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?
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?
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^{2 }+ y^{2}) and z = √(2 - x^{2} - y^{2})
Now as you calculated, div(F) = (∂/∂x)x^{2} + (∂/∂y) (z^{2} - x) + (∂/∂z) y^{3 }= 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 x^{2} + y^{2} = r^{2} so it consists of z = r and z = √(2 - r^{2})
The two parts meet where r = √(2 - r^{2}) so r^{2} = 2 - r^{2} so 2r^{2} = 2 so r = 1 because √(2 - r^{2}) > 0
Thus we get
∫∫∫_{Q} 2x dx dy dz = ∫_{0}^{1}∫_{0}^{2π}∫_{r}^{√(2 - r^2)} r * 2 r cos θ dz dθ dr
= (∫_{0}^{1}∫_{r}^{√(2 - r^2)} 2r^{2} dz dr) (∫_{0}^{2π} cos θ dθ)
= 0
because ∫_{0}^{2π} cos θ dθ = 0 and the double integral in r and z converges.
Comments
Thanks.
Thanks.