Integration problem... meh

[mooeypoo]

Hey guys,

 

I'm stuck.. I needed to integrate the surface of a "cap" (some part of a sphere). I got most of it, and got stuck at the actual integral.

 

What I have right now (after setting up the drawing and all limits, and everything else) is this integral:

 

[math]\int \int \sqrt{\frac{R^2}{R^2-x^2-y^2}} dxdy =[/math]

 

I rewrote it as

 

[math]R \int \int (R^2-x^2-y^2)^{-\frac{1}{2}} dxdy[/math]

My first thought was substitution, but..

 

[math] u=R^2-x^2-y^2[/math]

[math] du = -2x dx [/math]

[math] -\frac{1}{2}du = xdx[/math]

 

Which doesn't help me, because I don't have an xdx ... I don't have an extra x.

 

I thought of integration by parts, but got it all mixed up and repetitive.

 

At this point, I don't need a final answer or anything, just.. please.. some.. hint? anything? I'm stuck! have no clue how to go on

 

heeelp

 

~moo

[D H]

Trig substitution.

 

General suggestions:

 

∫ f(x²+a²) dx -- try x=a tanθ, motivated by identity tan²θ+1 = sec²θ

 

∫ f(x²-a²) dx -- try x=a secθ, motivated by identity sec²θ-1 = tan²θ

 

∫ f(a²-x²) dx -- try x=a sinθ, motivated by identity 1-sin²θ = cos²θ

[Mr Skeptic]

How about you use radial coordinates? [math]\int_0^{\theta_x} \int_0^{2\pi}r sin \theta d\phi d\theta[/math] Both integrals are very simple, but I think this would only work for a cap with a circular border (otherwise would become again painful). Draw a picture if you need, but I think this will be the area for a cap with a circular border.

[mooeypoo]

I thought of that, but since it's a cap, I'm having lots of problems with my r ... the cap base isn't really my original R.. and my limits are not working well... I might just have overly confused myself ... any idea how I can see the limits of r?

 

I thought of setting it up as my new r' = r sin(theta)

and then figure out my theta limits ... where.. I.. went crazy. Ideas?

[D H]

mooey, do you mean a true spherical cap (the part of a sphere to one side of the intersection of a plane and a sphere), or some more generic shape?

 

If you are indeed talking about a spherical cap, the answer is quite simple. I don't want to give it to you yet. Mr. Skeptic's equation will work quite well. Another approach: look at it is as a surface of revolution of a circular arc.

[mooeypoo]

No no, a spherical cap - a part of a sphere of radius R, and the "part"/"cap" is of height H.

 

 

My problem is mainly setting up the radius vs height or x/y/z..., or, if I take the spherical angle (polar coordinates), I got all mixed up in finding the proper "max" limit of theta :\

 

Okay, I need to run off to work, but I am going to take another fresh look on this and try it again. I might've just over-confused myself ....

I also have another question asking me to calculate the *volume* of the same shape using double integration... I know in theory what I want to do, but I am stuck in the same place (setting it up with all the r-h and limits of r or of theta).

 

I'll post another try a bit later

 

Thanks guys!

[D H]

No no, a spherical cap - a part of a sphere of radius R, and the "part"/"cap" is of height H.

So, just a plain old spherical cap.

 

The integral you have in the original post is not quite right. It is missing a factor of two: It yields the area of the half of the spherical cap that extends above the x-y plane.

 

Suppose the cap is the part of the sphere with [math]x>R-H[/math]. Let [math]z_+=\sqrt{R^2-(x^2+y^2)}[/math] and [math]z_-=-\,\sqrt{R^2-(x^2+y^2)}[/math]. The spherical cap comprises the two half-spherical caps defined by the functions z₊(x,y) and z_-(x,y). These two functions obviously have the equal areas over the x-y region of interest: determine one of them and multiply by two. For z₊(x,y), the area is

 

[math]A_{z_+}=\int_{R-H}^R \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \sqrt{\frac {R^2}{R^2-x^2-y^2}}\; dy\,dx[/math]

 

i.e., your integral in post #1 with limits specified. Note that I switched the integration order.

 

How to solve this: Hint: Define [math]Y=\sqrt{R^2-x^2}[/math] and make the trig substitution [math]y=Y\sin\theta[/math]. You will get a ridiculously simple result. Remember to multiply by 2.

 

An even easier way: This is a surface of revolution.

 

An even easier way: The area [math]dA[/math] of an infinitesimal chunk of a sphere is [math]r^2\sin\theta d\phi\,d\theta[/math] (Mr. Skeptic's post #3).