Wikipedia:Reference desk/Archives/Mathematics/2010 April 29

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April 29

area of the sphere above a plane

Suppose there is some sphere centred at the origin x^2 + y^2 + z^2 = r^2. Take an arbitrary plane/slice z=a between z=0 and z=r. Can I have some help with a general formula for the surface area of the sphere above a plane?

I've tried using a surface integral but I get a nasty dS formula ... and apparently using spherical coordinates won't give me the right answer. Help! John Riemann Soong (talk) 02:11, 29 April 2010 (UTC)

Archimedes answered this question. He found that if two parallel planes intersect a sphere, the surface area between them depends on the distance between them, but not on where they are otherwise. I'll see if I can remember or work out or find the details. Michael Hardy (talk) 03:30, 29 April 2010 (UTC)

The area of a slice of a sphere is equal to the area of the corresponding slice of the circumscribed cylinder of the sphere. Bo Jacoby (talk) 05:46, 29 April 2010 (UTC).

At an angle θ above the horizontal, the circle has a circumference of 2πrcosθ, so the area you want is $\int _{0}^{\alpha }2\pi r\cos \theta \,rd\theta$ where α = sin^-1(a/r). That evaluates to 2πr²sinα = 2πar. But I guess that's exactly what Bo Jacoby already said. Rckrone (talk) 07:43, 29 April 2010 (UTC)

Second limit comparisson test?

A textbook of a friend's briefly mentioned a theorem which I had never heard of before, and am therefore somewhat dubious of. Suppose $\left\{a_{n}\right\},\left\{b_{n}\right\}>0$, $\sum _{n=1}^{\infty }b_{n}$ converges, and the limit $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=0$, then $\sum _{n=1}^{\infty }a_{n}$ converges as well. I've taken a class on series, and have never run across this, nor did I see it in an internet search. However, it's been useful in answering a number of otherwise tricky series, so I have no idea why this wouldn't have been mentioned to me before. Either I had a bad teacher, or something's wrong with this book. Which is it? 173.179.59.66 (talk) 02:38, 29 April 2010 (UTC) Another possibility is that it's not so useful, and any series that I've tackled with the above "theorem" could have been handled by the regular limit comparisson test or something. If that's so, how would you have determined the convergence or divergence of a series like $\sum _{i=1}^{\infty }ln(i)/i^{3}$? —Preceding unsigned comment added by 173.179.59.66 (talk) 02:41, 29 April 2010 (UTC)

Under the hypotheses you state, for sufficiently large n, a_n must be less than b_n. That in itself is enough to get you the conclusion (combined with the fact that all the terms are nonnegative, of course).

So the test you state is just a little shortcut — you could always simply go through the fact that the a_n are eventually dominated by the b_n. It strikes me as being a little like l'Hospital's rule — a clever little gadget that's occasionally useful, but never a substitute for understanding what's going on. --Trovatore (talk) 03:28, 29 April 2010 (UTC)

Oh, by the way, you don't need the limit to be zero. It's enough that it be finite. If the limit is finite, then the a_n are eventually dominated by some constant multiple of the b_n, and that will still get you the conclusion. --Trovatore (talk) 03:37, 29 April 2010 (UTC)

And, it's not clear if you're assuming the a_n to be positive too; but in any case it's not needed, for a_n=O(b_n) implies that the series of a_n is absolutely convergent as explained above. --pma 11:12, 29 April 2010 (UTC)

Okay, thanks! 173.179.59.66 (talk) 14:01, 29 April 2010 (UTC)

1/49

Is 49 the smallest or somehow the simplest example of the phenomenon described at 49_(number)#Reciprocal? Lazily — Michael Hardy (talk) 03:59, 29 April 2010 (UTC)

Generalisation: If 10ⁿ is divisble by m then (10ⁿ−m)/m is an integer, and its reciprocal is

${\frac {m}{10^{n}-m}}=\sum _{k=1}^{\infty }{\frac {m^{k}}{10^{nk}}}$

49 is the case n=2, m=2. Taking n=1, m=2 we have

${\frac {1}{4}}=\sum _{k=1}^{\infty }{\frac {2^{k}}{10^{k}}}=0.2+0.04+0.008+0.0016+0.00032...=0.24999...$

For n=2, m=4 we have

${\frac {1}{24}}=\sum _{k=1}^{\infty }{\frac {4^{k}}{10^{2k}}}=0.04+0.0016+0.000064+0.00000256+0.0000001024...=0.0416666...$

For n=3, m=8 we have

${\frac {1}{124}}=\sum _{k=1}^{\infty }{\frac {8^{k}}{10^{3k}}}=0.008+0.000064+0.000000512+0.000000004096...=0.008064516...$

The pattern is clearest in the decimal expansion when m is small and n is large. For example, when n=3 and m=2 we have

${\frac {1}{499}}=0.002004008016032064128256513...$ Gandalf61 (talk) 09:22, 29 April 2010 (UTC)

But it's not clear that those other examples are examples of what I had in mind. If you write

${\frac {1}{49}}=0.02+0.0004+0.00000008+\cdots ,$

there's nothing that emerges from just staring at that expansion that suggests that it starts repeating after 42 digits, nor is it clear from any of the above how long the various repitends are. Michael Hardy (talk) 10:34, 29 April 2010 (UTC)

Drawing a Penis on a Ti-84

What function(s) will most easily create a penis-shaped graph (complete with 2 testes the sides) on a ti-84 graphing calculator between x=-10,10 and y=-10,10 ?

Thanks, Acceptable (talk) 04:49, 29 April 2010 (UTC)

Try y=sin(x)/x (Are you really sure you want to do this?) -- SGBailey (talk) 10:03, 29 April 2010 (UTC)

Well, that's one way to have fun with math. Here's the pic: [1].

Note that the portion of interest is approximately between x = -10 and x =10, but that the y value only ranges from (approximately) -0.25 to 1. To make it range from (approximately) -10 to 10, let's do some math.

We need to increase the range from 1.25 to 20, so we multiply by 20/1.25 or 16 to get y=16sin(x)/x. When we multiply this by 1 we get 16 and when we multiply by -0.25 we get -4. So, we now have a y range from (approximately) -4 to 16 and need to move this down by 6, or subtract 6 from the graph. This gives us y=(16sin(x)/x)-6.

Note that the lower y limit is only approximate, because the original graph didn't quite make it to -0.25. An interesting exercise might be to make the lower limit exactly y=-10. StuRat (talk) 14:11, 29 April 2010 (UTC)

You can also experiment with piecewise functions. Such as having semi circles, ellipses (sorry, you said functions, so you'll have to try two half ellipses) and two parallel lines. --Kvasir (talk) 15:53, 29 April 2010 (UTC)

Try $|\sin(x)/x|$, from x=-6 to +6 instead (that's the absolute value of the sinc function). Much better! HTH, Robinh (talk) 07:12, 30 April 2010 (UTC)

Perhaps you'll find one of these cubics erotic enough? They're not straightforward functions as such, I don't know about the Ti-84 but perhaps it'll be able to draw the graphs from the equations. Dmcq (talk) 09:48, 30 April 2010 (UTC)

Wow! these are wild! Never thought of these! Thx. --Kvasir (talk) 15:24, 30 April 2010 (UTC)

Jeez, xy² - x² + y² - 2x + 3 = 0 is downright pornographic. It should come in a plain brown wrapper. StuRat (talk) 22:40, 30 April 2010 (UTC)

I just had a look at the specs and it can't graph equations directly. It does seem to support polar mode graphing and I believe that should be good enough for your purpose with a little hacking, think about developing a polar plot of that sin(x)/x for instance should get halfway there. I'd go for something easy and pretty like a flower though. I'd have thought calculators would be far more powerful that that by now or perhaps they limit them so they are allowed into exams or so people can sell applications on them. Dmcq (talk) 10:20, 30 April 2010 (UTC)

That's because most of them are not true functions and do not past the vertical line test. Most people would not know how to graph Algebraic functions once out of highschool, beyond the typical circle, ellipse, hyperbola etc. --Kvasir (talk) 00:45, 1 May 2010 (UTC)

Changing the scale in a log log function

Say I have a function $\ln(y)=a+b\ln(x)$ and I change the scale of measurement of y (say from metres to kilometers), but dont change the scale of measurement of x, will the coefficient of the slope change, or just the intercept? —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 08:57, 29 April 2010 (UTC)

Just the intercept. -- Meni Rosenfeld (talk) 09:14, 29 April 2010 (UTC)

You're going to change ln(y) to ln(cy) which is the same as ln(y)+ln(c). So you're just adding a constant to ln(y). 69.228.170.24 (talk) 09:18, 29 April 2010 (UTC)

Understanding Sphere Volume/Area Formulas

Does anyone know of a way to prove or at least conceptualize why the formulas for the volume and area of a sphere are what they are for someone who hasn't learned any calculus? It's ok if it's hand-wavy. I couldn't come up with anything any good. Rckrone (talk) 22:17, 29 April 2010 (UTC)

The volume can be found by using Cavalieri's principle. Last time I looked, the account there was fairly terse, but nonetheless covered the ideas. Michael Hardy (talk) 22:49, 29 April 2010 (UTC)

...and now I've looked again. The argument showing what the volume of a sphere is is all there, provided you know the volume of a cone is (1/3)×base×height. The argument showing that the volume of a cone is (1/3)×base×height is the part that's sketchy. Michael Hardy (talk) 22:52, 29 April 2010 (UTC)

(ec) The best I can think of is to do some soft calculus and essentially do a Method of exhaustion with of a stack of cylinders of varying radii. This is, of course, just Riemann sums for solids of revolution, but you can explain it without having to actually do the calculus or the integrals (the method of exhaustion was known to the ancient Greeks, long before calculus was invented). In fact I see someone has given you a more direct reference to this idea. -- Leland McInnes (talk) 22:55, 29 April 2010 (UTC)

Did you see the article on sphere?. Bo Jacoby (talk) 23:06, 29 April 2010 (UTC).

The way I remember the sphere volume formula is that I imagine a cylinder with radius r, with height r as well. Hence the volume of that cylinder is πr²*r = πr³. A sphere with the same radius r is just a little bit bigger than that cylinder, and hence the magic number 4/3.

As for surface area, just remember it's 4 times the area of the circle you'd see when you look at a sphere (same as the circular cross section of the sphere through its centre). Or imagine an open cube (no top and bottom) with length 2r thus fitting the sphere inside. Draw circle the same size of the sphere on each of the four sides. When you cut out the 4 circles, their combined area will be enough to cover the surface of the sphere.

Can't you tell i'm a visual person? --Kvasir (talk) 23:07, 29 April 2010 (UTC)

Integrating for volume

Hey all. Is it possible to extend integration to three dimensions? By this, I mean finding the volume of a three dimensional figure whose equation is known. Thanks --76.229.165.96 (talk) 23:45, 29 April 2010 (UTC)

Certainly. If you have an expression for the cross-sectional area at any value along one axis, all you need to do is integrate the area along that axis.

For example, imagine the equation $y=1/x$ with $x=1...\infty$. It's a curve that starts at {1,1} and gradually approaches y=0. Now imagine that curve rotated around the x axis, to make a trumpet shape. You now have a 3-D figure, of radius $r=1/x$, cross-sectional area of $\pi /x^{2}$. The volume is simply the area of the cross-section integrated over the length, i.e.:

$V=\int _{1}^{\infty }{\frac {\pi }{x^{2}}}dx=\pi$

Interestingly, for this shape, the surface area is

$A=\int _{1}^{\infty }{\frac {2\pi }{x}}dx=\left[2\pi \ln x\right]_{1}^{\infty }=\infty$

In other words, you can fill it with paint (due to finite volume), but you can't paint it (due to infinite surface area). ~Amatulić (talk) 00:19, 30 April 2010 (UTC)

In general, the concept is multiple integral. See in particular Multiple_integral#Computing_a_volume. Staecker (talk) 02:13, 30 April 2010 (UTC)