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We get some expression in Cylindrical coordinates (r, ϕ, z ) like : expr := r*z^2*sin((1/3)*ϕ) we need to convert it into Cartesian coordinates and than back to Cylindrical coordinates. How to do such thing?

So I found something like this : eval(expr, {r = sqrt(x^2+y^2), z = z,ϕ= arctan(y, x)}) but it seems incorrect, how to correct it and how make eval to convert backwords from Cartesian to Cylindrical?

ϕ == ϕ

So I try:

R := 1; 

H := h; 

sigma[0] := sig0;

sigma := sigma[0]*z^2*sin((1/3)*`ϕ`);

toCar := eval(sigma, {r = sqrt(x^2+y^2), z = z, `ϕ` = arctan(y, x)});

toCyl := collect(eval(toCar, {x = r*cos(`ϕ`), y = r*sin(`ϕ`), z = z}), `ϕ`)

It looks close to true but look: enter image description here

why arctan(r*sin(ϕ), r*cos(ϕ)) is not shown as ϕ?

Actually it is only begining of fun time for me because I also need to calculate

Q := int(int(int(toCar, x = 0 .. r), y = 0 .. 2*Pi), z = 0 .. H)

and to get it back into Cylindrical coordinates...

Rella
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1 Answers1

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simplify(toCyl) assuming r>=0, `&varphi;`<=Pi, `&varphi;`>-Pi;

Notice,

arctan(sin(Pi/4),cos(Pi/4));
                          1   
                          - Pi
                          4   

arctan(sin(Pi/4 + 10*Pi),cos(Pi/4 + 10*Pi));
                          1   
                          - Pi
                          4   

arctan(sin(-7*Pi/4),cos(-7*Pi/4));
                          1   
                          - Pi
                          4   

arctan(sin(-15*Pi/4),cos(-15*Pi/4));
                          1   
                          - Pi
                          4   


arctan(sin(-Pi),cos(-Pi));
                           Pi

K:=arctan(r*sin(Pi/4),r*cos(Pi/4));
                      arctan(r, r)

simplify(K) assuming r<0;
                           3   
                         - - Pi
                           4   

simplify(K) assuming r>0;
                          1   
                          - Pi
                          4

Once you've converted from cylindrical to rectangular, any information about how many times the original angle" might have wrapped around (past -Pi) is lost.

So you won't recover the original &varphi; unless it was in (-Pi,Pi]. If you tell Maple that is the case (along with r>-0 so that it knows which half-plane), using assumptions, then it can simplify to what you're expecting.

acer
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