Surface area of a sphere

[hobz]

In the shower this morning, I suddenly got the idea of calculating the surface area of a sphere as an integral of circumferences.

 

My approach is as follows:

 

The integral is [math] 2\cdot \int_0^r 2\pi t dt [/math] where the "2" in front accounts for the surface area of whole sphere (equivalent to having [math]2r[/math] as upper limit in the integral - by symmetry).

 

The result is then [math]2\cdot 2\pi [\frac{1}{2}t^2]_0^r[/math] giving [math]4\pi \frac{1}{2}r^2[/math] or [math]2\pi r^2[/math] which is wrong.

 

Where did I go wrong?

Edited November 8, 2008 by hobz

[swansont]

You've described a circle, not a sphere.

[Skye]

You have two dimensions. You need three. To be able to do that you need to know have to do double integration and find the area of parametric surfaces.

[hobz]

I want to use circle circumferences to find an area. I don't see how I lack dimensions in the question.

[Skye]

Here's an example from wikipedia. I hadn't seen it done before. Zany.

 

http://en.wikipedia.org/wiki/Surface_area#Sphere

[D H]

In the shower this morning, I suddenly got the idea of calculating the surface area of a sphere as an integral of circumferences.

 

My approach is as follows:

 

The integral is [math] 2\cdot \int_0^r 2\pi t dt [/math] where the "2" in front accounts for the surface area of whole sphere (equivalent to having [math]2r[/math] as upper limit in the integral - by symmetry).

 

You are indeed calculating the area of something, but it's not a sphere.

You've described a circle, not a sphere.

 

Your idea is conceptually correct but wrong in implementation. Where you went wrong was in forming the integrand. The correct integral for any surface of revolution is

 

[math]A=\int_a^b 2\pi\,r(l)\,dl[/math]

 

where the dummy integration parameter is arc length, r(l) is the distance from the curve to the axis of rotation at an arc length of l, and a and b denote the start and end points of the rotated curve in terms of arc length; b-a is the total arc length of the curve being rotated to form a surface. This is Pappus' theorem.

 

A sphere is surface of revolution formed by rotating a semicircle about its chord. A point on a semicircle can be parameterized in terms of the angle between the line from the center of the circle to one of the endpoints and the line from the center of the circle to the point in question. This parameterization makes arc length an easy calculation: [math]l=r\theta[/math], or [math]dl=r d\theta[/math]. The distance from a point to the axis of rotation is simply [math]r(l)=r\sin\theta[/math]. Thus

 

[math]A=\int_0^{\pi} 2\pi\,r\sin\theta\,r d\theta = 4\pi r^2[/math]

 

Or you could do it the way described in the previously cited Wikipedia article.

[hobz]

Thanks for clarifying. I will look into the methods proposed.

 

It would seem that what I am calculating is related to the area of an annulus http://en.wikipedia.org/wiki/Annulus_(mathematics)

 

I follow you explanation, D H, but I am having trouble with visualizing what is meant by this on the wikipedia article:

 

The width of the ring can be determined by using Pythagoras' formula for a rectangular triangle with side lengths [math]dx[/math] and [math]f'(x) \cdot dx[/math], which leads to [math]\sqrt{1+f'(x)^2}\,dx[/math]

The infinitesimal surface area of the circular ring thus is equal to [math]2\pi f(x)\cdot \sqrt{1+f'(x)^2}\,dx[/math]

[Harlequinne]

In the shower this morning, I suddenly got the idea of calculating the surface area of a sphere as an integral of circumferences.

 

My approach is as follows:

 

The integral is [math] 2\cdot \int_0^r 2\pi t dt [/math] where the "2" in front accounts for the surface area of whole sphere (equivalent to having [math]2r[/math] as upper limit in the integral - by symmetry).

 

The result is then [math]2\cdot 2\pi [\frac{1}{2}t^2]_0^r[/math] giving [math]4\pi \frac{1}{2}r^2[/math] or [math]2\pi r^2[/math] which is wrong.

 

Where did I go wrong?

 

Integrate 2pi r you get pi r^2. Now you must start afresh from 4 pi r^2.

and integrate to 4/3 pi r^3.

 

i think.i failed my a-levels.

[mooeypoo]

Integrate 2pi r you get pi r^2. Now you must start afresh from 4 pi r^2.

and integrate to 4/3 pi r^3.

 

i think.i failed my a-levels.

 

 

You only need to integrate r once, and you get an "extra" power because the [math]d\tau[/math] integrand in spherical coordinates.

 

Please, if you're not sure of your answer, don't answer.

[Dave]

Please, if you're not sure of your answer, don't answer.

 

I wouldn't go that far - just put a label saying you're not sure about it.