In Mathematica, how can I convert an equation written in Polar coordinates
to Cartesian coordinates with x and y values?
You could use the usual substitution. I use ArcTan[y,x] rather than ArcTan[y/x] because it accounts for the quadrant.
*** Below edited to correct error ***
In[1]:= fromPolarRule = {r -> Sqrt[x^2 + y^2], \[Theta] ->
ArcTan[x,y]};
In[2]:= eq = r Cos[\[Theta]] == r Sin[\[Theta]];
In[3]:= eqxy = eq /. fromPolarRule
Out[3]= x == y
I can do it on paper. I worded my question poorly.
What I am looking for is a Mathematica command that will do this.
Some commands will convert to polar but I haven’t found one that
converts to Cartesian.
Thanks
Perhaps CoordinateTransform?
CoordinateTransform["Polar" -> "Cartesian", {r, \[Theta]}]
(* => {r Cos[\[Theta]], r Sin[\[Theta]]} *)
The built-in functions I’m aware of (CoordinateTransform (above) and FromPolarCoordinates) operate on vectors, not equations. Perhaps you could say more about the application.
TransformedField[
“Polar” →
“Cartesian”, [(r^2)* Cos[[Theta]]], {r, [Theta]} → {x, y}]
Syntax::sntxf: “TransformedField[Polar->Cartesian,” cannot be followed by “[(r^2)*Cos[[Theta]]],{r,[Theta]}->{x,y}]”.
I am taking a calculus course on “Great Courses”.
On the Polar part of the course he shows how to convert an expression written in Polar Coordinate to Cartesian Coordinates.
I still can’t do this in Mathematica. I took the suggestion to try
TransformedField. As you can see, I can’t execute the example shown on the Help
page. Do I have to load some kind of package before executing what I am trying to do. It doesn’t look like it. A few years age I used Mathematica a lot. But now I am rusty.
I my Trig book the show an example:
r=@/(1-Cos(theta)) Converts to Y^2=4x+4.
I have done polarplots ,successfully, in Mathematica on everything I am I am trying yo convert.
You help is appreciated.
Can I supply more info?
In regard to David Keith’s fromPolarRule, note that ArcTan[y, x] should read ArcTan[x, y].
Well, after 20+ year using Mathematica I still learn something new every day. 
The only thing wrong with your use of Transform field above is too many brackets, although code not in a code block sometimes gets mangled in a posting.
Below is a use of the built-in function you found, as well as an example of making your own, which is one of Mathematica’s real strengths.
*** and note that in my earlier post above I incorrectly used ArcTan[y,x] rather than ArcTan[x,y] ***
In[1]:= (* with the built-in function *)
TransformedField[
"Polar" -> "Cartesian", (r^2) Cos[theta], {r, theta} -> {x, y}]
Out[1]= x Sqrt[x^2 + y^2]
In[2]:= (* doing it with a Rule and Replace works also *)
fromPolarRule = {r -> Sqrt[x^2 + y^2], theta -> ArcTan[x, y]};
In[3]:= (* and we can use it to define a function *)
myTransformedField[expression_] := expression /. fromPolarRule
In[4]:= myTransformedField[(r^2) Cos[theta]]
Out[4]= x Sqrt[x^2 + y^2]
Thanks David. That worked!
I finally have the conversion.
Thanks, Eric – I edited it to correct the error.
David, I implemented your myTransformedField function.
It is great.
Years ago I bought a software package called Scientific Notebook by MacKichan
Software Inc. It is a very user friendly front end to the old Maple version 5.5.
It is no longer being upgraded. I wish someone would look at it to see the features and concept.
Maybe something like it could be done for Mathematica.
Thanks again.