MaplePrimes - answers and comments on Question, Coordinate System

Coordinate System

SandorSzabo — Wed, 16 Jan 2008 23:15:12 Z

(Very sorry, something is wrong. I use a lot of script, it can be the reason.) I want to (triple)integrate f(x,y,z):=7z on the cylinder

x^2+y^2 less than R, 0 less than z less than m.
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There are a lot of predefined region, but I haven't found Cylinder.
In fact, my question is not "how to solve this exercise", but
"does exist in Maple somewhere in a package where I can define the region Cylindrical which I could use in integration.
In VectorCalculus there is SetCoordinates('cylindrical'[r, theta,z]); but I don't know
it can be applied and how, or not.

Thanks, Sandor

Cylinder

Robert Israel — Thu, 17 Jan 2008 00:19:42 Z

Please note that to enter < in this forum, you have to use < ; (without the space), otherwise it will interpret your < as the start of an HTML tag. You seem to be working with int in the Student[VectorCalculus] package. There isn't a "built-in" Cylinder there, but you can use the generic Region.

> with(Student[VectorCalculus]):
  int(f(x,y,z),[x,y,z]=Region(-sqrt(R) .. sqrt(R),
    -sqrt(R-x^2) .. sqrt(R-x^2), 0 .. m));

Unfortunately this doesn't work well with coordinates other than cartesian. For example, I would have hoped that you could do it this way:

 SetCoordinates(cylindrical[r,theta,z]);
 int(f(r,theta,z),[r,theta,z]=Region(0 .. sqrt(R), 0 .. 2*Pi,
    0 .. m));

But that does not work: it gives you an integral of f(r,theta,z) dz dtheta dr, when what you need is f(r,theta,z) r dz dtheta dr.

Circle times the segment

jakubi — Thu, 17 Jan 2008 04:24:56 Z

I think that it is easier:

int(int(7*z,[x,y] = Circle( < 0,0 >, R ) ),z=0..m);

Integration

John Fredsted — Thu, 17 Jan 2008 13:11:02 Z

Do you mean x^2 + y^2 <= R or x^2 + y^2 <= R^2? The former is what Israel assumes, the latter what Jakubi assumes. I will assume the latter. Because f(x,y,z) = 7z is independent of x and y the integral separates and becomes int(7z,z = 0..m) times the area of the base of the cylinder. So, the result is

> int(7*z,z=0..m)*Pi*R^2;
                           7  2     2
                           - m  Pi R 
                           2

PS: Reading your post again I can see that a solution is not actually what you ask for. Anyway, here it is.

Integration on cylinder

SandorSzabo — Thu, 17 Jan 2008 14:05:54 Z

Thanks for all. I wanted to solve this problem in a similar way as Robert. (I'm thinking of a modification that works well. If I find a solution I will post it here.) At this moment jakubi's solution is the simplest (that works also if integral is not separated). Anyway, I learned some useful methods from you. Thanks again. Sandor