Surface Integration of a Cone (Sloped Surface)

[Wildcat04]

Homework Statement

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.

Homework Equations

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

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?

 

[HallsofIvy]

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 \phi, and \Omegais 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

 

[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:

[ (\Pia²)/2 ] (cot²\Omega + 1)^.5

Which simplifies down to

[(\Pia²)/2] csc\Omega

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

 

[HallsofIvy]

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

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:

[ (\Pia²)/2 ] (cot²\Omega + 1)^.5

Which simplifies down to

[(\Pia²)/2] csc\Omega

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