Verified answer

Note that the normal to S is given by...

Since z = ±√(1 - x^2 - y^2), we have normal vector <-z_x, -z_y, 1>

= <±x/√(1 - x^2 - y^2), ±y/√(1 - x^2 - y^2), 1>.

= <x/z, y/z, 1>.

This has length √((x/z)^2 + (y/z)^2 + 1) = √((x^2 + y^2 + z^2)/z^2) = 1/z, since we are on the unit sphere. So, a unit vector in the direction of n is <x, y, z>.

So, ∫∫s (9x + 3y + z^2) dS

= ∫∫s <9, 3, z> · <x, y, z> dS

= ∫∫s <9, 3, z> · dS (with dS now as a vector)

Now, we can apply the Divergence Theorem, obtaining

∫∫∫ div<9, 3, z> dV

= ∫∫∫ 1 dV

= 4π/3, the volume of the unit sphere.

I hope this helps!

The progression is going as such (from my information): Pre-Calculus, Calc AB (Calc I) Calc BC (Calc II) Multi-variable Calc (Calc III) ...then, Vector diagnosis or some thing comparable to that.