Simplifying the Work Done Equation: x(e^y) + (z+1)(e^z) - (e^z) + k

[chetzread]

Homework Statement

for the work done θ , after combining all three , the work done should be = 2x(e^y) + (z+1)(e^z) - (e^z) + k , am i right ?
why the author stated it is x(e^y) + (z+1)(e^z) - (e^z) + k ?
There's x(e^y) in equation (i) and (ii)

Homework Equations

The Attempt at a Solution

θ= (x)(e^y) + f(y,z) + (x)(e^y) + g(x,z) + (z+1)(e^z) - (e^z) + h(x,y )
= 2(x)(e^y) + (z+1)(e^z) - (e^z) + k

 

[beamie564]

the work done should be = 2x(e^y) + (z+1)(e^z) - (e^z) + k , am i right ?
why the author stated it is x(e^y) + (z+1)(e^z) - (e^z) + k ?
There's x(e^y) in equation (i) and (ii)

No, the author is right. You shouldn't simply add the RHS of the equations. The expressions which are common don't need to be added. In your example, if you had $2xe^y$ ( like you thought), you wouldn't get $\frac {\partial \phi}{\partial x} = e^y$ and $\frac {\partial \phi}{\partial y}= xe^y$

 

[Chestermiller]

I would never have solved it this way. I would have expressed everything in terms of t.

 

[beamie564]

I would never have solved it this way. I would have expressed everything in terms of t.

Exactly, I too would have done the same.