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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 fieldGinto 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 ofG, 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

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# Homework Help: Conservative Vector Fields - Is this right?

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