Consider the vector field vec$(F)(x,y,z)=($:$-xcos(xz),-z{e}^{xz}+1,zcos(xz)-1$:$).$ Suppose we wanted to find the flux of this vector field upward across $S-$ the surface of the unit hemisphere $z=\sqrt[\phantom{2}]{1-x^2-y^2}.$ Note that this vector field is horrible $-$ we do not want to find the flux with a surface integral.

$($a$)$ Let vec$(A)(x,y,z)=($:$y+z,sin(xz),e^{xz}$:$).$ Show directly $($by computing the curl of vec$(A))$ that grad$$vec$(A)=$vec$(F).$

$($b$)$ Assuming the previous part is true, note that since grad$$vec$(A)=$vec$(F),$ the flux of vec$(F)$ upward across $S$ is equal to the surface integral ${}_S(grad$vec$(A))*hat(n)dS.$ Use Stokes' Theorem to calculate this integral and find the flux of vec$(F)$ upward across $S.$ Hint: You may find the trig identity $si{n}^2()=\frac{1}{2}(1-cos(2))$ useful in integrating here.