Using Green's theorem and the divergence theorem

Question

I'm exploring the divergence theorem and Green's theorem, but I seem to be lacking some understanding. I have tried this problem several times, and I am wondering where my mistake is in this method.

For one example, I am trying to find the divergence of some vector field from a hemisphere. Let the hemisphere be given by $x^2 + y^2 + z^2 = 9$. Also, the vector field in question is given by $$ \textbf{V} = \bigg(y,\hspace{2mm} xz,\hspace{2mm} 2z-1\bigg) $$

Now, I want to evaluate the integral over the surface: $$\iint\textbf{V}\cdot\textbf{n}\hspace{2mm}d\sigma$$

Here is how I try to solve it. I instead use (by Green's theorem, where $\tau$ is a volume element) $$\iiint\nabla\cdot\textbf{V}\hspace{2mm}d\tau.$$

Taking the gradient of the the vector field, I get 2 (only the $\hat{z}$ component of the field will contribute). And since it is a simple hemisphere, I can integrate over the volume in spherical coordinates with the following limits:

$$r \hspace{1mm}\epsilon\hspace{1mm}[0,3]$$ $$\phi \hspace{1mm}\epsilon\hspace{1mm}[0,2\pi]$$ $$\theta \hspace{1mm}\epsilon\hspace{1mm}[0,\pi/2]$$

The Jacobian is standard for going from Cartesian to spherical coordinates: $r^2 \hspace{1mm}sin(\theta)$.

Lastly, evaluating this integral (and not forgetting to include the gradient of the vector field in the integral), I get $36\pi$.

The answer given in the text is $27\pi$. This is not a hard problem, and I am most certain that my integration and arithmetic is correct. There must be some fundamental step that I am missing.

Answer 1

i think you understand the theory fine. reading back from the answer and the data, i guess that the question is to estimate the flux across the curved surface of the hemisphere.

note that because the divergence in constant: $$ \nabla\cdot\textbf{V} = 2 $$ you may integrate this over the interior of the hemisphere simply by multiplying this by the volume of the hemisphere. i.e. $$ \iiint\nabla\cdot\textbf{V}\hspace{2mm}d\tau = 2 \frac{2\pi}3 3^3 = 36\pi $$ now the outward flux through the disc bounding the hemisphere below is given by the integral of scalar product of $\textbf{V}.\textbf{n}$ over the disc, where $\mathbf{n}$ is the unit normal pointing in the negative $z$-direction.

since the $z$-component of $\mathbf{V}$ on the $xy$-plane has a constant value of $-1$ the outward flux is therefore just $1$ times the area of the disc, i.e. $9\pi$

hence the outward flux across the curved surface of the hemisphere is: $$ 36\pi - 9\pi = 27\pi $$