Chapter 15 - Section 15.7 - Stokes' Theorem - Exercises - Page 895: 4

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Applying Stoke's Theorem, we have $\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$ Here, we have $r(x,y)=xi+yj+(1-x-y)k$ $r_x= i-k$ and $r_y=j-k$ Thus, $|r_x \times r_y|=i+j+k$ Now, $curl F ds=\int_0^{1} \int_0^{1-x} (2y-2z+2z-2x+2x-2y) (\dfrac{1}{\sqrt 3}) dy dx $ or, $\int_0^{1} \int_0^{1-x} (0) dy dx=0$