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.