Flux of the vector field F = (1,1,1)

[squenshl]

I'm studying for a test.
How do I find the flux of the vector field F = (1,1,1) down through the surface \sigma, given by z = \sqrt{x^2+y^2} and 1 < z < 2. The answer is 3pi but have no idea how to get it. I got it down to\int\int_R x+y/\sqrt{x^2+y^2} +1 dA. Now what?

 

[tiny-tim]

Hi squenshl!

(have a sigma: σ and a square-root: √ and an integral: ∫ and a pi: π and try using the X² tag just above the Reply box )

Use the divergence theorem (Gauss' theorem)

(oh, and it's a cone )

 

[HallsofIvy]

As tiny-tim says, the simplest thing to do is to use Gauss' theorem and integrate over the volume instead. If you want to do it directly, write the cone in parametric equations, x= r cos(\theta), y= r sin(\theta), z= r.

Now we can write each point as the vector \rho(r,\theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k}.

The derivatives, \rho_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k} and \rho_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j} are in the tangent plane at each point and the cross product of the two vectors, r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k} is normal to the surface. The "vector differential of surface area" is d\vec{S}= (r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k})drd\theta).

(I chose the order of cross product to give a negative z component so the vector is oriented downward since the problem said "down through the surface".)

Since the vector function is \vec{i}+ \vec{j}+ \vec{k} the flux is given by \int\int (\vec{i}+ \vec{j}+ \vec{k})\cdot(r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k})drd\theta)) or

\int_{\theta= 0}^{2\pi}\int_{r= 1}^2 r( cos(\theta)+ sin(\theta)+ 1) drd\theta
(\theta runs from 0 to 2\pi to go all the way around the cone and r runs from 1 to 2 because z= r and we are told that "1< z< 2".)

 

[LCKurtz]

I'm afraid both of the previous answers have mistakes. First, the Gauss theorem does not apply directly since there is no enclosed volume. So you need the method suggested by Halls, but his rk term in dS should be -rk. His integral leads to -3\pi, which I think is correct vs. 3\pi.

 

[tiny-tim]

First, the Gauss theorem does not apply directly since there is no enclosed volume.

Yes there is … it's the frustrum of a cone.

 

[LCKurtz]

Yes there is … it's the frustrum of a cone.

The top and bottom surfaces are not included so it is not a surface enclosing a volume. Of course the answer would be 0 for a constant vector and a closed surface.

 

[squenshl]

Cheers.

 

[tiny-tim]

The top and bottom surfaces are not included so it is not a surface enclosing a volume.

oh i see now … I've been reading it as z = 1 and z = 2 …

… through the surface \sigma, given by z = \sqrt{x^2+y^2} and 1 < z < 2.

 

[squenshl]

Actually it was 1 \leq z \leq 2