Integral of a closed surface over a general region

AI Thread Summary
The discussion revolves around proving the equation \(\frac{1}{3}\oint\oint_{S}\vec{r} \cdot d\vec{s} = V\), where V represents the volume enclosed by the closed surface S. The original poster struggles with the problem despite initial attempts using Gauss' theorem, leading to confusion on how to proceed. A participant suggests visualizing the surface as composed of pyramids formed by connecting points on the surface to a central point, illustrating that the sum of these pyramids equals the enclosed volume. For a more rigorous mathematical approach, they recommend seeking assistance in a dedicated math forum. The conversation emphasizes the relationship between surface integrals and volume calculation through geometric interpretation.
Kwandae
Messages
1
Reaction score
0
I have been working on this problem for a few hours and am completely stuck. It seems like a simple problem to me but when I attempt it I get nowhere. The problem is:

Show that

\frac{1}{3}\oint\oint_{S}\vec{r} \cdot d\vec{s} = V

where V is the volume enclosed by the closed surface S= \partial V

I have tried to use Gauss' theorem to get as far as

\frac{1}{3}(\int\int\int_{V}(r_{x}+r_{y}+r_{z})dxdydz)

But am completely stuck on what to do from this point or even if I started this correctly. It's been about 2 years since I have done any surface integrals so I was hoping if someone here could maybe give me a helping push in the right direction
 
Physics news on Phys.org
I see no one has commented for several hours, so I'll at least indicate I'm interested.
I just think of ds a square element of surface and r a radius from it to some center. The dot product takes only the component of r that is perpendicular to ds. If we imagine more lines from the center to the vertices of the square ds, we'll have a pyramid. It's volume will be 1/3*r.ds. If we continue to picture an infinite number of these pyramids to an infinite number of ds elements making up the whole surface area, we will see that all the pyramids make up exactly the volume enclosed by the surface. So the integral is indeed equal to the volume.

If you need a more mathematical analysis, I suggest you take it over to the math forum.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top