# Surface Area of of an area - parametric surface

1. Nov 25, 2008

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?

Last edited: Nov 25, 2008
2. Nov 25, 2008

### 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.

3. Nov 25, 2008

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

4. Nov 25, 2008

### 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.

5. Nov 25, 2008

### HallsofIvy

Staff Emeritus
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 $\pi/4$.

6. Nov 25, 2008

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

7. Nov 25, 2008

### HallsofIvy

Staff Emeritus
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 $2\pi$.

8. Nov 25, 2008

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

9. Nov 25, 2008

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?

10. Nov 25, 2008

### Dick

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

11. Nov 25, 2008

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

12. Nov 25, 2008

### Dick

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

13. Nov 26, 2008

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...

14. Nov 26, 2008

### Dick

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.

15. Nov 26, 2008

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

16. Nov 26, 2008

### Dick

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

17. Nov 27, 2008

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

18. Nov 28, 2008

### Dick

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

19. Nov 28, 2008

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

20. Nov 28, 2008

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