Calculate the line integral of vector field $F = <e^z, e^{x-y}, e^y>$ over the given path

Question

This was a problem on my exam which I got wrong. I'm still trying to figure out where I went wrong.

What I did was split it into three integrals, one for C1, one for C2, and one for C3. Then I added all the solutions together.

I'm assuming my mistake was in parameterizing C1, C2, and/or C3; and if so, how would I find the appropriate Curves?

Thanks!

Answer 1

The curve $C_1$ should be a straight line between $(2,0,0)$ and $(0,4,0)$. Such a line is given by

$$ r(t) = (1-t)(2,0,0) + t(0,4,0) = (2(1-t), 4t, 0) = (2 - 2t, 4t, 0). $$

Your parametrization of "$C_1$" is

$$ r(t) = (2t + 2, 4t, 0) = (1-t)(2,0,0) + t(4,4,0) $$

which describes a straight line between $(2,0,0)$ and $(4,4,0)$.

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The general formula for a (constant speed) parametrization of a line segment that starts at $t = 0$ at the point $p_1$ and ends at $t = 1$ at the point $p_2$ is given by

$$ \gamma(t) = (1 - t)p_1 + t p_2. $$

Answer 2

Hint for C_1

the parametric equations are

$$x=2+(0-2)t\;,\; dx=-2dt$$

$$y=0+(4-0)t\;,\; dy=4dt$$

$$z=0+(0-0)t\;,\;dz=0$$

$$I_1=\int_0^1(e^0(-2dt)+e^{2-2t-4t}(4dt)+e^{4t}.0)$$

$$=\int_0^1(-2+4e^{2-6t})dt$$

Yes, you can finish.