Surface Area of of an area - parametric surface

[Damascus Road]

Greetings,

I'm trying to find the surface area of the part of the sphere x^{2}+y^{2} + z^{2}=1 above the cone z=\sqrt{x^{2}+y^{2}}.

I know, that a surface area of a surface r(u,v) = x(u,v) + y(u,v) + z(u,v) can be given by,
A(S) = \int\int | r_{u} \times r_{v} | dA

A function, z=f(x,y) can utilize

A(S) = \int \int \sqrt{1 + (\frac{d z}{d x})^{2}+(\frac{dz}{dy})^{2}} dA

It seems part of my surface requires the first equation and the second part requires the second equation. (One is an equation z=f(xy) and the other is x y z = 1, how do I approach this?

Thanks in advance!

 

[Dick]

Those two equations are really forms of the same equation. Try just using the second one. The sphere is z=sqrt(1-(x^2+y^2)). It looks like it would be a good idea to use polar coordinates.

 

[Damascus Road]

Just use the equation for the sphere?
But my shape is not a sphere, its kind of a rounded cone...

 

[Dick]

You are only supposed to find the area of the spherical part. You intersect the sphere with the cone to get the limits of integration.

 

[HallsofIvy]

No, it's not. The surface is, as you say in your first post, that part of the sphere that is above that cone. The conical sides are not part of the area you are trying to find. Since, in spherical coordinates, z= \rho cos(\phi) and x^2+ y^2= \rho^2 sin^2(\phi) so z= \sqrt{x^2+ y^2} is \rho cos(\phi)= \rho sin(\phi) which tells you that \phi runs from 0 to \pi/4.

 

[Damascus Road]

Can this be done in cylindrical? Going from 0 to z (the cone) and 0 to 1?

 

[HallsofIvy]

Sure, but it is simpler in cylindrical equations. In cylindrical coordinates the sphere is r^2+ z^2= 1 and the cone z= r^2. The cone intersects the sphere at r^2+ z^2= r^2+ r^4= 1. That's a quadratic equation for r^2. There is only one positive root for r² and only one positive square root of that:
\sqrt{\frac{-1+\sqrt{5}}{2}}
The integral would be over r from 0 to that value and theta from 0 to 2\pi.

 

[Damascus Road]

That is precisely why I want to use cylindrical! :)

 

[Damascus Road]

I think I am making a mistake somewhere.

My first integration leaves me with

\int [ -\frac{1}{3} (1-r^{2})^{3/2} - \frac{1}{3} ] d \theta

Doing yet another substitution, my second integration leaves me with

-\frac{2}{15}(1-r^{2})^{5/2} - \frac{1}{3} (1-r^{2})

But I get errors when evaluating this, is there an error here?

 

[Dick]

Yes, you are making big mistakes. What are you getting for the integrand dA? You need to show more of your work.

 

[Damascus Road]

dA is what I have shown, but not yet evaluated from 0 to 2 pi

 

[Dick]

Then it's wrong. Can you explain how you got it? What's sqrt(1+(dz/dx)^2+(dz/dy)^2)?

 

[Damascus Road]

I was making a big error, making a u substitution in stead of using a trig sub. However, I'm still getting the wrong answer using software...

 

[Dick]

I was making a big error, making a u substitution in stead of using a trig sub. However, I'm still getting the wrong answer using software...

Can't debug your software for you but you don't need it. It's not that hard. And you don't need a trig substitution either.

 

[Damascus Road]

I can't think of a simpler way to integrate \sqrt{1 - r^{2}} then with a trig substitution... maybe I should start there.

 

[Dick]

I can't think of a simpler way to integrate \sqrt{1 - r^{2}} then with a trig substitution... maybe I should start there.

I can if it's r/sqrt(1-r^2). Don't forget what dA is in polar (or cylindrical) coordinates.

 

[Damascus Road]

I'm attempting spherical coordinates now, cylindrical is giving me no success.

 

[Dick]

I have no idea why not. I worked for me. Why don't you try showing some of your work?

 

[Damascus Road]

I'll try to clean it up, right now all these pages of mess into latex would take me a few years...

 

[Damascus Road]

Greetings,

I'm trying to find the surface area of the part of the sphere x^{2}+y^{2} + z^{2}=1 above the cone z=\sqrt{x^{2}+y^{2}}.

I know, that a surface area of a surface r(u,v) = x(u,v) + y(u,v) + z(u,v) can be given by,
A(S) = \int\int | r_{u} \times r_{v} | dA

A function, z=f(x,y) can utilize

A(S) = \int \int \sqrt{1 + (\frac{d z}{d x})^{2}+(\frac{dz}{dy})^{2}} dA

It seems part of my surface requires the first equation and the second part requires the second equation. (One is an equation z=f(xy) and the other is x y z = 1, how do I approach this?

Thanks in advance!

Ok, in order to do this in cylindrical coordinates. using the method bolded above,
I need to integrate the sphere, which is

z=\sqrt{ 1- r^{2}}.

Now, the above formula requires me to calculate the partial derivatives of dz/dx and dz/dy. Since, I am in cylindrical coordinates, what partial derivatives do I need? dz/dz (which doesn't make any sense I realize, but I don't need dz/dr for a surface area) and dz/d (theta) ?

 

[Dick]

The derivatives are in cartesian coordinates. Fine. Take the derivatives in cartesian coordinates. THEN simplify and convert to cylindrical coordinates.

 

[Damascus Road]

Ok, my next problem is this

dz/dx = \frac{-2x}{2\sqrt{1-x^{2}-y^{2}}}

and dz/dy is the same but with y's.

switching to cylindrical:

\frac{-rcos(\theta)}{2\sqrt{1-r^{2}}} (for dz/dx)

and I'm integrating in the end by dz and d\theta.
So, what happens to my r's ?

 

[Dick]

So inside of the square root you've got 1+x^2/(1-r^2)+y^2/(1-r^2). Simplify that. You'll find that it's a function only of r. And why are you integrating dz,dtheta in the end? You've set the problem up as z(x,y). You want to integrate over a circle in the x-y plane. That's integrating dr, dtheta.

 

[Damascus Road]

Sorry, I'm not sure why I want to integrate dr... as I understand it, dz (the height) and d/theta will give me the surface area, and dr "fills in" the shape, which I would use if I wanted a volume...

 

[Dick]

If you are going to use the second formula, and I recommend that you do, the dA is over an x-y region. Once you've set up the integral, convert that region to polar coordinates to simplify the integration. If you want to parametrize the region using z and theta, you can, but then you have to use the first formula. Why don't you just do it the simpler way first?

 

[Damascus Road]

Ok, this is where I'm going wrong. I've been trying to use the second formula, AND do so with z and theta, I'm beginning to see that it doesn't make sense to do so... but I don't understand how using r can give me a surface area at all.

 

[Dick]

I don't know what kind of picture you've got in your head that's bothering you so much, but set up the integral using the second formula in cartesian coordinates, ok? Then we can discuss how the r comes in. It's not the spherical coordinate r.

 

[Damascus Road]

Ok, (thanks for the help by the way :) )

Here is the second formula in cartesian coors:

A(S) = \int \int \sqrt{1 + (\frac{-2x}{2\sqrt{1-x^{2}-y^{2}}})^{2}+(\frac{-2y}{2\sqrt{1-x^{2}-y^{2}}})^{2}} dA

switching to cylindrical

A(S) = \int \int \sqrt{1 + (\frac{-rcos(\theta)}{\sqrt{1-r^{2}}})^{2}+(\frac{-rsin(\theta)}{\sqrt{1-r^{2}}})^{2}} dA

Yeah?

 

[Dick]

Square the numerators out and use that sin(theta)^2+cos(theta)^2=1. You can write that more simply.

 

[Damascus Road]

How's this look?

A(S) = \int \int \sqrt{1 - \frac{r^{2}}{(1-r^{2})} dA

 

[Dick]

I got a + sign in there. (-1)^2=(+1). You can still simplify some more.

 

[Damascus Road]

Woops, your right:

A(S) = \int \int \sqrt{\frac{1}{(1-r^{2})} dA

 

[Dick]

Ok, now you want to integrate over the values of x and y that will cover the spherical cap. That's a circular region, right? That's a good reason to switch to polar coordinates. What is dA in polar coordinates?

 

[Damascus Road]

r^2 sin (phi) dr dtheta dphi

 

[Dick]

A is a region in the x-y plane. Use polar coordinates. Not spherical.