The integral of $x^{1/33}+y^{1/27}+z^{1/39}$ of the line segment $(161, 283, 73)$ to $(168, 361, 145)$. I tried to do it on my own but my answer $-2873.78$ seems extremely wrong.
Originally I tried to brute-force it. I did $(161-7t)^{1/33}+\ldots(j+k)$, took the integral of that from $0$ to $1$, and tried to solve.
My integral for $i$ for example looked like $-\frac{33}{238}(161-7t)^{34/33}$. I substituted in $1$ and added them together. In retrospect though my mistake may have been not properly subtracting the $0$ and automatically assuming it would cancel out the numbers when it doesn't in this case.
We parameterize $\gamma(t) = (161,283,73)^T + (7,78,72)^T t$. Then $\|\gamma'(t)\|_2 = \sqrt{11317}$ and $$\int_L f(x,y,z) \ \mathrm dS = \int_0^1 f(\gamma(t)) \cdot \|\gamma'(t)\|_2 \ \mathrm dt\\ = \sqrt{11317} \int_0^1 (161+7t)^{1/33} + (283+78t)^{1/27} + (73+72t)^{1/39}\ \mathrm dt \\ = \sqrt{11317} \left( \frac17 \int_{161}^{168} x^{1/33} \ \mathrm dx + \frac1{78} \int_{283}^{361} y^{1/27} \ \mathrm dy + \frac1{72} \int_{73}^{145} z^{1/39} \ \mathrm dz \right)$$ The last are standard one-dimensional integrals wich you can find by using the primitives and were obtained using the canonical substitutions ($x = \gamma_1(t)$, $y = \gamma_2(t)$, $z=\gamma_3(t)$ where $\gamma_i$ is the $i$-th component of $\gamma$).