Gauss's theorem to find flux of field

[schmiggy]

Homework Statement

F(x,y,z) = x^3zi + y^3zj + xyk
across the surface x^2 + y^2 + z^2 = 9

Homework Equations

See attached image

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

 

[dx]

You did the last integral wrong.

Use the substitution u = sin φ

In terms of u the integral becomes

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

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

Yes you can pay and become a Gold member. No ads, recognition graphic, set invisible, custom title, signature, avatars, profile photo, 200pm limit, who's online, user notes, boolean searching...

There is an "Upgrade" button at the top.

 

[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? )

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

there used to be an upgrade button, but now you have to click "My PF", then go down to "Paid subscriptions" near the bottom

 

[dx]

hi schmiggy!

your 3x²z + 3y²z was right,

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

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

 

[tiny-tim]

oops! i was thinking of cylindrical coordinates!

thanks, dx ​

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

 

[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!

 

[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

 

[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