# Using Green's theorem and the divergence theorem

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.

• Is it possible you have the wrong hemisphere with your choice for the ranges for $\phi, \theta$? What were the original specifications of the hemisphere? Commented Nov 25, 2014 at 13:56
• It seems you have done everything correct. Maybe textbook's answer is wrong. By the way in the last step you don't even need to introduce spherical coordinates, you just need to multiply the hemisphere's volume by $2$. Commented Nov 25, 2014 at 14:15

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: $$36\pi - 9\pi = 27\pi$$