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Find the flux of F over Q where Q is bounded?

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1 Answer

We can apply the divergence theorem, ∫∂Q F·n dA = ∫...∫Q div(F) dV (left hand integral is over the entire boundary ∂Q) as follows.

Here you mention Q as bounded by z = √(x+ y2) and z = √(2 - x2 - y2)

Now as you calculated, div(F) = (∂/∂x)x2 + (∂/∂y) (z2 - x) + (∂/∂z) y= 2x

So by the divergence theorem the answer equals  ∫∫∫Q 2x dx dy dz

Before we evaluate, note, it is convenient to convert to cylindrical coordinates.

Note that the conversion to cylindric coordinates is

∫∫∫ f(x,y,z) dx dy dz = ∫∫∫ r * f(r cos θ, r sin θ, z) dr dθ dz

For the boundary, note that x2 + y2 = r2 so it consists of z = r and z = √(2 - r2)

The two parts meet where r = √(2 - r2) so r2 = 2 - r2 so 2r2 = 2 so r = 1 because √(2 - r2) > 0

Thus we get

∫∫∫Q 2x dx dy dz = ∫010r√(2 - r^2) r * 2 r cos θ dz dθ dr

= (∫01r√(2 - r^2)  2r2 dz dr) (∫0 cos θ dθ)

= 0

 because ∫0 cos θ dθ = 0 and the double integral in r and z converges.