Find surface of maximum flux given the vector field's potential

[Addez123]

Homework Statement

A vector field has the potential
$$\Phi = x^2 + y^2 + z^2 - (x^2 + y^2 + z^2)^2$$
Find the surface where the flux is at maximum.

Relevant Equations

Gauss theorem

The vectorfield is
$$A = grad \Phi$$ $$A = x^2 + y^2 + z^2 - (x^4 + y^4 + z^4 + 2x^2y^2 + 2x^2z^2 + 2y^2z^2)$$
The surface with maximum flux is the same as the volume of maximum divergence, thus:
$$div A = 6 - 20(x^2 + y^2 + z^2)$$

This would suggest at the point 0,0,0 the flux is at maximum.

But the book says its the surface where
$$div grad \Phi > 0$$
aka:
$$x^2 + y^2 + z^2 < 3/10$$

 

[pasmith]

Homework Statement:: A vectorfield has the potential
$$\Phi = x^2 + y^2 + z^2 - (x^2 + y^2 + z^2)^2$$
Find the surface where the flux is at maximum
Relevant Equations:: Gauss theorem

The vectorfield is
$$A = grad \Phi$$ $$A = x^2 + y^2 + z^2 - (x^4 + y^4 + z^4 + 2x^2y^2 + 2x^2z^2 + 2y^2z^2)$$

A = \nabla \Phi is a vector. What are its components? Your expression appears to be a scalar.

The surface with maximum flux is the same as the volume of maximum divergence, thus:
$$div A = 6 - 20(x^2 + y^2 + z^2)$$

This would suggest at the point 0,0,0 the flux is at maximum.

But the book says its the surface where
$$div grad \Phi > 0$$
aka:
$$x^2 + y^2 + z^2 < 3/10$$

You're not choosing a point such that \nabla \cdot A is maximal; you are choosing a volume V such that \int_V \nabla \cdot A\,dV = \int_{\partial V} A \cdot dS is maximal.

In this case, \nabla \cdot A is positive for r &lt; \frac{3}{10}. Therefore you can increase the flux across a surface within this volume by increasing the volume within the surface. The largest you can make the volume is by having it consist of everything inside r &lt; \frac{3}{10}.

Outside of this sphere \nabla \cdot A &lt; 0, so including any portion of this region will reduce the flux.

 

[Orodruin]

Also note that you really don't need the divergence theorem for this. In spherical coordinates the scalar field is $\Phi = r^2(1-r^2)$ so that it is clear that everything has spherical symmetry and therefore so will the sought surface. Because of the radial symmetry $\vec A = d\Phi/dr \vec e_r = (2r-4r^3) \vec e_r$ and so the flux $F$ out of the sphere of radius $r = R$ is
$$
F = \oint_{r = R} (2R - 4R^3) R^2 d\Omega = 8\pi (R^3 - 2 R^5).
$$
This leads to
$$
\frac{dF}{dR} = 8\pi (3R^2 - 10 R^4) = 0
$$
for the extreme value and therefore $R^2 = 3/10$. We obtain the maximal flux from this by insertion.

The largest you can make the volume is by having it consist of everything inside r &lt; \frac{3}{10}.

Nitpicking, but $r^2 < 3/10$.

 

[Addez123]

Very nice explained @Orodruin I just have one small problem.

$$F = \int A * n dS = \int r^2(1 - r^2) * ? dS$$
dS = r^2 (jacobian) dr dv du

But what's my normal vector?

 

[Orodruin]

Very nice explained @Orodruin I just have one small problem.

$$F = \int A * n dS = \int r^2(1 - r^2) * ? dS$$
dS = r^2 (jacobian) dr dv du

But what's my normal vector?

There is no normal left at that (the last) step. The normal of a sphere is $\vec e_r$ and this has been dotted with the $\vec e_r$ from $\vec A$ for a result of 1.

 

[Addez123]

F=∮r=R(2R−4R^3)R^2dΩ=8π(R^3−2R^5).

What is your limit of integrations on this?

Also I don't see how your integral is correct:
$$\int (2r-4r^3)*r^2 = \int 2r^3 - 4r^5 = | r^4/2 - 2/3*r^6$$
not $R^3−2R^5$

 

[Orodruin]

What is your limit of integrations on this?

Also I don't see how your integral is correct:
$$\int (2r-4r^3)*r^2 = \int 2r^3 - 4r^5 = | r^4/2 - 2/3*r^6$$
not $R^3−2R^5$

The integral is over the unit sphere. In other words $R$ is constant.