You need exp(t), not exp^(t) or exp(^t) or e^t.

Each time assume is called it creates a new version of the variable, and assigns that as the value of the original variable.  So here's what goes on step-by-step in your example.

restart :
expr := x :

expr has value x

assume( x > 0 ) :

x has value (first new x)
expr has value x

expr := x :

x has value (first new x)
expr has value (first new x)

assume( x > 0 ) :

x has value (second new x)
expr has value (first new x)

simplify( x - expr ); 
is( x - expr, 0 ) ;

x - expr evaluates to (second new x) - (first new x), which is not 0.

I think this is just like any other solve command: if you don't specify the variables to solve for, Maple will make its own choice.

I think this is just like any other solve command: if you don't specify the variables to solve for, Maple will make its own choice.

It works for me in Maple 11.02 under Windows XP.  Have you updated to the latest
version of Maple 11?

It works for me in Maple 11.02 under Windows XP.  Have you updated to the latest
version of Maple 11?

My Maple (11) didn't like the y=1..1-x specification in the implicit plot

I don't understand this.  I'm also using Maple 11, which had no trouble with y = 1 .. 1-x.  And I have no idea why y = 1 .. x would be better.  Can you upload a worksheet where the error occurs?

As for plotting outside the simplex: yes, of course that will happen if you use something like  x=0..1, y=0..1,  because the transformation T maps the point  [1,1] to [1, 1, -1].

My Maple (11) didn't like the y=1..1-x specification in the implicit plot

I don't understand this.  I'm also using Maple 11, which had no trouble with y = 1 .. 1-x.  And I have no idea why y = 1 .. x would be better.  Can you upload a worksheet where the error occurs?

As for plotting outside the simplex: yes, of course that will happen if you use something like  x=0..1, y=0..1,  because the transformation T maps the point  [1,1] to [1, 1, -1].

f:= (x,y,z) -> x^2 + y^2 + z^2 - 1/2;
with(plots): 
T:= plottools[transform]((x,y) -> [x,y,1-x-y]):
P := implicitplot(f(x,y,1-x-y),x=0..1,y=0..1-x):
Simplex:= polygonplot3d([[1,0,0],[0,1,0],[0,0,1]], colour="Beige"):
display([T(P),Simplex]);

f:= (x,y,z) -> x^2 + y^2 + z^2 - 1/2;
with(plots): 
T:= plottools[transform]((x,y) -> [x,y,1-x-y]):
P := implicitplot(f(x,y,1-x-y),x=0..1,y=0..1-x):
Simplex:= polygonplot3d([[1,0,0],[0,1,0],[0,0,1]], colour="Beige"):
display([T(P),Simplex]);

interface(rtablesize=10);

interface(rtablesize=10);

It's worth thinking about some special cases.  In principle, the general change of variables can be broken down into individual steps, where each step either changes the "inside" variable only or interchanges the order of two integrations.  Thus going from int_R f(x,y,z) dx dy dz to an integral ds dt du might be done in the following steps:

dx dy dz -> du dy dz -> dy du dz -> dt du dz -> dt dz du -> dz dt du -> ds dt du

> plot([1+cos(t), sin(t), t=0..2*Pi]);

> plot([1+cos(t), sin(t), t=0..2*Pi]);