# Proving the formula for the surface area of a sphere

• Dec 6th 2008, 12:21 PM
chella182
Proving the formula for the surface area of a sphere
Okay. (I've not posted on here before so forgive me if I'm not typing things properly or whatever).

In one of my assignment questions (I'm a 1st year Maths degree student, by the way) I have to prove the formula for the surface area of a sphere using the surface of revolution formula. So basically, my answer should be 4*Pi*r². I'm having difficulty with it, though. So far within the integral I have this horrible looking thing...

http://i198.photobucket.com/albums/a...leintegral.jpg
(I thought posting a picture would be easier than trying to type it all out). Those should be x²s on the ends of the fraction by the way - the bracket kinda cuts that off. That has to simplify a little, otherwise, I've done something wrong in my calculations prior to arriving at that.

Any ideas?

Cheers.
• Dec 6th 2008, 02:23 PM
mr fantastic
Quote:

Originally Posted by chella182
Okay. (I've not posted on here before so forgive me if I'm not typing things properly or whatever).

In one of my assignment questions (I'm a 1st year Maths degree student, by the way) I have to prove the formula for the surface area of a sphere using the surface of revolution formula. So basically, my answer should be 4*Pi*r². I'm having difficulty with it, though. So far within the integral I have this horrible looking thing...

(I thought posting a picture would be easier than trying to type it all out). Those should be x²s on the ends of the fraction by the way - the bracket kinda cuts that off. That has to simplify a little, otherwise, I've done something wrong in my calculations prior to arriving at that.

Any ideas?

Cheers.

Using the formula for rotating about the x-axis (and using some symmetry) you have

$S = 4 \pi \int_{0}^{R} \sqrt{R^2 - x^2} \sqrt{1 + \frac{x^2}{R^2 - x^2}} \, dx$

$= 4 \pi \int_{0}^{R} \sqrt{R^2 - x^2} \, \sqrt{\frac{R^2}{R^2 - x^2}} \, dx$

$= 4 \pi \int_{0}^{R} R \, dx = 4 \pi R \int_{0}^{R} dx$.
• Dec 7th 2008, 08:50 AM
chella182
That's not the formula I use for the surface of revolution (Worried)
I use $2\pi\int_{-R}^{R}y\sqrt{1+y'^2},dx$
• Dec 7th 2008, 09:01 AM
Krizalid
Quote:

Originally Posted by chella182

How have you gone from your second step to the last one?

Algebra.

Quote:

Originally Posted by chella182

And also, doesn't the formula for 'S' have $2\pi$ not $4\pi$? (Worried)

Oh, and the limits in the integral in my question are $-R$ to $R$ I think.

mr fantastic claims for symmetry. Have you ever heard about even functions?

(I see you edited your post. Glad you find it by yourself.)
• Dec 7th 2008, 09:05 AM
chella182
Sorry, I edited again. You've used a different formula to me for the surface of revolution.
• Dec 7th 2008, 09:08 AM
Krizalid
He just applied the formula for $y=\sqrt{R^2-x^2}$ an used symmetry.
• Dec 7th 2008, 09:17 AM
chella182
Ah okay, I think I'm getting it now.
So when you have $2\pi\int_{-R}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx$ the $\sqrt{R^2-x^2}$ terms just cancel to leave $\sqrt{R^2}$ in the integral, yes?
• Dec 7th 2008, 03:13 PM
mr fantastic
Quote:

Originally Posted by chella182
Ah okay, I think I'm getting it now.
So when you have $2\pi\int_{-R}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx$ the $\sqrt{R^2-x^2}$ terms just cancel to leave $\sqrt{R^2}$ in the integral, yes?

Yes.

And $2\pi\int_{-R}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx = 4\pi\int_{0}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx$ by symmetry (since the integrand is clearly an even function).

Note: Questions at this level will assume an ability to perform basic algebraic manipulation and to recognise the application symmetry.