Let S be the surface 16=x^2+y^2+z^2. x>=0;y<=0;z<=0. (>=)=greater than or equal to

Let F be the vector field F(x,y,z)=. Calculate the surface area of S and the flux across S (double integral F*ds with bounds S).

Using Cartesian Coordinates, we have z = -√(16 - x^2 - y^2).

So, the surface area equals
∫∫ √(1 + (z_x)^2 + (z_y)^2) dA
= ∫∫ √(1 + (x/√(16 - x^2 - y^2))^2 + (y/√(16 - x^2 - y^2))^2) dA
= ∫∫ √(1 + (x^2 + y^2)/(16 - x^2 - y^2)) dA
= ∫∫ √(16/(16 - x^2 - y^2)) dA
= ∫∫ 4 dA / √(16 - x^2 - y^2).

Since the region of integration is x^2 + y^2 = 16 with x ≥ 0 and y ≤ 0, converting to polar coordinates yields
∫(θ = 3π/2 to 2π) ∫(r = 0 to 4) 4 * (r dr dθ) / √(16 - r^2)
= (π/2) * ∫(r = 0 to 4) 4r(16 - r^2)^(-1/2) dr
= (π/2) * -4(16 - r^2)^(1/2) {for r = 0 to 4}
= 8π.
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As for the flux:
∫∫s F · dS
= ∫∫ F · <-z_x, -z_y, 1> dA, using Cartesian Coordinates
= ∫∫ · dA
= ∫∫ [(x^2 + y^2)/√(16 - x^2 - y^2) - 2√(16 - x^2 - y^2)] dA

Convert to polar coordinates (as above):
∫∫ [(x^2 + y^2)/√(16 - x^2 - y^2) - 2√(16 - x^2 - y^2)] dA
= ∫(θ = 3π/2 to 2π) ∫(r = 0 to 4) [r^2/√(16 - r^2) - 2√(16 - r^2)] * (r dr dθ)
= (π/2) * ∫(r = 0 to 4) [r^2 * r/√(16 - r^2) - 2r√(16 - r^2)] dr
= (π/2) * [∫(r = 0 to 4) r^2 * (r/√(16 - r^2)) dr - ∫(r = 0 to 4) 2r√(16 - r^2) dr]
= (π/2) * [-r^2√(16 - r^2) {for r = 0 to 4} + ∫(r = 0 to 4) 2r√(16 - r^2) dr} - ∫(r = 0 to 4) 2r√(16 - r^2) dr]
by using integration by parts
= 0.

I hope this helps!