The divergence theorem states that `int_S (F*hatn) dS = int_V (grad*F) dV,` where `S` is a closed surface, `V` is the volume inside it and `F` is a good enough vector field defined inside `S` and on it. The symbol `grad` d the divergence operator.
In our case `S` is an ellipsoid, a smooth closed surface, the vector field `F` is defined everywhere and is infinitely differentiable. The divergence of `F` is
`(del/(del x), del/(del...
The divergence theorem states that `int_S (F*hatn) dS = int_V (grad*F) dV,` where `S` is a closed surface, `V` is the volume inside it and `F` is a good enough vector field defined inside `S` and on it. The symbol `grad` d the divergence operator.
In our case `S` is an ellipsoid, a smooth closed surface, the vector field `F` is defined everywhere and is infinitely differentiable. The divergence of `F` is
`(del/(del x), del/(del y), del/(del z))*(F_x, F_y, F_z) = del/(del x) F_x + del/(del y) F_y + del/(del z) F_z`
where `F_x, F_y, F_z` are the components of `F.`
In our case `F_x=9x, F_y=y*cosh^2(x), F_z = z*sinh^2(x),` thus
`grad F = 9 + cosh^2(x) - sinh^2(x) = 9` and the integral becomes a very simple one:
`int_S (F*hatn) dS = int_V (grad*F) dV = int_V (9) dV = 9*vol(V) = 12 pi abc.`
The equation of our ellipsoid is equivalent to `x^2/3^2 + y^2/6^2 + z^2/2^2 = 1,` therefore the semiaxes are `3,` `6` and `2` and the final answer is `12 pi abc = 12*3*6*2 pi = 432 pi.`
I used WolframAlpha to sketch the ellipsoid.
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