Answer: To apply the divergence theorem, we need to find the divergence of the vector field F.
∇ · F = ∂/∂x (x − z) + ∂/∂y (y − x) + ∂/∂z (z^2 − y)
= 1 − 0 + 2z
= 2z + 1
Now we need to find the surface integral of F over the closed surface S that bounds the region R.
We can decompose the surface S into two smooth pieces: the top surface S1, given by z = 0, and the curved surface S2, given by z = 16 − x^2 − y^2.
For the top surface S1, the unit normal vector is k, so the surface integral is:
∬S1 F · dS = ∬D F(x, y, 0) · k dA
= ∬D (x − 0)i + (y − x)j + (0^2 − y)k · k dA
= ∬D −y dA
= −∫0^4 ∫0^(2π) r sin θ dθ dr (using polar coordinates)
= 0
For the curved surface S2, we can parameterize it using cylindrical coordinates:
x = r cos θ, y = r sin θ, z = 16 − r^2
The unit normal vector is given by:
n = (∂z/∂r)i + (∂z/∂θ)j − k
= (−2r cos θ)i + (−2r sin θ)j − k
So the surface integral over S2 is:
∬S2 F · dS = ∬D F(x, y, 16 − x^2 − y^2) · ((−2r cos θ)i + (−2r sin θ)j − k) dA
= ∬D [(r cos θ − (16 − r^2))·(−2r cos θ) + (r sin θ − r cos θ)·(−2r sin θ) + (16 − r^2)^2 − (r^2 sin^2 θ − (16 − r^2))] r dr dθ
= ∬D (−16r^3 cos^2 θ − 16r^3 sin^2 θ + 16r^5 − 2r^2 sin^2 θ) r dr dθ
= ∫0^2π ∫0^4 (−16r^3) r dr dθ
= −2048π/3
Therefore, by the divergence theorem:
∬S F · dS = ∭R ∇ · F dV
= ∭R (2z + 1) dV
= ∫0^4 ∫0^(2π) ∫0^(16 − r^2) (2z + 1) r dz dθ dr
= ∫0^4 ∫0^(2π) (16r^2 + 8r) dθ dr
= 512π/3
So the left-hand side and right-hand side of the divergence theorem are equal:
∬S F · dS = ∭R ∇ · F dV
= 512π/3
Therefore, the divergence theorem is verified for the vector field F over the region R.