Use Stokes' Thm to evaluate. F(x,y,z)=(x^2)(e^yz) i +(y^2)(e^xz) j +(z^2)(e^xy) k S is the hemisphere x^2 +y^2 +z^2 =4 z is greater than or equal to 0 oriented upward.
Do you want to compute the flux of the field through the hemisphere?
Take cross product of F and then take the double integral
The flux is equal to the work done by the field over the boundary.
take cross product of F That will give you a new vector Call it V2 then use this equation to integrate
It is very easy to see that this flux is zero. The field restricted to the plane z = 0 becomes \[F(x,y)=\{x^2 , y^2, 0\} \] This field is conservative since it is the gradient of \[ f(x,y)= \frac {x^3} 3 + \frac {y^3}3 \] The work done by this field over a closed path as \[x^2 + y^2=4 \] is zero.
Of course in the above, we used Stokes's Theorem to conclude the the flux thru the hemisphere is equal to the work done by the field on the circle of center 0 and radius 2 in the xy-plane.
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