# Calculating Flux: Homework Statement

• oteggis
In summary,-On the top surface of the cylinder z = 5, ##\hat n##= ##\hat k##-Using Cylindrical coordinates,-DivF = ∇.F = -3(x^2+y^2-2z)-IintF.\hat n\mathrm dS = -3\iint_{S} F.\hat n\mathrm dS-IintF.\hat n\mathrm dS = 0-Since z=0, then-IintF.\hat n\mathrm dS = -3\iint_{V} F.\hat n\
oteggis

## Homework Statement

Let S be the surface of a solid R , which lies inside the cylinder:
##x^2+y^2=16##
and between the plane

where x=0 and z=5

There is also defined a vector field F by:
##\begin{align}F(x,y)=(-x^3i-y^3j+3z^2k)\end{align}##

(a) Calculate : $$\iint_{T} F.\hat n\mathrm dS$$with T = {##(x,y,5)∈ℝ | x^2+y^2≤16##}(b) Calculate DivF and $$\iint_{S} F.\hat n\mathrm dS$$

with n the outward pointing unit normal.

(c) Calculate: $$\iint_{V} F.\hat n\mathrm dS$$

with V = {##(x,y,z)∈ℝ |x^2+y^2≤16 and 0≤z≤5##} and the unit normal ##\hat n\ ##points out of the solid R

## Homework Equations

$$\iint_{V} F.\hat n\mathrm dS$$

## The Attempt at a Solution

(a) On the top surface of the Cylinder z = 5, ##\hat n##= ##\hat k##
##F.\hat n = [-x^3i +-y^3j +3z^2k].[k]=3z^2 ##
##\iint_{S} F.\hat n \mathrm dS = \iint_{S} 3z^2\mathrm dS##
##\iint_{S} F.\hat n \mathrm dS = \iint_{S} 3(5^2)\mathrm dS##
##\iint_{S} F.\hat n \mathrm dS = 75\iint_{S}\mathrm dS##
The area enclosed by the circle is ##16\pi##
since the radius of the circle is 4.
Therefore
##\iint_{S} F.\hat n \mathrm dS = 75(16\pi) = 1200\pi ##(b) ##DivF = \nabla.F = [i\frac{\partial }{\partial x}+ j\frac{\partial }{\partial y} +k\frac{\partial }{\partial z}].[-x^3i +-y^3j +3z^2k]= -3x^2-3y^2 +6z##
DivF = ∇.F = -3(x^2+y^2-2z)
##\iint_{S} F.\hat n \mathrm dS =\iiint_{V} \operatorname{div} F dV -\iint_{S_1} F.\hat n \mathrm dS##
## \iint_{S_1} F.\hat n \mathrm dS = 0 ## since z=0, then
## \iint_{S} F.\hat n \mathrm dS = \iiint_{V} \operatorname{div} F dV ##**How do I get ##\hat n## in this case?**

(c) From my understanding, I have to use Divergence Theorem here
$$\iint_{V} F.\hat n\mathrm dS = \iiint_{V} \nabla.F \mathrm dV$$
$$\iint_{V} F.\hat n\mathrm dS = \iiint_{V} -3(x^2+y^2-2z) \mathrm dV$$
$$\iint_{V} F.\hat n\mathrm dS = -3\iiint_{V} (x^2+y^2-2z) \mathrm dV$$

Using Cylindrical coordinates
$$\iint_{V} F.\hat n\mathrm dS = -3\iiint_{V} [(r^2\cos^2\theta+r^2\sin^2\theta-2z)] \mathrm rdzdrd\theta$$
$$\iint_{V} F.\hat n\mathrm dS = -3\int_0^{2\pi} \int_0^4 \int_0^5 [(r^2\cos^2\theta+r^2\sin^2\theta-2z)] \mathrm rdzdrd\theta$$
$$\iint_{V} F.\hat n\mathrm dS = 432\pi$$

Last edited:
Hello oteggis,

Could you read what you posted and fix the typesetting ? for in-line math you want to use ## ... ## delimiters and for displayed math $$...$$

I take it the solid ##R## completely fills the cylinder ? ('lies inside' can apply to a very tiny fraction of the space). And not ##x=0## but ##z=0## ?

On the curved cylinder surface, $$\hat{n}=\hat{r}=\frac{x\hat{i}+y\hat{j}}{4}$$

Chestermiller said:
On the curved cylinder surface, $$\hat{n}=\hat{r}=\frac{x\hat{i}+y\hat{j}}{4}$$
so in Polar coordinates: ##\hat n## = ##cos\theta i + sin\theta j##
But I got ##\hat n## = ##-cos\theta i-sin\theta j## instead

oteggis said:
so in Polar coordinates: ##\hat n## = ##cos\theta i + sin\theta j##
But I got ##\hat n## = ##-cos\theta i-sin\theta j## instead
That's the inward-directed normal, not the outward normal required by the divergence theorem.

## 1. What is flux and why is it important?

Flux is a measure of the flow of a physical quantity through a given surface. It is important in many scientific fields, including physics, engineering, and mathematics, as it allows us to understand and quantify how much of a quantity is passing through a specific surface.

## 2. How is flux calculated?

Flux is calculated by taking the dot product of the vector field and the surface normal vector. This calculation involves taking the integral of the vector field over the surface, which can be done using various methods such as the divergence theorem or Stokes' theorem.

## 3. What units is flux measured in?

The units of flux depend on the specific physical quantity being measured. For example, electric flux is measured in volts, while magnetic flux is measured in webers. It is important to pay attention to the units when calculating and interpreting flux.

## 4. How is flux related to Gauss's Law?

Gauss's Law states that the total flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. This relationship allows us to use flux to calculate the electric field produced by a charge distribution.

## 5. What are some real-world applications of flux calculations?

Flux calculations have a wide range of applications in various fields. In physics, they are used to understand and predict the behavior of electric and magnetic fields. In engineering, they are used in designing circuits and determining heat transfer. They are also used in meteorology to study the flow of air and water currents.

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