Flux Integral Help through abnormal cone

In summary, the problem requires the calculation of the flux of a given vector field out of a closed cone, which is defined as x= sqrt(y^2 + z^2) with x between 0 and 1. The student is struggling with approaching the problem because most problems they have done involve a surface given by z=f(x,y) rather than x=f(y,z). They are seeking help in modifying and applying the given formula for flux through a surface to this scenario. A suggestion is given to swap x and z and parametrize the cone using polar coordinates in the yz-plane. The vector differential of surface area and the vector field in this parameterization are also provided for an easy integral. The reminder to integrate over the base
  • #1
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1. The problem statement

The problem requires me to calculate the flux of F=x^2 i + z j + y k out of the closed cone, x=sqrt(y^2 + z^2) with x between 0 and 1.

I am having trouble approaching this problem because most of the problems I have done give the curve as z=f(x,y) instead of x=f(y,z) and I am therefore confused as to how to apply the below equation.

Homework Equations



For the flux through a surface given by z=f(x,y)

Flux = int(F . dA) = int( [ F(x,y, f(x,y)) dot (-df/dx i - df/dy j + k) ]dxdy

where df/dx is the partial derivative of f with respect to x and df/dy is the partial derivative of f with respect to y.

How can I modify/apply this formula (if I even can) when given a surface as a function of x=f(y,z) as opposed to z=f(x,y) to find the flux through the horizontally opening cone?

Any help would be greatly appreciated! Thanks so much.


The Attempt at a Solution



I tried putting f in terms of z and going that route but ran into some nasty integrals.

I tried replacing z with x and x with z (for F and f) as to simulate the same vector field and cone in a way that better applied to the given formula but once again ran into some nasty integrals.

Suggestions?
 
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  • #2
If you have x= f(y,z) rather than z= f(x,y), then just swap x and z!
With x= f(y,z), x- f(y,z)= 0 can be treated as a level curve for a function whose gradient is then normal to the surface. That is, [itex](\vec{i}- \partial f/\partial y \vec{j}- \partial f/\partial z\vec{k})dydz[/itex] is the vector differential of surface area.

You can parametrize [itex]x= \sqrt{y^2+ z^2}[/itex] by taking [itex]y= r cos(\theta), z= r sin(\theta)[/itex]- in other words, use polar coordinates but in the yz-plane rather than the xy-plane. The equation of the cone becomes x= r so the position vector for any point on the plane is [itex]\vec{r}= r\vec{i}+ r cos(\theta)\vec{j}+ r sin(\theta)\vec{k}[/itex].

Then [itex]\vec{r}_r= \vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}[/itex], [itex]\vec{r}_\theta= -r sin(\theta)\vec{j}+ r cos(\theta)\vec{k}[/itex] and the "fundamental vector product, their cross product, is [itex]r\vec{i}- r cos(\theta)\vec{j}- r sin(\theta)\vec{k}[/itex]. Since you said the "flux out of the closed cone", you want this oriented by the outward pointing normals and so the vector differential of surface area is [itex]d\vec{\sigma}= (-\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k})r dr d\theta[/itex].

In this parameterization, [itex]F(x,y,z)= x^2\vec{i}+ z\vec{j}+ y\vec{k}= + r sin(\theta)\vec{j}+ r cos(\theta)\vec{k}[/itex]. I think that gives a very simple integral.

Oh, and since this is a "closed cone", don't forget to integrate over the x= 1 base.
 

1. What is flux integral?

Flux integral is a mathematical concept used in physics and engineering to calculate the flow of a vector field through a surface. It is represented as the surface integral of the dot product of the vector field and the surface's normal vector.

2. How is flux integral related to abnormal cone?

The concept of abnormal cone is used in the calculation of flux integral when the surface is not smooth. It refers to the collection of all vectors perpendicular to the tangents of the surface at a given point. This helps in determining the direction of the normal vector and thus, the direction of the flow.

3. What is the formula for calculating flux integral through abnormal cone?

The formula for calculating flux integral through abnormal cone is given by Flux = ∫∫S F · n dS, where F is the vector field, n is the unit normal vector to the surface, and dS is the differential surface area element.

4. Can flux integral be negative?

Yes, flux integral can be negative. This indicates that the flow of the vector field is in the opposite direction of the normal vector of the surface. It is important to pay attention to the sign of the flux integral to accurately determine the direction of the flow.

5. How is flux integral used in real-world applications?

Flux integral has various real-world applications, such as calculating the flow of fluids through a surface in fluid mechanics, determining the electric field flux through a surface in electromagnetism, and evaluating heat transfer through a surface in thermodynamics. It is also used in fields like geology, meteorology, and oceanography to study fluid dynamics.

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