Compute the flux of a vector field through the boundary of a solid

[DottZakapa]

Homework Statement

Compute the outward flux of the vector field F(x,y,z) = 2x,−2y,z2 through the boundary of the solid
Ω= (x,y,z)∈R3: x2+y2≤z≤4 .

Relevant Equations

flux through a surface

is it correct if i use Gauss divergence theorem, computing the divergence of the vector filed,
that is :

div F =2z
then parametrising with cylindrical coordinates
$x=rcos\alpha$
$y=rsin\alpha$
z=t

1≤r≤2
0≤$\theta$≤2π
0≤t≤4

$\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{4} 2tr \, dt \, dr \,d\theta$

but i guess there is something missing because the result is not correct

 

[Orodruin]

1≤r≤2
0≤θθ\theta≤2π
0≤t≤4

This describes a hollowed out cylinder, which is not your volume. What is your reasoning behind theses bounds?

 

[HallsofIvy]

$z= x^2+ y^2$ is a paraboloid. The volume lies above that paraboloid and below z= 4. Of course, $x^2+ y^2= 4$ is the circle in the x,y plane with center at the origin and radius 2.

In cartesian coordinates, the integral of any function, f(x, y, z) over that region would be $$\int_{x= -2}^2 \int_{y= -\sqrt{4- x^2}}^{\sqrt{4- x^2}}\int_{z= x^2+ y^2}^4 f(x,y,z)dzdydx$$.

In cylindrical coordinates, $$\int_{r= 0}^2\int_{\theta= 0}^{2\pi}\int_{z= r^2}^4 f(r,\theta,z) dzd\theta dr$$.