# Triple Integral for Divergence Theorem

1. Dec 21, 2014

### checkmatechamp

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

3. 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

∫∫∫σ2sin(φ)dσdφdθ

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

∫∫∫rdzdrdθ

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:

∫∫(-8)(-cos(3φ)/6)+((cosφ)/2))

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

-4π√3

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.

Last edited: Dec 21, 2014
2. Dec 21, 2014

### LCKurtz

Since $z=1$ is $\rho\cos\phi = 1$ or $\rho = \sec\phi$, why not integrate in the $\rho$ direction from $\rho$ on the plane to $\rho = 2$ instead of 0 to 2? Then you don't have to subtract the cone.

Also, I haven't, and don't intend to, carefully read your post because it's hard to read and follow.

3. Dec 22, 2014

### checkmatechamp

So you mean change the limits as follows?

θ from 0 to 2π
φ from 0 to π/3
ρ from secφ to 2

4. Dec 22, 2014

### Zondrina

You wish to evaluate:

$$\iint_S \vec F \cdot d \vec S$$

In two ways. Once using the flux integral above and once using the divergence theorem.

In the case of the flux integral:

$$\iint_S \vec F \cdot d \vec S = \iint_S \vec F \cdot \vec n \space dS = \iint_D \vec F(x, y, z(x,y)) \cdot (\vec r_x \times \vec r_y) \space dA$$

Where $\vec r(x,y) = x \hat i + y \hat j + 1 \hat k$ and $z(x,y) = 1$.

You can obtain the $xy$ limits by looking at the sphere and noticing $x^2 + y^2 + z^2 = 4 \Rightarrow x^2 + y^2 = 3$. Switching to polar co-ordinates should finish it off.

In the case of the divergence theorem:

$$\iint_S \vec F \cdot d \vec S = \iiint_E \text{div}(\vec F) \space dV = 3 \iiint_E \space dV$$

Where you have already found the divergence. Now the solid region $E$ is bounded by the plane and the hemisphere, so a switch to spherical co-ordinates would finish off the integral nicely.

5. Dec 22, 2014

### LCKurtz

Yes. And, for what it's worth, I get $5\pi$ for the triple integral using the divergence of $3$.

Last edited: Dec 22, 2014