Wikipedia:Reference desk/Archives/Mathematics/2012 August 4

= 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.
 * $$\int\int\int_V div (x^3,y^3,0) dV$$
 * $$ = \int\int\int_V 3x^2+3y^2 dV$$
 * $$= 5 \int\int_{Disc} 3x^2+3y^2 \, d(Disc)$$
 * $$ = 5 \int_0^{2\pi}\int_0^2 (3(r\cos\theta)^2+3(r\sin\theta)^2)r \, dr \, d\theta$$
 * $$ = 15 \int_0^{2\pi}\int_0^2 r^3 dr \, d\theta $$
 * $$ = 60 \int_0^{2\pi}d\theta $$
 * $$ = 120 \pi $$


 * $$\int\int_S (x^3,y^3,0)\cdot \vec{n} \, dS$$
 * $$ = \frac{1}{2}\int\int_S x^4+y^4 dS$$
 * $$=\frac{5}{2} \int_C x^4+y^4 dC$$
 * $$=\frac{5}{2} \int_0^{2\pi} 2^4\cos^4\theta+2^4\sin^4\theta \, d\theta$$
 * $$=40 \int_0^{2\pi} \cos^4\theta+\sin^4\theta \, d\theta$$
 * $$=60\pi$$
 * 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: Bo Jacoby (talk) 05:00, 4 August 2012 (UTC).
 * 1) Try first the cylinder of radius 1 and height 1.
 * 2) Remember the parentheses: $$\int\int\int_V (3x^2+3y^2) \, dV$$
 * 3) Take a nap and your mind will solve the problem while you sleep.
 * 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, $$dC=r\,d\theta$$.  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
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. 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)
 * 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.
 * 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 $$\{ \emptyset, \{h\}, \{h,e\}\}$$. --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 wordsdeeds 23:52, 4 August 2012 (UTC)