August 4 Information

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

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

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

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

${\displaystyle =5\int _{0}^{2\pi }\int _{0}^{2}(3(r\cos \theta )^{2}+3(r\sin \theta )^{2})r\,dr\,d\theta }$

${\displaystyle =15\int _{0}^{2\pi }\int _{0}^{2}r^{3}dr\,d\theta }$

${\displaystyle =60\int _{0}^{2\pi }d\theta }$

${\displaystyle =120\pi }$

${\displaystyle \int \int _{S}(x^{3},y^{3},0)\cdot {\vec {n}}\,dS}$

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

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

${\displaystyle ={\frac {5}{2}}\int _{0}^{2\pi }2^{4}\cos ^{4}\theta +2^{4}\sin ^{4}\theta \,d\theta }$

${\displaystyle =40\int _{0}^{2\pi }\cos ^{4}\theta +\sin ^{4}\theta \,d\theta }$

${\displaystyle =60\pi }$

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) reply

Three pieces of advice:

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

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

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) reply

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) reply

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

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) reply

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) reply

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) reply

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) reply

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

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

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) reply

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) reply

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) reply