Answer:
[tex]4\sqrt 3/9[/tex]
Explanation:
As it is given that F is radial, by definition, some function A(x, y) can be rewritten as follow:
[tex]F(x, y) = <x, y, z> * A(|<x, y, z>|) = <x, y> * A(\sqrt{x^2 + y^2})[/tex]
[tex]|F(x, y, z)|[/tex] which is proportional to [tex]1/(x^2 + y^2 + z^2)[/tex] is given, so that for random constant [tex]t[/tex]:
[tex]F(x, y) = t<x, y, z> / (x^2 + y^2 + z^2)^{3/2}[/tex]
Where the magnitude of [tex]<x, y, z>[/tex] is [tex]\sqrt{x^2 + y^2 + z^2}[/tex].
Hence,
[tex]|t<x, y, z>/(x^2 + y^2 + z^2)^{3/2}| = t\sqrt {x^2 + y^2 + z^2}/(x^2 + y^2 + z^2)^{3/2} =\\= t/(x^2 + y^2 + z^2)[/tex]
The sphere of radius 8 and 16 can be parametrized by spherical coordinates: as follow:
[tex]i) s_8(a, b) = <8cos(a)sin(b), 8sin(a)sin(b), 8cos(b)>\\ii) s_{16}(a, b) = <16cos(a)sin(b), 16sin(a)sin(b), 16cos(b)>[/tex]
For a in [tex][0, 2\pi][/tex] and b in [tex][0, 2\pi][/tex].
From above, it is seen that [tex]s_{16}(a, b) = 2s_8(a, b)[/tex], thus the partial derivative and normal vectors of [tex]s_{16}(a, b)[/tex] equals to the twice of [tex]2s_8(a, b)[/tex]'s.
Moreover, x, y and z components of [tex]s_{16}(a, b)[/tex] equals to the twice of s_8(a, b)'s, so doubling x, y, and z will rise [tex]x/(x^2 + y^2 + z^2)^{3/2}, y/(x^2 + y^2 + z^2)^{3/2},[/tex] and [tex]z/(x^2 + y^2 + z^2)^{3/2}[/tex] by a factor of [tex]2/(2^2 + 2^2 + 2^2)^{3/2} = \sqrt 3/36[/tex].
Thus,
[tex]F[s_{16}(a, b)][/tex] . (normal of [tex]s_{16}(a, b)) = (\sqrt3/36)F[s_8(a, b)][/tex] . (2 * normal of [tex]s_8(a, b)) = \sqrt 3/18 . F[s_8(a, b)] [/tex] . (normal of [tex]s_8(a, b))[/tex]
As a result,
[tex]\int\int S_2 (F . n) dS = (\sqrt3/18)\int\int S_1 (F . n) dS = (\sqrt3/18)*(8) = 4\sqrt 3/9[/tex]
