# Surface Integration of a Cone (Sloped Surface)

1. Nov 12, 2008

### Wildcat04

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

Given Parameterization:
x = u cos $$\phi$$
y = sin $$\phi$$
z = u cot $$\Omega$$

Find the sloping surface of a right cone with semi-angle $$\Omega$$ with a base radius of a.

2. Relevant equations

Surface area of a cone = $$\pi r\sqrt{r^2 + h^2}$$

3. The attempt at a solution

Solid angle:
$$\Omega = \int(r dS)/ (r^3)$$

semi angle = (1/2) $$\Omega$$

Cartesian Equation of a cone:

(x2 + y2) / (r / h)2 = z2

I understand the concepts of surface integration but I have not run across a problem where F was not given. I have a feeling that I am much more likely to run into this issue in the future and I would like to know what the process of determining F is.

Should I start by taking the div of the cartesian equation and then plugging in the given parameters (x,y,z) and integrating?

Last edited: Nov 12, 2008
2. Nov 12, 2008

### HallsofIvy

Staff Emeritus
It doesn't make sense to tell us that "F" is not given, when you don't tell us what "F" is!

In any case, if $x= u cos(\phi)$, $y= u sin(\phi)$, and $z= u cot(\Omega), where u and [itex]\phi$, and $\Omega$is a constant, $\vec{r}= u cos(\phi)\vec{i}+ u sin(\phi)\vec{j}+ u cot(\Omega)\vec{k}$ so $\vec{r}_u= cos(\phi)\vec{i}+ sin(\phi)\vec{j}+ cot(\Omega)\vec{k}$ and $\vec{r}_\phi=- -u sin(\phi)\vec{i}+ u cos(\phi)\vec{j}$ are tangent vectors to the surface in the direction of "coordinate lines". The "fundamental vector product" $\vec{r}_u\times \vec{r}_\phi$ is $-ucos(\phi)cot(\Omega)\vec{i}- usin(\phi)cot(\Omega)\vec{j}+ u\vec{k}$. The "differential of surface area" is the length of that vector times $du\d\phi$: $u\sqrt{cot^2(\Omega)+ 1}dud\phi$

3. Nov 12, 2008

### Wildcat04

The "F" I was referring to is the function inside a double or triple integral when doing a surface integration

ie

$$\int\int\int F dV$$

With your help I believe that I have come up with the answer, however I am having a little trouble showing that it is correct.

A = (2$$\Pi$$)(Slope Height) / 2

after my integration I end up with the following:

[ ($$\Pi$$a2)/2 ] (cot2$$\Omega$$ + 1).5

Which simplifies down to

[($$\Pi$$a2)/2] csc$$\Omega$$

Can someone give me a push (or shove) in the right direction to prove this equals A from above?

4. Nov 12, 2008

### HallsofIvy

Staff Emeritus
That's a volume integral, not a surface integral! The "F" you want to find the surface area, when dS is the differential of surface area is just "1":
$$\int\int 1 dS= S$$.

[/quote]With your help I believe that I have come up with the answer, however I am having a little trouble showing that it is correct.

A = (2$$\Pi$$)(Slope Height) / 2

after my integration I end up with the following:

[ ($$\Pi$$a2)/2 ] (cot2$$\Omega$$ + 1).5

Which simplifies down to

[($$\Pi$$a2)/2] csc$$\Omega$$

Can someone give me a push (or shove) in the right direction to prove this equals A from above?[/QUOTE]