Conservative Vector Fields -- Is this right?

EngageEngage

1. The problem statement, all variables and given/known data

G = <(1 + x)e[tex]^{x+y}[/tex], xe[tex]^{x+y}[/tex]+2z, -2y>
Evaluate [tex]\int[/tex][tex]_{C}[/tex]G.dR
where C is the path given by:

x = (1 - t)e[tex]^{t}[/tex], y = t, z = 2t, 1=>t>=0
2. Relevant equations



3. The attempt at a solution

First, i noticed that there is a scalar potential associated with several terms (not all) in
the above vector field:

[tex]\varphi[/tex] = xe[tex]^{x + y}[/tex]
[tex]\nabla[/tex][tex]\varphi[/tex] = <(1 + x)e[tex]^{x + y}[/tex], xe[tex]^{x + y}[/tex],0>

So, i then separated the initial vector field G into two parts:

G=<0, 2z, -2y> + <(1 + x)e[tex]^{x + y}[/tex], xe[tex]^{x + y}[/tex],0>

First, I compute some derivatives and then evaluate the first line integral:
[tex]\frac{dc}{dt}[/tex] =
{dx = -te[tex]^{t}[/tex]dt
dy = 1dt
dz = 2dt}

so,

[tex]\int[/tex][tex]_{C}[/tex]G[tex]_{1}[/tex] . [tex]\frac{dc}{dt}[/tex]dt =
[tex]\int[/tex][tex]^{1}_{0}[/tex]<0, 2(2t), -2(t)> . <-te[tex]^{t}[/tex], 1, 2> =

[tex]\int[/tex]0dt = 0

I then computed the second part of G, keeping in mind that this portion of the vector field is conservative. Therefore, i just made a path from point t = 0 to t = 1 with a straight line r, parametrized like so:

r = { x = 1-t, y = t, z = 2t}
[tex]\frac{dr}{dt}[/tex] = {dx = -1, dy = 1, dz = 2}

my second integral then becomes;

[tex]\int[/tex][tex]^{1}_{0}[/tex]<(1 + 1 - t)e[tex]^{1 - t + t}[/tex], (1-t)e[tex]^{1-t + t}[/tex],0> . <-1, 1, 2> dt =


[tex]\int[/tex][tex]^{1}_{0}[/tex]<(2-t)e, (1-t)e,0> . <-1, 1, 2> dt =

So, is my final answer simply this integral evaluated, or am I doing this completely wrong? Sorry if this is a bad question, my book just has no example problems at all, and integrating the theory can be confusing. Any feedback would be appreciated greatly. Thank you

Dick

That looks like a reasonable approach to me. I didn't check all of the details, though.

EngageEngage

The final answer that i got was -e, by computing this integral and adding the two portions (o & -e)

EngageEngage

Great, thanks. I don't want anyone to check the details, I just wanted to see if what I was doing was reasonable, or if i was doing it completely wrong.

Dick

I get -e as well.

EngageEngage

Thanks a lot, I appreciate the help!