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

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SUMMARY

The discussion centers on calculating the flux of the vector field F(x,y,z) = zk across a portion of the sphere defined by x² + y² + z² = a² in the first octant. The user initially computed the flux using a double integral over the x-y plane, resulting in an incorrect value of πa³/8, while the correct answer is πa³/6 as stated in the textbook. The user employed the divergence theorem and symmetry to derive the correct integral value, confirming that the divergence of F over the volume yields the expected result. The discussion highlights the importance of correctly applying the divergence theorem and understanding the geometry of the problem.

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cos(e)
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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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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?
 
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
 

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