# Flux over sphere between 2 planes

• sandy.bridge
In summary, the conversation is about finding the flux over a part of a sphere with a given radius and between two planes, using polar coordinates and splitting the sphere into two parts. The participants are discussing the limits of integration and the correct answer, which involves taking the sine of the total flux. They also mention the possibility of an error in the given answer choices.
sandy.bridge

## Homework Statement

$\vec{F}=(4x, 3y, 0)$
Find the flux over the part of the sphere centered at (0, 0, 0) with radius 4 between the planes z=-1.19 and z=0.87.

## The Attempt at a Solution

$\hat{N}=(x, y, z)/4$
$\int\int_{S}(4x, 3y, 0)\bullet([(x, y, z)/4])dxdy$
$0.25\int\int_{S}(4x^2+3y^2)dxdy$

The issue I am having is the limits of integration. I converted to polar but got a bit confused because there was three values for r due to the locations of the planes that intersect the sphere. Any recommendations?

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sandy.bridge said:

## Homework Statement

$\vec{F}=(4x, 3y, 0)$
Find the flux over the part of the sphere centered at (0, 0, 0) with radius 4 between the planes z=-1.19 and z=0.87.

## The Attempt at a Solution

$\hat{N}=(x, y, z)/4$
$\int\int_{S}(4x, 3y, 0)\bullet([(x, y, z)/4])dxdy$
$0.25\int\int_{S}(4x^2+3y^2)dxdy$

The issue I am having is the limits of integration. I converted to polar but got a bit confused because there was three values for r due to the locations of the planes that intersect the sphere. Any recommendations?

Split the sphere into two parts, one with z>=0 and another z<0. Integrate over the two and add the result. Now you only have two limits for r in each half sphere.

Another quick question: does the lower limit coincide with r at the origin, and the upper limit would be r at the intersection of the plane?

sandy.bridge said:
Another quick question: does the lower limit coincide with r at the origin, and the upper limit would be r at the intersection of the plane?

The flux should be positive in each half sphere, right? I would say in each case the lower limit should be intersection with the plane and the upper limit should be 'at the origin' i.e. r=4.

Alright, yeah that was what I was doing. I must be messing up somewhere because I am not getting the right result.

After converting to polar I get:
$$0.25\int\int_{D}(4r^3cos^2\theta+3r^3sin^2\theta)drd\theta$$
which after evaluating theta from 0 -> 2pi I get
$$0.25\pi\int(4r^3+3r^3)dr$$
and the limits for r are determined via
$$r=\sqrt(4-z^2)$$

Is this alright so far?

sandy.bridge said:
Alright, yeah that was what I was doing. I must be messing up somewhere because I am not getting the right result.

After converting to polar I get:
$$0.25\int\int_{D}(4r^3cos^2\theta+3r^3sin^2\theta)drd\theta$$
which after evaluating theta from 0 -> 2pi I get
$$0.25\pi\int(4r^3+3r^3)dr$$
and the limits for r are determined via
$$r=\sqrt(4-z^2)$$

Is this alright so far?

Seem ok so far.

Alright, when I finish the rest of the problem I don't get the right answer though.
$$7/8\pi(r^4)$$
I then solved for both the values of r at which the planes intersect the sphere, then added the two fluxes together.

sandy.bridge said:
Alright, when I finish the rest of the problem I don't get the right answer though.
$$7/8\pi(r^4)$$
I then solved for both the values of r at which the planes intersect the sphere, then added the two fluxes together.

I get the antiderivative to be 7*pi*r^4/16.

Nvm, got it. Thanks!

sandy.bridge said:
Nvm, got it. Thanks!

Ooopsa. Sorry, I did overlook something. r=sqrt(16-z^2), not sqrt(4-z^2), right?

Yeah, I caught that as well. I am still getting the wrong answer, though.
They both share r=4.
r1=3.819, r2=3.904.
so we have
14*pi*4^4/16 - 7*pi*(3.819^4)/16 - 7*pi*(3.904^4)/16
which evaluates to be not the right answer

sandy.bridge said:
Yeah, I caught that as well. I am still getting the wrong answer, though.
They both share r=4.
r1=3.819, r2=3.904.
so we have
14*pi*4^4/16 - 7*pi*(3.819^4)/16 - 7*pi*(3.904^4)/16
which evaluates to be not the right answer

Hmm. Seems to be the same thing I'm getting. If there's an error we are both making the same one. Do you know what the correct answer is supposed to be?

Dick said:
Hmm. Seems to be the same thing I'm getting. If there's an error we are both making the same one. Do you know what the correct answer is supposed to be?
I have multiple choice answers, and my answer is not one of them.

I'm going to try one more time. The question has z=-1.1912, and z=0.8768 (I simplified it for the question).

Using these values I get r1=3.818513, r2=3.90272
I get the flux to be 92.642, which is not right.

sandy.bridge said:
I have multiple choice answers, and my answer is not one of them.

I'm going to try one more time. The question has z=-1.1912, and z=0.8768 (I simplified it for the question).

Using these values I get r1=3.818513, r2=3.90272
I get the flux to be 92.642, which is not right.

Just out of curiousity, what are the choices?

Alright, so the final answer is attained by taking the sine of the total flux. For example, let F be the flux, then the value of sin( F ) =,
a) -0.489343, b) 0.306737, c) -0.540973, d) -0.528764, e) -0.499915, f) 0.552286, g) 0.775165, h) 0.0326167

sandy.bridge said:
Alright, so the final answer is attained by taking the sine of the total flux. For example, let F be the flux, then the value of sin( F ) =,
a) -0.489343, b) 0.306737, c) -0.540973, d) -0.528764, e) -0.499915, f) 0.552286, g) 0.775165, h) 0.0326167

Nah. Doesn't help. Taking the sin of a flux is a pretty strange thing to do. But it sure makes it hard to guess what they are doing. For one thing, just changing the z values from the approximate ones you have before to the more accurate ones changes sin(F) from -0.7978 to -0.9994.

I agree that it's a weird thing to do... hmmm

sandy.bridge said:
I agree that it's a weird thing to do... hmmm

Yeah. Fluxes of fields (like an electric field) usually have dimensions associated with them. You basically never take the sin of a dimensionful quantity.

Either way, I wonder where there is an error. The department has been using these questions for 3 years now, so I would assume if there was an error in the answers itself that it would be known by now.

sandy.bridge said:
Either way, I wonder where there is an error. The department has been using these questions for 3 years now, so I would assume if there was an error in the answers itself that it would be known by now.

If there is an error, I'm sure not seeing where it is. Lemme ask somebody else to look at this and get a third opinion.

Doh! I see it. You don't want the first integral dxdy. That should be dS, the surface element of the sphere. I've really been blind to that. Sorry! You want dS=sqrt(1+(dz/dx)^2+(dz/dy)^2)*dx*dy. Right?!

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Fwiw, here's what I got:
I shall keep r to mean the radius of the sphere, =4.
Consider a horizontal band at height z. In polar, z=r sin(ϕ), dz = r cos(ϕ) dϕ. The radius of the band is r cos(ϕ).
Since the field has no z component, we can treat the surface of the sphere in this band as being a cylinder of height dz. A surface element of the band, dA, has area r2 cos2(ϕ) dθdϕ and is in the direction (cos(θ), sin(θ), 0).
F = (4x, 3y, 0) = (4cos(θ), 3sin(θ), 0)rcos(ϕ).
F.dA = (4cos2(θ)+3sin2(θ))r3 cos3(ϕ) dθ dϕ = (cos2(θ)+3)r3 cos2(ϕ) dθ dsin(ϕ)
= (1/2)(cos(2θ)+7)r3 (1-sin2(ϕ)) dθ dsin(ϕ)
Integrating
[(1/2)(sin(2θ)/2+7θ)r3 (sin(ϕ) - sin3(ϕ)/3)]
θ = 0 to 2π, sin(ϕ)=z/r:
[(7π)(r2z - z3/3)] = 675.65
sine of that (!) is -.205, still none of the offered answers.

Edit: error in final step. Flux is 710.31, sine being 0.30674

Last edited:
haruspex said:
Fwiw, here's what I got:
I shall keep r to mean the radius of the sphere, =4.
Consider a horizontal band at height z. In polar, z=r sin(ϕ), dz = r cos(ϕ) dϕ. The radius of the band is r cos(ϕ).
Since the field has no z component, we can treat the surface of the sphere in this band as being a cylinder of height dz. A surface element of the band, dA, has area r2 cos2(ϕ) dθdϕ and is in the direction (cos(θ), sin(θ), 0).
F = (4x, 3y, 0) = (4cos(θ), 3sin(θ), 0)rcos(ϕ).
F.dA = (4cos2(θ)+3sin2(θ))r3 cos3(ϕ) dθ dϕ = (cos2(θ)+3)r3 cos2(ϕ) dθ dsin(ϕ)
= (1/2)(cos(2θ)+7)r3 (1-sin2(ϕ)) dθ dsin(ϕ)
Integrating
[(1/2)(sin(2θ)/2+7θ)r3 (sin(ϕ) - sin3(ϕ)/3)]
θ = 0 to 2π, sin(ϕ)=z/r:
[(7π)(r2z - z3/3)] = 675.65
sine of that (!) is -.205, still none of the offered answers.

Edit: error in final step. Flux is 710.31, sine being 0.30674

Thanks for checking! Yes, I'm not quite following everything there, but I'd agree with the final answer. sandy.bridge, check your notes again. You are missing an important component of dS. I'm so ashamed I missed seeing it while just worrying about the r limits.

## What is "Flux over sphere between 2 planes"?

"Flux over sphere between 2 planes" refers to the measurement of the flow of a vector field through a spherical surface located between two parallel planes. This is a calculation used in physics and engineering to understand the flow of energy or particles through a given area.

## How is "Flux over sphere between 2 planes" calculated?

The calculation for "Flux over sphere between 2 planes" involves integrating the dot product of the vector field and the unit normal vector of the surface over the surface of the sphere. This results in a scalar value that represents the amount of flow through the surface.

## What factors affect the value of "Flux over sphere between 2 planes"?

The value of "Flux over sphere between 2 planes" is affected by the strength and direction of the vector field, as well as the size and orientation of the sphere and the distance between the two planes. The value will also be affected by any boundaries or obstacles in the vector field that may alter the flow.

## What is the significance of calculating "Flux over sphere between 2 planes"?

Calculating "Flux over sphere between 2 planes" helps scientists and engineers understand the flow of energy or particles through a specific area. This can be applied to various fields, such as fluid dynamics, electromagnetism, and thermodynamics, to analyze and predict the behavior of systems and processes.

## Are there any real-world applications of "Flux over sphere between 2 planes"?

Yes, "Flux over sphere between 2 planes" has many practical applications. For example, it can be used to calculate the amount of air or water that flows through a specific area, the amount of heat transferred in a system, or the amount of light that passes through a surface. It is also used in engineering design to optimize the flow of fluids or particles in various systems.

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