Let f = ay i + bz j + cx k where a, b, and c are positive constants. let c be the triangle obtained by tracing out the path from (7, 0, 0) to (7, 0, 2) to (7, 6, 2) to (7, 0, 0). find f · drc.

You can either compute three line integrals, or use Stokes' theorem and compute one surface integral. I prefer the latter.

The curl of the given vector field is

[tex]\nabla\times\mathbf f(x,y,z)=-b\,\mathbf i-c\,\mathbf j-a\,\mathbf k[/tex]

Parameterize the triangular surface bounded by [tex]\mathcal C[/tex] - I'll call it [tex]\mathcal S[/tex] - by

[tex]\mathbf s(u,v)=7\,\mathbf i+6v\,\mathbf j+2(u+v-uv)\,\mathbf k[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. By Stokes' theorem, we have

[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y,z)\cdot\mathrm d\mathbf r=\iint_{\mathcal S}\nabla\times\mathbf f(x,y,z)\cdot\mathrm d\mathbf S[/tex]

where [tex]\mathcal S[/tex] is positively oriented; that is, every vector normal to [tex]\mathcal S[/tex] is pointed in the positive [tex]x[/tex] direction. The surface element is given by

[tex]\mathrm d\mathbf S=\mathbf s_u\times\mathbf s_v\,\mathrm du\,\mathrm dv[/tex]

So our integral is

[tex]\displaystyle\int_{v=0}^{v=1}\int_{u=0}^{u=1}(-b\,\mathbf i-c\,\mathbf j-a\,\mathbf k)\cdot((12v-12)\,\mathbf i)\,\mathrm du\,\mathrm dv[/tex]
[tex]\displaystyle=12b\int_{v=0}^{v=1}(1-v)\,\mathrm dv=6b[/tex]