This is the same result as the back of the book. So your solution is correct.

In summary, the conversation discusses finding the flux of a given vector field across a portion of a sphere in the first octant directed away from the origin. The attempt at a solution involves using the formula for flux and solving for the integral, with some algebra and use of polar coordinates. However, there may be a simple error in the calculations as the final answer does not match the solution in the book. The conversation also mentions using the divergence theorem to find the integral over the entire surface of the sphere, which results in the correct answer of pi*a^3/6.
  • #1
cos(e)
27
0

Homework Statement


find the flux of the field F(vector) across the portion of the sphere x^2+y^2+z^2=a^2 in the first octant directed away from the origin



Homework Equations


F(x,y,z)=zk(hat)



The Attempt at a Solution


i used Flux=double integral over x-y plane F.n(unit normal)dsigma
where dsigma=abs(grad(g))/abs(grad(g).p(unit normal to surface))
i let g=x^2+y^2+z^2-a^2=0
and grad(g)=(2x,2y,2z)
p=n=(x,y,z)/sqrt(x^2+y^2+z^2)

and after some algebra i got:
double integral( (a^2-x^2-y^2)/a dA)
which i used polar co-ordinates
to get 1/a double integral( a^2-r^2)r dr dtheta
which gave me pi*a^3/8 yet the solutions in back of book is pi*a^3/6

Looks like iv made a simple error somewhere(iv looked like 3 times but can't find one), but i just need to check I am using the right method.
sorry bout the way i worte out the maths but I am unaware how to put it in nice maths form
Can any1 help please?

cheers,
cos(e)
 
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  • #2
ok i get the answer with p=k(hat)
now I am sure why I am letting p=k(hat), is it because p is the normal to the area of the double intergral i take, in this case the x-y plane with theta varying from 0 to pi/2 adn r varying from 0 to a?
 
  • #3
If we call the integral I, then by symmetry, the integral over the entire surface of the sphere is 8 I. By the divergence theorem, we have:

8 I = Integral of div F over volume = 4/3 pi a^3 -------->

I = pi a^3/6
 

1. What is an easy flux surface integral?

An easy flux surface integral is a mathematical calculation that measures the amount of a vector field that passes through a given surface. It is often used in physics and engineering to calculate the flow of a fluid or the strength of an electric field.

2. How is an easy flux surface integral different from a regular flux surface integral?

An easy flux surface integral is a simplified version of a regular flux surface integral that is used when the vector field is constant and the surface is a simple shape, such as a rectangle or circle. In these cases, the integral can be calculated easily without using more complex mathematical methods.

3. What are some real-world applications of easy flux surface integrals?

Easy flux surface integrals have many practical applications, such as calculating the flow of air through a ventilation system, determining the strength of an electric field around a charged object, or measuring the amount of heat transfer in a thermal system.

4. How is an easy flux surface integral calculated?

To calculate an easy flux surface integral, you first need to determine the vector field and the surface you want to measure. Then, you can use simple formulas to calculate the integral, such as multiplying the magnitude of the vector field by the area of the surface. If the surface is curved, you may need to use more advanced methods, such as parametric equations.

5. What are some common mistakes when using easy flux surface integrals?

One common mistake when using easy flux surface integrals is forgetting to account for the direction of the vector field. It is important to make sure the vector field and the surface are aligned properly to get an accurate calculation. Another mistake is using the wrong formula for a non-rectangular surface, which can lead to incorrect results. It is also important to check the units of measurement to ensure consistency throughout the calculation.

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