Answer
$\dfrac{\sqrt {46}}{4} (e^{6 (\ln 8)^2}-1) $
Work Step by Step
In order to compute the integral we will use the following formula:
$\oint F \ dr=\int_m^{n} F[r(t)] r'(t) \ dt$
Here, the curve is given as: $r(t)= (t, 3t, -6t)$
This implies that $r'(t) = (1, 3, -6) \implies |r'(t)|=\sqrt {46}$
Therefore, the integral is:
$\oint F \ dr=\int_0^{\\ln (8)} (3t) \times e^{-6 (-6t)} | r'(t)| \ dt\\=3 \sqrt {46} \int_0^{\ln (8)} t e^{6t^2} \ dt$
Suppose that $a=6t^2 \implies da=12t dt$
or, $=\dfrac{3 \sqrt {46}}{12} \times \int_0^{6 (\ln 8)^2} e^x \ dx \\=\dfrac{3 \sqrt {46}}{12} [e^x]_0^{6 (\ln 8)^2} \\=\dfrac{\sqrt {46}}{4} (e^{6 (\ln 8)^2}-1) $
