Proving the formula for the surface area of a sphere

**chella182** · December 6th 2008, 11:21 AM

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.

**mr fantastic** · December 6th 2008, 01:23 PM

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

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

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

$= 4 \pi \int_{0}^{R} R \, dx = 4 \pi R \int_{0}^{R} dx$ .

**chella182** · December 7th 2008, 07:50 AM

That's not the formula I use for the surface of revolution

I use $2\pi\int_{-R}^{R}y\sqrt{1+y'^2},dx$

**Krizalid** · December 7th 2008, 08:01 AM

Originally Posted by chella182

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

Algebra.

Originally Posted by chella182

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

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.)

**chella182** · December 7th 2008, 08:05 AM

Sorry, I edited again. You've used a different formula to me for the surface of revolution.

**Krizalid** · December 7th 2008, 08:08 AM

He just applied the formula for $y=\sqrt{R^2-x^2}$ an used symmetry.

**chella182** · December 7th 2008, 08:17 AM

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?

**mr fantastic** · December 7th 2008, 02:13 PM

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.

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