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August 6

Parabolic Problem

Greetings. I have another integration problem, involving surface integrals. We are asked to evaluate the surface area of

z=f(x,y)=x^{2}+y^{2}\,

where 2 ≤ z ≤ 6.

Now I understand this to be what is known as a parabloid, where from the side it is shaped like a parabola, and from above it looks like a circle - kind of in all rather similar in 3D to a tornado. Now we are told to integrate over a region $S$ , which I assume is the circle

x^{2}+y^{2}\,

,

and I understand then that the surface area we want is that of the upper and lower circles, from $z=2$ up to $z=6$ , as well as the cut off tornado ( or carrot like ) shape in between, and how much surface area it all has combined, but now I am not too sure as to how to do this. I believe the ∂f/∂x- partial wrt x - will be $2x$ , and wrt to y should be $2y$ , so that all of $dS$ should be ${\sqrt {4x^{2}+4y^{2}+1}}$ , but then what is the function I am integrating, as well as the limits thereof ?

The second problem is to work out the area of the portion of the surface $x^{2}-2z=0$ . that lies above the triangle bounded by the line x = ${\sqrt {3}}$ , $y=0$ and $y=x$ in the $xy-$ plane.

Any assistance on this matter will be of great value. Thank You. Chris the Russian Christopher Lilly 07:21, 6 August 2011 (UTC)[reply]

If I'm not mistaken, the easiest way to approach this problem is to consider the surface as the result of rotating the curve y = √x around the x-axis for 2 ≤ x ≤ 6. You'll need to express the arc length differential dS in terms of dx and dy using Pythagoras' theorem. The integral to consider is

\int _{2}^{6}2\pi ydS

.

—Anonymous Dissident^Talk 12:00, 6 August 2011 (UTC)[reply]

I'd use polar coordinates. Michael Hardy (talk) 13:07, 6 August 2011 (UTC)[reply]

More specifically: If you want

\int \int {\sqrt {4x^{2}+4y^{2}+1}}\,dx\,dy

over the region in which

2<x^{2}+y^{2}<6,\,

then recall that when you transform to polar coordinates,

x^{2}+y^{2}{\text{ becomes }}r^{2}\,

and

dx\,dy{\text{ becomes }}r\,dr\,d\theta .\,

So you get

\int _{0}^{2\pi }\int _{\sqrt {2}}^{\sqrt {6}}{\sqrt {4r^{2}+1}}\ r\,dr\,d\theta .

Since the thing inside of

\int _{0}^{2\pi }\cdots d\theta

actually doesn't depend on θ, this becomes $2\pi \int _{\sqrt {2}}^{\sqrt {6}}{\sqrt {4r^{2}+1}}\ r\,dr.$ The substitution

u=4r^{2}+1,\quad du=8r\,dr

handles this neatly, with u going from 4·(√2)² = 8 to 4·(√6)² = 24. Michael Hardy (talk) 13:18, 6 August 2011 (UTC)[reply]

Thank You all very much, this has helped a lot. If anyone has any idea about the second problem, which to repeat, is to work out the area of the portion of the surface $x^{2}-2z=0$ . that lies above the triangle bounded by the line x = ${\sqrt {3}}$ , $y=0$ and $y=x$ in the $xy-$ plane. Now I assumed that the surface $x^{2}-2z=0$ was the surface $f(x,y):x^{2}=2z$ or also the surface $-z=1/2x^{2}$ , which formed a ceiling over the triangle formed by the three lines given, so that the solid volume formed under the ceiling was kind of like a cubic block, but with a parbolic shape cut out for the lid, and cut diagonally across it, and we were to find what ever surface area that was. The shape I was thinking of is rather hard to describe. Now I then went and evaluated

\int _{0}^{\sqrt {3}}\int _{0}^{\sqrt {3}}\ x{\sqrt {x^{2}+1}}\,dx\,dy.

,

which reduces to 7 times root three all over three, which equals just over four. The trouble is, when I look at what I believe to be the shape, and work out the areas of some of faces thereof, I do only half of them, and the surface area is already over seven, so do I even have the correct function, or have I gone amiss ? Thanks. Chris the Russian Christopher Lilly 14:16, 6 August 2011 (UTC)[reply]

Differentiability

The Weierstrass function famously challenged the notion that every continuous function is differentiable; with this in mind, what other properties are necessary for a, real valued, function to be differentiable? If such an answer exists, I'd like something more instructive than being reminded that a differentiable function is one whose derivative exists at each point in its domain. Thanks. asyndeton talk 18:49, 6 August 2011 (UTC)[reply]

I think non-differentiable functions were known before this, |x| for example. What made the Weierstrass function different was that it doesn't have derivative anywhere. My impression is that mathematicians had an intuitive notion as to what a 'curve' is. Namely, that a curve was something that you could approximate with a string or a pencil mark, you just had to imagine the string or mark being infinitely thin. The 'pathological' curves violate that intuition, there is no arclength for example. The intuition leads to the idea of a tangent to a curve, specifically that at smaller and smaller scales a curve stops wiggling and looks more and more like a straight line, at least if you ignore corners. The tangent can be defined in terms of derivatives but really the notion of a tangent has been around for much longer. Anyway, kind of a non-mathematical answer but you said you didn't want the mathematical one.--RDBury (talk) 20:56, 6 August 2011 (UTC)[reply]

I don't think there really is an answer other than to give the definition of the differentiable function, as you have done. We can give an intuitive form of that definition, as RDBury has done: A function is differentiable at a point if it is approximately straight around that point at small enough scales. I don't think you'll get anything different than that. We can give examples of types of functions that will always be differentiable (eg. rational functions, including polynomials, away from singularities), but I don't think that's really what you are after. --Tango (talk) 21:42, 6 August 2011 (UTC)[reply]

Imaginary Numbers

I understand what imaginary numbers are, and why they are not "real". I have read that there are practical applications of imaginary numbers, but it seems to me that whenever someone is supposedly using imaginary numbers, all they're doing is using the letter i as a variable. Am I misunderstanding this, or is there ever an actual use for the square root of a negative number? Shui9 (talk) 23:54, 6 August 2011 (UTC)[reply]

The difference between i and any old indeterminant, call it x, is that if you mulitply x*x you just get x², but i*i = -1. So for example if I multiply two complex numbers: (2 + 3i)(1 - 2i) = 8 - i, I get another complex number, rather than having an i² term. It's this multiplicative structure that makes complex numbers interesting. However it turns out that one way to define the complex numbers is to take the polynomials in one variable x and then set x² + 1 = 0, so there is some connection there. Rckrone (talk) 00:21, 7 August 2011 (UTC)[reply]

Everyone is comfortable with the natural numbers 0, 1, 2, ... . These allow us to solve simple arithmetic problems, but you run into trouble trying to solve the general equation ax + b = 0. For that you need to introduce the rational numbers. The introduction of the non-real numbers solves a similar problem when trying to solve the general ax² + bx + c = 0 – and, as it turns out, polynomial equations of all higher degrees. Many people are reluctant to accept the idea of non-real numbers because they are based on i, which is not a familiar numeral. But, mathematically, they fill a gap just like −5 and 7/2 and are just as "real" in that sense. —Anonymous Dissident^Talk 00:49, 7 August 2011 (UTC)[reply]

The symbol for the square root of -1 may be used to denote a transformation rather than a variable, for example reactive electrical circuit theory would be much harder without the use of j (i being used for electric current) to show an anticlockwise rotation of 90 degrees.86.155.185.195 (talk) 13:29, 7 August 2011 (UTC)[reply]

Imaginary numbers are "real", you can represent i in terms of real numbers as:

$i={\begin{pmatrix}0&-1\\1&0\end{pmatrix}}$

Count Iblis (talk) 14:51, 7 August 2011 (UTC)[reply]

i is not a variable because (1) it only ever has one value, and can't take other values, (2) it has special properties that variables generally don't have, especially the property that i² = -1. McKay (talk) 03:12, 8 August 2011 (UTC)[reply]

I don't know that I've ever had a use for the square root of a negative number, as such, but I often find it ferociously convenient to treat plane geometry as manipulation of complex numbers. (E.g., a current project.) —Tamfang (talk) 20:44, 11 August 2011 (UTC)[reply]