Wikipedia:Reference desk/Archives/Mathematics/2011 August 6

= 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)
 * 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
 * $$\int_{2}^{6} 2 \pi y dS$$.
 * — Anonymous Dissident  Talk 12:00, 6 August 2011 (UTC)

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

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 &theta;, 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&middot;(&radic;2)2 = 8 to 4&middot;(&radic;6)2 = 24. Michael Hardy (talk) 13:18, 6 August 2011 (UTC)

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)

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)
 * 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)
 * 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)

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)
 * 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)
 * 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 &minus;5 and 7/2 and are just as "real" in that sense. — Anonymous Dissident  Talk 00:49, 7 August 2011 (UTC)
 * 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)

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)

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)

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)