Question

Let F = (z − y) i + (x − z) j + (y − x) k . Let C be the rectangle of width 2 and length 5 centered at (7, 7, 7) on the plane x + y + z = 21, oriented clockwise when viewed from the origin. (a) Find curlF . curlF = ⟨2,2,2⟩ (b) Use Stokes' Theorem to find F · dr C . F · dr C = −60 √3​

Answer 1

I suppose the rectangle's width and length refer to the dimensions of the rectangle that the plane $x+y+z=21$ projects onto the $x,y$ plane, so that $2|x-7|\le2$ and $2|y-7|\le5$, or equivalently $6\le x\le8$ and $\dfrac92\le y\le\dfrac{19}2$.

The part of the plane bounded by $C$ can be parameterized by

$\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+(21-x-y)\,\vec k$

with the bounds on $x$ and $y$ listed above. Call this surface $S$. Take the normal vector to $S$ to be

$\vec s_x\times\vec s_y=\vec\imath+\vec\jmath+\vec k$

a. Your answer for the curl is correct, $\nabla\times\vec F(x,y,z)=2(\vec\imath+\vec\jmath+\vec k)$.

b. By Stokes' theorem,

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

$=\displaystyle\int_{9/2}^{19/2}\int_6^82(\vec\imath+\vec\jmath+\vec k)\cdot(\vec\imath+\vec\jmath+\vec k)\,\mathrm dx\,\mathrm dy$

$=\displaystyle6\int_{9/2}^{19/2}\int_6^8\mathrm dx\,\mathrm dy=\boxed{60}$