Let $\sigma$ be the surface $8 x + 10 y + 6 z=3$ in the first octant, oriented upwards. Let C be the oriented boundary of $\sigma$. Compute the work done in moving a unit mass particle around the boundary of $\sigma$ through the vector field $F = \left(3x-6y\right)\boldsymbol{i}+\left(6y-2z\right)\boldsymbol{j}+\left(2z-3x\right)\boldsymbol{k}$ using line integrals, and using Stokes' Theorem.

LINE INTEGRALS
Parameterize the boundary of $\sigma$ positively using the standard form, tv+P with $0\leq t \leq 1$, starting with the segment in the xy plane.

$C_1$ (the edge in the xy plane) is parameterized by

$C_2$ (the edge following $C_1$) is parameterized by

$C_3$ (the last edge) is parameterized by

$\displaystyle\int_{C_1}F\cdot dr =$

$\displaystyle\int_{C_2}F\cdot dr =$

$\displaystyle\int_{C_3}F\cdot dr =$

$\displaystyle\int_{C}F\cdot dr =$

STOKES' THEOREM

$\sigma$ may be parameterized by $r(x,y)=x\boldsymbol{i}+y\boldsymbol{j}+f(x,y)\boldsymbol{k}=$

$\mbox{curl } F =$

$\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y} =$

$\displaystyle\iint\limits_\sigma (\mbox{curl } F)\cdot {\bf n}\hspace{1pt} dS=$

$\displaystyle\int$

$\displaystyle\int$

$\hspace{3pt} dy\,dx$

$\phantom{\displaystyle\iint\limits_\sigma (\mbox{curl } F)\cdot {\bf n}\hspace{1pt} dS} =$