Question

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = i + (x + yz)j + (xy − sqrt(z )k, C is the boundary of the part of the plane 7x + 6y + z = 1 in the first octant.

Answer 1

By Stokes' theorem, the line integral of $\vec F$ over $C$ is equivalent to the surface integral of the curl of $\vec F$ over $S$, where $S$ is the part of the plane $7x+6y+z=1$ in the first octant, with $S$ having positive/upward orientation.

Parameterize $S$ by

$\vec s(u,v)=\dfrac{(1-u)(1-v)}7\,\vec\imath+\dfrac{u(1-v)}6\,\vec\jmath+v\,\vec k$

with $0\le u\le1$ and $0\le v\le1$.

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=\dfrac{1-v}6\,\vec\imath+\dfrac{1-v}7\,\vec\jmath+\dfrac{1-v}{42}\,\vec k$

The curl of $\vec F$ is

$\nabla\times\vec F(x,y,z)=(x-y)\,\vec\imath-y\,\vec\jmath+\vec k$

Then the line integral is equivalent to

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^1\int_0^1\left(\frac{(6-13u)(1-v)}{42}\,\vec\imath-\dfrac{u(1-v)}6\,\vec\jmath+\vec k\right)\cdot\left(\dfrac{1-v}6\,\vec\imath+\dfrac{1-v}7\,\vec\jmath+\dfrac{1-v}{42}\,\vec k\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\frac1{252}\int_0^1\int_0^1(12-6v-19u+19uv)(1-v)\,\mathrm du\,\mathrm dv=\boxed{\frac{11}{1512}}$