x}^4+y^{4}dC$

$=\frac{5}{2}\underset{0}{\overset{2\pi }{\int }}{2}^4{\cos }^4\theta +2^4{\sin }^4\theta d\theta$

$=40\underset{0}{\overset{2\pi }{\int }}{\cos }^4\theta +{\sin }^4\theta d\theta$

$=60\pi$

I'm ashamed I keep getting different answers. You don't have to worry about the top and bottom of the cylinder since they are orthogonal to the vector field.Widener (talk) 03:20, 4 August 2012 (UTC)

Three pieces of advice:

1. Try first the cylinder of radius 1 and height 1.
2. Remember the parentheses: $\int \int \underset{V}{\int }(3x^2+3y^2)dV$
3. Take a nap and your mind will solve the problem while you sleep.

Bo Jacoby (talk) 05:00, 4 August 2012 (UTC).

The height of the cylinder is parallel to the z-axis; its base lies on the xy plane. --Widener (talk) 05:37, 4 August 2012 (UTC)

Of course, the height is irrelevant really; in each calculation you just multiply by the height (due to the vector field being independent of the z component). --Widener (talk) 05:39, 4 August 2012 (UTC)

For a circle of radius r, $dC=rd\theta$. Sławomir Biały (talk) 14:27, 4 August 2012 (UTC)

sigh. Of course. I knew I had forgotten something simple. I guess I was getting used to circles of radius 1.--Widener (talk) 23:59, 4 August 2012 (UTC)

Quotient Spaces

Template:Resolved Consider the unit sphere in three-space. I want to identify the equator, and then identify the open upper- and lower-hemispheres. I've tried two approaches, but get stuck both times.

1. Identifying the equator first gives me a wedge of two spheres. Both of these spheres need to be identified. It's tempting to think that it all collapses to a point but then the equator and the hemispheres would be the same point, and that's not how it's set up. I'm also tempted to think that I end up with two points, but then that's a disconnected space and the sphere isn't.
2. Press the two open hemispheres together leaving a closed disk whose boundary was the equator of the sphere. Next, I identify the equator, i.e. the disk's boundary. This leaves me with a sphere. I now need to identify all but one point of this new sphere. Again, I'm also tempted to think that I end up with two points.

I'd be interested to hear people's opinions on how to do this, and what they end up with. In reality, I'm doing this as a warm up problem. I intend to consider RP² eventually, under a much more complicated identification. — Fly by Night (talk) 19:50, 4 August 2012 (UTC)

It's the indiscrete topology on two points. There's no rule that says the identification of a Hausdorff space must be Hausdorff. --COVIZAPIBETEFOKY (talk) 22:45, 4 August 2012 (UTC)

Not the indiscrete topology. The singleton non-equator point is an open set. It's the (unique) topology on 2 points which is neither discrete nor indiscrete.--121.73.35.181 (talk) 22:49, 4 August 2012 (UTC)

Yeah, you beat me to correcting myself. If $h$ and $e$ are the points representing the hemispheres and equator respectively, then it's the topology $\{\mathrm{\varnothing },\{h\},\{h,e\}\}$. --COVIZAPIBETEFOKY (talk) 22:51, 4 August 2012 (UTC)

• That's great. Thanks guys! — Fly by Night (talk) 00:47, 6 August 2012 (UTC)

This quite important space has a name, and of course we have an article on it. Sierpiński space. Once went to a well-attended lecture by Eilenberg which he began by asking "What is the most fundamental topological space?" Too many (>50) people going OOH OOH pick me & shouting wrong answers for anyone to hear the correct answer. :-) John Z (talk) 21:14, 7 August 2012 (UTC)

Percentage of silver and its weight in a mid-20th century US nickel

The silver nickel that was produced in the US between 1942 and 1945 contained 35% silver, 56% copper, and 9% manganese. It weighed a total of 5.00 grams. I know one troy ounce is 31.1 grams. I am unsure how to factor the weight of the silver. I know 36% of 5.00 grams is 1.8, but this doesn't factor in the weight of the other two materials (of which I am unsure). How would you go about answering this? I'm trying to figure out how many of the aforementioned nickels it would take to make one troy ounce of silver. --Ghostexorcist (talk) 22:56, 4 August 2012 (UTC)

The ratios should be by weight, or mass: see the second paragraph of alloy, for example. It's the only thing that makes sense for precious metals: if you know something is '50%' silver you want to be able to weigh it, divide by 2 and know that's how much silver there is from the point of view of valuing it. So 31.1 ÷ 1.8, or 31.1 ÷ 1.75 for 35%.--JohnBlackburne^words_deeds 23:52, 4 August 2012 (UTC)