Find the flux of the field F(x) = <x,y,z> across the hemisphere x^2 + y^2 + z^2 = 4 above the plane z = 1, using both the Divergence Theorem and with flux integrals. (The plane is closing the surface)
The Attempt at a Solution
Obviously, the divergence is 3, so the tricky part is setting up the integral. If it was above the plane z = 0, it would be easy (matter of fact, you could just use the formula V = (2/3)(pi)(r^3) (since it's only half the sphere)), but it's not.
So what I did was say that since z = (rho)*cos(phi), I tried to figure out what the angle phi would be when z = 1. Substituting that in meant that 1 = 2*cos(phi), which means that cos(phi) = 0.5, which means that phi = pi/3
So my integral is as follows:
rho from 0 to 2
phi from 0 to pi/3
θ from 0 to 2pi
Integrating with respect to σ gets me sin(φ)*σ3/3, and punching in my limits for σ gets me sin(φ)*(8/3).
Integrating with respect to φ gets me -cos(φ)*(8/3), and then punching in my limits for φ gets me (8/3)*(-cos(pi/3) - (-cos(0)), which is (8/3)(-0.5-(-1)), which is (0.5)(8/3) = 4/3.
Then integrating with respect to θ gets me (4/3)θ, and then substituting my limits gets me (8/3)*pi.
But then if I'm visualizing it correctly, I'm left with an ice cream cone-shaped object. So to get rid of the bottom of the cone, I substitute 1 in for z:
x^2 + y^2 + 1^2 = 4
x^2 + y^2 = 3
r^2 = 3
So r = sqrt(3)
So then I use cylindrical coordinates to subtract out the cone, using the following integral.
z from 0 to 1
r from 0 to sqrt(3)
θ from 0 to 2pi
Integrating with respect to z gets me rz, and punching in my limits gets me r.
Then integrating with respect to r gets me 0.5r^2, and punching in my limits gets me 1.5.
Then integrating with respect to θ gets me 1.5θ from 0 to 2π, which is 3π.
So then (8/3)π - 3π is (-1/3)π. But if everything is above the x-axis, how can it be negative? (If I multiply by 3 to account for the divergence, that's still -π)
Alright, so doing the flux integral, I have the sphere parametrized as <2sinΦcosθ, 2sinΦsinθ, -2sinΦ>. Taking the cross product of the partials gives me <-4sin2φcosθ, -4sin2Φsinθ, 4sinφcosΦ>
The parametrization of the vector field in spherical coordinates is simply <2sinφcosθ, 2sinφsinθ, 2cosφ>. Taking the dot product of the two gives me -8sinφ(sin2φ-cos2φ), which becomes 8sinφcos(2φ)
Then I have to integrate that over the region in the (φ, θ) domain.
φ goes from 0 to π/3, and θ goes from 0 to 2π
So ∫∫(-8sinφ(sin2φ-cos2φ) dφdθ
Using a trig-sub:
Integrate with respect to φ first.
-8(-sin(3φ)/18) + -8(sin(φ)/2)
(4/9)(sin(3φ)) - 4sinφ
((4/9)(0) - 2*sqrt(3) - 0) - 0
-2sqrt(3) integrate with respect to φ from 0 to 2π
And then you just take the area of the disk on the bottom.
r goes from 0 to sqrt(3), and θ goes from 0 to pπ
∫∫ r dr dθ
0.5r2 from 0 to sqrt(3) is just 1.5
Then basically multiply by 2π gives 3π
So it's 3π - 4π√3
But I'm getting two different answers for solving the flux integral vs. the divergence integral.