# Vector Calculus Problem, cylinderical co-ords.

1. Nov 26, 2008

### Spoony

For divergance thereom, say i have a volume integral to calculate of form $$\iiint_V \nabla F dV$$

i can relate it to the form:
$$\iint_S F.dS = \iiint_V \nabla F dV$$
and calculate using the left hand side,

$$\iint_S F.n dS$$

where n is the unit vector normal to the surface of integration.

for my surface I have a cylinder.
of form x^2 + y^2 =< 1 for 0 =< z =< 3

So splitting this up brings me 3 surfaces, the top 'cap' the bottom 'cap' and the part around the middle. i can compute by intergrating the vector field seperately over each surface (i think) and summing over each suurface.

question is do i use the same normal vector for all surfaces ie the normal to the main surfaces, or do i for each seperate surface use a different normal vector?

i switched to cylnderical co-ords to solve this.

i assumed at first it was a seperate normal for each vector. but this gave me a dot product of zero for the area of the strip aroundthe cylinder.

Last edited: Nov 26, 2008
2. Nov 26, 2008

### Office_Shredder

Staff Emeritus
Assuming your cylinder looks like this:

you break it up into the top, the bottom and the curved side surfaces. For each surface, you have to pick a normal direction, and you have to be careful. The divergence theorem uses the outward pointing normal, so the normal has to point out of the cylinder. Hence on the top the normal points up, on the bottom the normal points down, and on the sides the normal points away from the center.

3. Nov 26, 2008

### Spoony

Yup thanks i got that all :) cleared up aib gproblem for me thanks!

so my vector field F is (-yi + xj + zk) * f(x,y)

in cartesian co-ords
My intergral looks like this:

$$\iint F.-k dS + \iint F.-k dS + \iint F.n$$

where the parts are top, bottom, bit that goes round respectively.

on the top of the cylinder the circle's normal points up so its normal vector is positive k on the bottom the unit vector points down so its normal vector is negative k, (not worried about the bit that goes round yet)

so surely due to the laws of dot product:
$$\iint F.-k dS$$ = $$\iint -F.k dS$$ = $$- \iint F.k dS$$

therefore the top part is of form: $$\iint F.-k dS$$
and the bottom is of form: $$- \iint F.k dS$$

so surely these 2 should cancel? leaving only the round bit?

4. Nov 26, 2008

### Office_Shredder

Staff Emeritus
First, the top part shouldn't have that - sign... and they won't cancel anyway, because you're integrating over different domains

5. Nov 26, 2008

### Spoony

Yeah i see that they wont cancel now, different z values.
That was a mistake by me copying and pasting latex code :\$

The bottom part will have 0 value for its intergrand as z=0 there and it'll be -k.(0k-yi+xj)f(x,y)) this correct?

The problem is far easier to solve in cylndrical coordinate system, so area for a disc in this coordinate system is r dr dϕ.

but the vector field simplifies to eϕ/16 + zez/16

i treat the z value as 3 for the top circle, dot product with positive ez gives 3/16 (when working in cylnderical coords).
intergrate over $$\iint (3/16) r dr d\phi$$
for r = 1 unit circle range is 1 to 0, for ϕ range is 0 to 2 $$\pi$$

so awnser is 6 $$\pi$$ / 16 = 3 $$\pi$$ / 8

(thats not meant to be to the power of pi btw :S but multiplied)

anything wrong here? im so shaky on these type of computations

6. Nov 26, 2008

### Spoony

o btw the function f(x,y) = (x^2+y^2) / [1+(x^2+y^2)^4]^4
should this repeated x^2+y^2 = sqrt(r) or should it equal 1 as
$$x = pcos\phi$$ and $$x = psin\phi$$
where ive used the trigonemtric identity cos^2(t) + sin^2(t) = 1
im very confused here.

7. Nov 26, 2008

### Spoony

ok for the vector function i now have
$$\frac{p^{6} (\vec{e}_{\phi} + z \vec{e}_{z})}{(1 + p^{8})^{4}}$$
for the unit disk i have the intergral of $$3 \int \frac{p^{7}}{(1 + p^{8})^{4}}dp \int d_{\phi}$$

for the range p, 0 to 1 and phi 0 to 2*pi

is this correct?

8. Nov 26, 2008

### Spoony

Solving for the unit disk i get 7/32 * pi
using the substitution u = 1+p^8

How do i solve for the round part of the cylinder it has normal unit vector $$\vec{e} _{p}$$

As vector equation of field reads:
$$\frac{p^{6} (\vec{e}_{\phi} + z \vec{e}_{z})}{(1 + p^{8})^{4}}$$
But this is surely wrong as then the dot product with the normal is zero,
but taking back to co-ord's in cartesian then i could call the unit vector $$\vec{e} _{p} = \vec{i}cos _{\phi} + \vec{j}sin _{\phi}$$

Which would result in something hurrendus (sp?!) is this a valid method to take?

Last edited: Nov 26, 2008
9. Nov 26, 2008

### Office_Shredder

Staff Emeritus
On the actual cylinder portion (as opposed to the top/bottom) x2+y2=1 everywhere on the cylinder, so f(x,y) just becomes a constant for that integration (making things a lot easier). Although it doesn't look like your parametrization is right... parametrization should be

$$( \theta , z$$ where theta is the angle in the x-y plane (as per polar coordinates) of the vector, and z is the height of the vector.... so

$$x = cos( \theta )$$
$$y = sin( \theta )$$
$$z = z$$

are the equations of the parametrization

Last edited: Nov 26, 2008
10. Nov 26, 2008

### Spoony

well in cylinderical polar co-ords z (cartesian) = z (cylinderical polar).
(nevermind you corrected)

also what parametrization do you refer to being wrong? the calculation of the disk?

Ahh yes treating f(x,y) as a constant makes this a WHOLE lot simpler, f(x,y) now equals 1/16
for this intergrand and the dot product of the vector field -yi + xj with k terms redundant as the dot product is for the unit vector icos(phi) + jsin(phi).

so dot product is: $$\frac{1}{16} * ( -ycos(\phi) + xsin(\phi) )$$

since r = 1, x = cos(phi), y=sin(thi)

so dot product is

$$\frac{1}{16} * ( -sin(\phi)cos(\phi) + cos(\phi)sin(\phi) )$$
which is zero :S

i know is 3 to 0 for z and 0 to 2pi for phi.

trouble is though i still think that this should have a value, tearing my hair out here im so clsoe :P

Last edited: Nov 26, 2008
11. Nov 26, 2008

### Office_Shredder

Staff Emeritus
I forgot latex doesn't like spaces... I reformatted it so it makes more sense

With respect to

I wasn't sure whether you were referring to the unit disk or the cylindrical portion (I thought the cylindrical portion, but now I think I was wrong))... on the cylindrical portion this just becomes a constant. On the unit disk part, this becomes (if the parametrization is r, theta)

f(x,y) = r2/[1+r4]4

So your integration before was wrong. But notice when I multiply this by r, it's a nice easy substitution u=r4

12. Nov 26, 2008

### Spoony

i was inaccurate with my function f(x,y) let me correct myself, i aplogise for how long it was wrong. :(.

$$f(x,y) = \frac{(x^{2} + y^{2})^{3}}{(1+(x^{2} + y^{2})^{4})^{4}}$$

not very pretty at all, but when field is dotted with k or unit vector on spherical co-ords pointing up ez
is gives...

$$3 \frac{p^{6}}{(1 + p^{8})^{4}}$$

as (x^2+y^2) = r^2

Last edited: Nov 26, 2008
13. Nov 26, 2008

### Spoony

Can no-one help me? i appreciate the help very much from office_shredder, but he cant be the only one that knows about the divergance thereom and intergration of a vector field over a surface. I'm pretty desperate now as i just want to solve the damn thing, but i feel like i just need someone to check over all my work.

14. Nov 26, 2008

### Office_Shredder

Staff Emeritus
That looks fine (actually, I had an error in my previous post when calculating f(x,y) now that I look at it again; the old version didn't let you do integration by parts)). Remember you have an additional p from the change of coordinates when integrating over the disk, so you can still do the substitution u=p8 and get something easily integrable. So you end up getting....

$$\int_0^{2 \pi } \int_0^1 \frac{3}{16} \frac{p^7}{(1+p^8)^4} dp d \theta$$

which you should be able to do.

For your side component, that looks right... intuitively, (-y,x) in the x-y plane (we ignore z since the normal doesn't have a z component) is perpendicular to the vector (x,y). So F basically was the tangent vector of the circular slice of the cylinder, and hence you wouldn't expect it to contribute to how much stuff flows in and out of the cylinder

Mathematically speaking, your way's a better description of course, but this is why you would expect that kind of result

15. Nov 26, 2008

### Spoony

For the last latex expression, the 1/16 part is incorrect, you onl get 1/16 when you take x^2+y^2=1 ie when y and x are constants, that does not happen on the unit disk on z = 3.