# Flux through a Hemisphere.

I need to calculate the flux of a vector field :

$H = (y-z)\hat{i} + (z-x)\hat{j} + (x-y)\hat{k}$ outside the Hemisphere given by the equation :

$(x-1)^2 +y^2 +z^2=1$ with $z\ge0$

Now i used the divergence theorem where i deduced that the flux throughout the solid enclosed by the hemisphere and a disk in the plane $z=0$, is $0$ since $\operatorname{div}\vec H=0$. And so the flux through out the hemisphere equals minus that throughout the disk, and so it is equal to $\iint\vec H \cdot \vec n\,ds_1$ where $S_1$ is the surface of the disk in the $xy$-plane.

Now since $\vec n=-\vec k$ hence the integral $\iint\vec H \cdot \vec n\, ds_1$ will look like $\iint (x-y)dxdy$, but I need help knowing the boundaries after transforming the coordinates into spherical coordinates.

We have $x=r\cos\theta+1$ and $y=r\sin\theta$. Hence, $x-y=r(\cos\theta-\sin\theta)+1.$ This shouldn't be too difficult to integrate.
• I previously done so, But i need the bounds of $\theta$ and $r$ – Mistos May 31 '15 at 19:24
• @MohamadMisto Since it's a disc of radius $1$, we take $r\in[0,1]$ and $\theta\in[0,2\pi]$. – Amitai Yuval May 31 '15 at 19:26
• Mmmm i dont guess so where its center is not at the origin of the orthonormal system, so isn't $\theta$ $\in$[$-\pi\over2$,$\pi\over2$]? – Mistos May 31 '15 at 19:37
• @MohamadMisto No. What you're talking about is the reason for the "$+1$" in the expression for $x$. – Amitai Yuval May 31 '15 at 19:45