# Homework Help: Gauss's theorem to find flux of field

1. Feb 3, 2013

### schmiggy

1. The problem statement, all variables and given/known data
F(x,y,z) = x^3zi + y^3zj + xyk
across the surface x^2 + y^2 + z^2 = 9

2. Relevant equations
See attached image

3. The attempt at a solution
See attached image.

I'm fairly certain there's either something wrong with my initial integral, also I don't think I integrated correctly with respect to phi.. but I'm not really sure how to fix it.

According to my exercise book solutions, the answer is 0, however if I were to continue with my working out I would get 729pi.

Any help would be greatly appreciated, thank you.

On an unrelated note, this forum has been a great help to me in the past, and all free of charge.. is there somewhere I can make a donation? or a paid service I can sign up for? Thanks

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Last edited: Feb 3, 2013
2. Feb 3, 2013

### dx

You did the last integral wrong.

Use the substitution u = sin φ

In terms of u the integral becomes

$$\int_0^\pi \sin^3\phi \cos \phi \ d\phi = \int_0^0 u^3 du = 0$$

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Last edited: Feb 3, 2013
3. Feb 3, 2013

### tiny-tim

hi schmiggy!

your 3x2z + 3y2z was right,

but then you only converted the 3x2z into polar coordinates!

(btw, that's 3r2z, which is odd in z, so isn't it obvious that its integral over that sphere will be zero? )
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4. Feb 3, 2013

### dx

No tiny-tim, I think he did it correctly.

3x2z + 3y2z = 3rcosφ(x2 + y2) = 3rcosφ(r2 - z2) = 3rcosφ(r2 - r2cos2φ) = 3r3cosφsin2φ

which is what schmiggy got

5. Feb 3, 2013

### tiny-tim

oops! i was thinking of cylindrical coordinates!

thanks, dx

(but my comment on "odd in z" still stands )

6. Feb 4, 2013

### schmiggy

Firstly, thanks to both of you for the help!

I got the integral to result in
$$\int_0^\pi \sin^4\phi/4 = 0$$
(sorry, not sure how to use that code to format the integral properly, basically it's sin^4(phi) multiplied by (1/4))

I embarrassingly sat for about 2 minutes afterwards going.. "but when I integrate with respect to theta, I'll end up with 2pi..." before realising I was integrating 0. /facepalm

Anyway, thanks again!

Last edited: Feb 4, 2013
7. Feb 4, 2013

### dx

Hi schmiggy,

I think there's something wrong either with the latex or quoting. I was talking about the integral over phi at the end

∫ sin3φ cosφ dφ (from 0 to π)

which can be performed with the substitution u = sin φ to give

(1/4)sin4π - (1/4)sin40 = 0

8. Feb 4, 2013

### schmiggy

I accidentally edited your quote when I was playing around with the latex seeing how it works, my bad! I also accidentally left the d(phi) at the end. I've edited my previous post and that should be correct.. I hope.. or else I'll have a serious case of egg on my face