Yes, I'd use the Divergence Theorem, obtaining

∫∫∫ (z + 0 + 2z) dV = ∫∫∫ 3z dV.

The region of integration is best expressed in cylindrical coordinates:

∫(θ = 0 to 2π) ∫(r = 0 to 2) ∫(z = 0 to 4 - r²) 3z * (r dz dr dθ)

= 2π ∫(r = 0 to 2) (3/2)rz² {for z = 0 to 4 - r²} dr

= 3π ∫(r = 0 to 2) r(4 - r²)² dr

= 3π ∫(w = 4 to 0) w² dw/(-2), letting w = 4 - r², dw = -2r dr

= (3π/2) ∫(w = 0 to 4) w² dw

= (π/2)w³ {for w = 0 to 4}

= 32π.

I hope this helps!

# Please help with Calc III, Divergence Theorem question...will award best answer. "Use the Divergence Theorem to calculate the net outward...?

Use the Divergence Theorem to calculate the net outward of the vector field F = <xz, z, z^2> across the surface S that is the union of the parabaloid z = 4 - x^2 - y^2 for z>= 0 and the disk x^2 + y^2 <= 4 on the xy plane.

So the divergence theorem is ∬(over S) F dot n dS = ∭(over D) ∇ dot F dV where n is the outward unit normal vector on S.

if i use the "volume * the ∇F form...my ∇F = < z, 0, 2z>...but then how would i evaluate that over the volume?

Would it be more convenient to use the F dot n form? ...how would you then find the normal vector?

Thanks for any help...I will award best answer and really appreciate the time

So the divergence theorem is ∬(over S) F dot n dS = ∭(over D) ∇ dot F dV where n is the outward unit normal vector on S.

if i use the "volume * the ∇F form...my ∇F = < z, 0, 2z>...but then how would i evaluate that over the volume?

Would it be more convenient to use the F dot n form? ...how would you then find the normal vector?

Thanks for any help...I will award best answer and really appreciate the time

## 1 Answers

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