Let $F(x,y,z) = (y,0,0)$ and $$S = \{ (x,y,z);\,\,{x^2} + {y^2} + {z^2} = 1,\,\,z \ge 0\} $$ Let $\widehat N$ be the unit normal field to $S$ pointing away from the origin. Calculate $$\iint\limits_S {{\text{Curl }}F \cdot \widehat N}\,dS$$
My solution:
Let $$G = {\text{Curl }}F = \nabla \times F = \left| {\begin{array}{*{20}{c}} i&j&k \\ {\frac{d}{{dx}}}&{\frac{d}{{dy}}}&{\frac{d}{{dz}}} \\ y&0&0 \end{array}} \right| = - k$$ $$\iint\limits_S {{\text{Curl }}F \cdot \widehat N}\,dS = \iint\limits_S {G \cdot \widehat N}\,dS$$
$$\widehat N = \frac{{(2x)i + (2y)j + (2z)k}}{{\sqrt {4{x^2} + 4{y^2} + 4{z^2}} }} = (x,y,z)$$
Projecting the $dS$ element onto $xy$ plane:
$$dx\,dy = dS\left| {\widehat N \cdot \overrightarrow k } \right| = \left| {(x,y,z) \cdot (0,0,1)} \right| = z\,dS \Leftrightarrow dS = \frac{1}{z}\,dx\,dy$$
$$G \cdot \widehat N\,dS = (0,0, - 1) \cdot (x,y,z)\,\frac{1}{z} = - 1$$ $$\iint\limits_S {{\text{Curl }}F \cdot \widehat N}\,dS = - 1 \cdot \iint\limits_{{x^2} + {y^2} \le 1} {dx\,dy} = - \pi $$
Am i correct that the answer should be $ - \pi $ and not $ + \pi $?