Flux of a vector and parametric equation

[Xsnac]

Homework Statement

Compute the flux of a vector field $\vec{v}$ through the unit sphere, where

$$ \vec{v} = 3xy i + x z^2 j + y^3 k $$

Homework Equations

Gauss Law:
$$ \int (\nabla \cdot \vec{B}) dV = \int \vec{B} \cdot d\vec{a}$$

The Attempt at a Solution

Ok so after applying Gauss Law, one gets
$$ \int 3y dV $$
and after converting it into a spherical integral I get
$$3 \int_0^{ \pi} \sin^2 \theta d \theta \int_0^{2 \pi} \sin \phi d \phi = 0$$ since integral of sin over a full period is. Is this correct? or if not, where did I go wrong?

 

[Orodruin]

It looks fine, but you do not need to actually do the spherical parametrisation. You could just use that the integrand is asymmetric over a symmetric region so the result must be zero.

Edit: Also, you are missing the radial part of the integral, but it will not matter.

 

[alejandromeira]

Xsnac I've been making the transformation right now and seems me that your work is ok.

 

[Xsnac]

It looks fine, but you do not need to actually do the spherical parametrisation. You could just use that the integrand is asymmetric over a symmetric region so the result must be zero.

Edit: Also, you are missing the radial part of the integral, but it will not matter.

Thanks. Regarding the radial part, it specified it's unit sphere therefore constant, and = 1 so I neglected it.

 

[alejandromeira]

It looks fine, but you do not need to actually do the spherical parametrisation. You could just use that the integrand is asymmetric over a symmetric region so the result must be zero.

Edit: Also, you are missing the radial part of the integral, but it will not matter.

Yes, visualizing the field in the space there are a symetry in the plane XZ, then like the volume has also spherical symetry around the point (0,0,0) we don't need to calculate anything. So is more fun!

 

[Orodruin]

Thanks. Regarding the radial part, it specified it's unit sphere therefore constant, and = 1 so I neglected it.

This is conceptually wrong, even if the result is the same. Your original surface of integration was the unit sphere. In using the divergence theorem, you rewrite the closed surface integral as a volume integral over the enclosed volume, in this case the unit ball.