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

Question

Find the flux of F over Q where Q is bounded by z=sqrt(x^2+y^2) and z=sqrt(2-x^2-y^2), F=<x^2, z^2-x, y^3>. (Answer: 0)

 

I know the integrand is 2x and I need to integrate 2x dV but how do I find the points of the integral?

Answer 1

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 = √(x² + y²) and z = √(2 - x² - y²)

Now as you calculated, div(F) = (∂/∂x)x² + (∂/∂y) (z² - x) + (∂/∂z) y³ = 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 = ∫₀¹∫₀^2π∫_r^{√(2 - r^2)} r * 2 r cos θ dz dθ dr

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

= 0

 because ∫₀^2π cos θ dθ = 0 and the double integral in r and z converges.