through the hemisphere 
SOLUTION:
Hard Way:
The unit normal vector to the sphere is 
,
where 
in this case. Now, let’s
change over to spherical coordinates:
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Our unit normal vector now becomes 
.
Now we apply the rule for a general vector field 
in 
its surface integral through
S is the scalar surface integral over S of the function 
So this means:
We now multiply this by the element of surface area on the sphere (dS),
which is 
and integrate over
. Therefore, the flux is
We use the trigonometric identity to convert to:
Even Harder Way:
We do a full-scale parametrization of this problem as in Example 2 in Sec.
7.6. Again we change to cylindrical coordinates as above. We see that the
flux integral can be found through this equation:
.
Each term is calculated like so: 
and everything else follows from that.
Therefore,
We substitute this into our equation:
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We end up with the same integrand as before which checks
out and we end up with the same answer of 
.