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Minchanka [31]
3 years ago
14

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 · dr
c.
Mathematics
1 answer:
oee [108]3 years ago
3 0
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

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

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

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

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

\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

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

\mathrm d\mathbf S=\mathbf s_u\times\mathbf s_v\,\mathrm du\,\mathrm dv

So our integral is

\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
\displaystyle=12b\int_{v=0}^{v=1}(1-v)\,\mathrm dv=6b
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