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​

I suppose the rectangle's width and length refer to the dimensions of the rectangle that the plane [tex]x+y+z=21[/tex] projects onto the [tex]x,y[/tex] plane, so that [tex]2|x-7|\le2[/tex] and [tex]2|y-7|\le5[/tex], or equivalently [tex]6\le x\le8[/tex] and [tex]\dfrac92\le y\le\dfrac{19}2[/tex].The part of the plane bounded by [tex]C[/tex] can be parameterized by[tex]\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+(21-x-y)\,\vec k[/tex]with the bounds on [tex]x[/tex] and [tex]y[/tex] listed above. Call this surface [tex]S[/tex]. Take the normal vector to [tex]S[/tex] to be[tex]\vec s_x\times\vec s_y=\vec\imath+\vec\jmath+\vec k[/tex]a. Your answer for the curl is correct, [tex]\nabla\times\vec F(x,y,z)=2(\vec\imath+\vec\jmath+\vec k)[/tex].b. By Stokes' theorem,[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex][tex]=\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[/tex][tex]=\displaystyle6\int_{9/2}^{19/2}\int_6^8\mathrm dx\,\mathrm dy=\boxed{60}[/tex]