Question

Q2 (10 points) Use the Green-Stokes formula in order to compute the vector line integral: (z + y') dx + (2x - z) dy + (x- - y) dz where T is the triangle with vertices (1, 0, 0), (0, 1,0) and (0, 0, 2) traversed in the direction implied by the order in which the vertices were given (i.e. from the 1st to the 2nd, to the 3rd and then back to the 1st one).

Answer 1

`\displaystyle\int_C(z+y)\,\mathrm dx+(2x-z)\,\mathrm dy+(x-y)\,\mathrm dz`

has the underlying vector field

`\vec F(x,y,z)=(z+y)\,\vec\imath+(2x-z)\,\vec\jmath+(x-y)\,\vec k`

with curl

`\\abla*\vec F(x,y,z)=\vec k`

Parameterize
`T` by

`\vec s(u,v)=(1-v)((1-u)\,\vec\imath+u\,\vec\jmath)+v(2\,\vec k)`

`\vec s(u,v)=(1-v)(1-u)\,\vec\imath+u(1-v)\,\vec\jmath+2v\,\vec k`

with
`0\le u\le1`,
`0\le v\le1`.

The implied orientation of the curve is counter-clockwise when viewing
`T` from the first octant (each of
`x,y,z` greater than 0), so that the normal vector to
`T` is pointing upward and away from the origin. Take this vector to be

`\vec n=(\partial\vec s)/(\partial u)*(\partial\vec s)/(\partial v)=2(1-v)\,\vec\imath+2(1-v)\,\vec\jmath+(1-v)\,\vec k`

Then by Stokes' theorem, the line integral is equivalent to

`\displaystyle\iint_T(\\abla*\vec F(x,y,z))\cdot\mathrm d\vec S=\iint_T\vec k\cdot\vec n\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_0^1\int_0^1(1-v)\,\mathrm du\,\mathrm dv=\boxed{\frac12}`