# Gauss's TheoremNet Force due to Uniform Pressure on a Closed Surface = 0

1. Oct 18, 2009

1. The problem statement, all variables and given/known data

In my fluid mechanics text, it states that the Net Force due to a uniform pressure acting on a
closed surface is zero or:

$$\mathbf{F} = \int_{surface}p(\mathbf{-n})\,dA = 0 \,\,\,\,\,\,\,(1)$$

where n is the unit normal vector and is defined as positive pointing outward from the surface.

The text then states that "...this result is independent of the shape of the surface. This can be proved by using Gauss's Theorem from Vector Analysis..."

I would like to prove this, but it has been over a year since I took multivariate calculus. When I google "Gauss's Theorem" I keep getting the divergence theorem, but fail to see how this is helpful?

The divergence theorem states that:

$$\int_{Volume}(\nabla\cdot\mathbf{F})\,dV = \int_{surface}\mathbf{F}\cdot\mathbf{n}\, dS \,\,\,\,\,\,\,\,(2)$$

where F is a differentiable vector field.

So Gauss's Theorem relates a volume integral to a surface integral. In (1) I am clearly dealing with a surface integral.

Can I get a nudge here? Also I am dealing with a pressure field (constant at that). This is a scalar field; why does Gauss's Theorem still apply?

Thanks,
Casey

2. Oct 19, 2009

Okay, I think that I was mistaken when I said

Clearly pressure has a direction, namely inwards and normal to the control surface.

So if pressure is uniform, I would write Gauss's Theorem as

$$\int_{Vol}(\nabla\cdot\mathbf{p})\,dV = \int_{surface}p(\mathbf{-n})\,dA$$

or

$$\int_{Vol}(\nabla\cdot(p\mathbf{-n}))\,dV = \int_{surface}p(\mathbf{-n})\,dA \,\,\,\,\,\,(3)$$

This look better?

EDIT: I feel like I am almost there. But I don't know if the form of (3) helps me..

3. Oct 20, 2009