Wikipedia:Reference desk/Archives/Mathematics/2008 August 23

= August 23 =

Pyramid Volume section contains wrong derivation of volume: a negative sign in the integral...
Hi,

Just noticing in the Pyramid_(geometry) following "The volume is given by the integral"

There's a negative sign (-A / 3h^2) and there shouldn't be! Right? link title

How would I fix this? I don't know how to generate the math images... Oh wait, it's

$$\frac{A}{h^2} \int_0^h (h-y)^2 \, dy = \frac{A}{3h^2} (h-y)^3 \bigg|_0^h = \frac{1}{3}Ah.$$

Right? Thanks! InverseSubstance (talk) 18:12, 23 August 2008 (UTC)
 * No, the sign is correct. Algebraist 18:16, 23 August 2008 (UTC)
 * More explicitly, the sign gets flipped once by the (implied) substitution for $$-y$$, and it gets flipped again because you evaluate at $$y=h$$ where the antiderivative is 0 and subtract the positive evaluation at $$y=0$$. Hope this helps. --Tardis (talk) 18:26, 23 August 2008 (UTC)

Wait. I understand the evaluation part. But not the integration part. So it seems to me what your saying is, fex:

$$\int (1-x)dx = -\frac {1}{2}(1-x)^2$$ Right? What rule of integration does this fall under? Thanks... InverseSubstance (talk) 19:25, 23 August 2008 (UTC)
 * There is no rule for that, but it's useful to note that $$\int(a+bx)^n\,dx=\frac{(a+bx)^{n+1}}{(n+1)b}+c$$ for complex a and b. x42bn6 Talk Mess  20:15, 23 August 2008 (UTC)
 * It's called the substitution rule. You write $$u=1-x \Rightarrow du=-dx,dx=-du$$ so $$\int(1-x)dx=\int u(-du)=-\int u\,du=-\frac{u^2}2+C=-\frac{(1-x)^2}2+C$$.  For definite integrals, you have to transform the limits too, unless you immediately (as I did here, and as the original example does) switch back to the original variable before evaluating. --Tardis (talk) 04:34, 24 August 2008 (UTC)
 * You can see it's true by differentiating, the key rule there being the chain rule, the derivative of -x (and, therefore, 1-x) is -1. --Tango (talk) 20:50, 23 August 2008 (UTC)

How to find the shorter path?
What algorithms can you use to find the shorter path in a net (like a very complex subway net with thousands of nodes)? —Preceding unsigned comment added by Mr.K. (talk • contribs) 19:16, 23 August 2008 (UTC)


 * Is Shortest path problem what you're looking for? -- BenRG (talk) 19:44, 23 August 2008 (UTC)

Geometric figure for xn
If x is a segment, x² is a square, x³ a cube, what geometric figure is x4, and is there a figure for every xn? —Preceding unsigned comment added by Mr.K. (talk • contribs) 19:17, 23 August 2008 (UTC)


 * x4 is the hypervolume of a tesseract, and x5 a five-dimensional hypercube, and so on. There can't be a three-dimensional figure with a volume of x4, because the units come out wrong. For example if x is 2 meters, then x4 is 16 meters4, but a volume has to have units of meters3. -- BenRG (talk) 19:49, 23 August 2008 (UTC)


 * It is interesting to note that the first solution to the quartic equation was published together with the first solution to the cubic equation (by Cardano, I think), but Cardano downplayed the solution of the quartic equation, because at that time raising things to the power of 4 was commonly regarded as meaningless, as there is no corresponding geometrical object in three dimensions. Eric.  213.173.233.179 (talk) 09:48, 24 August 2008 (UTC)