Triple Integral for Divergence Theorem

In summary: Does that help?In summary, the conversation discusses finding 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 conversation includes a detailed attempt at solving the problem using both methods, with some confusion around setting up the integral for the Divergence Theorem. The final solution for the divergence integral is found to be 5π.
  • #1

Homework Statement


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)

Homework Equations



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:
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  • #2
checkmatechamp said:

Homework Statement


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)

Homework Equations



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


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
LCKurtz said:
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.

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
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
checkmatechamp said:
So you mean change the limits as follows?

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

Yes. And, for what it's worth, I get ##5\pi## for the triple integral using the divergence of ##3##.
 
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1. What is the purpose of using a triple integral for the Divergence Theorem?

The Divergence Theorem is used to relate the surface integral of a vector field to the volume integral of its divergence over a three-dimensional region. The triple integral is used to calculate the volume integral, allowing us to evaluate the surface integral using a simpler and more efficient method.

2. How is a triple integral for the Divergence Theorem set up?

A triple integral for the Divergence Theorem is set up as the volume integral of the divergence of a vector field over a three-dimensional region. It is typically written as ∭F · dV, where F is the vector field and dV represents an infinitesimal volume element.

3. What is the relationship between the triple integral and the divergence of a vector field?

The triple integral for the Divergence Theorem allows us to calculate the volume integral of the divergence of a vector field. This relationship is important because it helps us to understand the behavior of the vector field within a three-dimensional region.

4. What are the key concepts to understand when using a triple integral for the Divergence Theorem?

Some key concepts to understand when using a triple integral for the Divergence Theorem include the concept of flux, which is the flow of a vector field through a surface, and the concept of divergence, which represents the net flow of a vector field through an infinitesimal volume element. It is also important to understand how to set up and evaluate a triple integral.

5. What are some real-world applications of the triple integral for the Divergence Theorem?

The triple integral for the Divergence Theorem has many real-world applications, particularly in the fields of physics and engineering. It is commonly used in fluid mechanics to calculate the flow of fluids through a given region, and in electromagnetism to calculate the electric and magnetic fields within a three-dimensional region. It is also used in other areas such as heat transfer, aerodynamics, and geophysics.

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