# Homework Help: Surface of recolution of a curve integrating wrong :-(

1. Nov 15, 2009

### darkwolfe5

1. The problem statement, all variables and given/known data

the curve y= sqrt(x² +1) , 0 ≤ x ≤ sqrt(2) is revolved about the x-axis to generate a surface. Find the area of the surface of revolution.

2. Relevant equations

A = 2π ∫ f (x) * sqrt[1+f ' (x)] dx

3. The attempt at a solution

I've gotten down to 2π ∫ sqrt(2x²+1) dx
and I know that's correct thus far (confirmed using my calculator)
calc says: 13.14153998

but once I integrate, I get 2π[ ⅔ * (2x²+1)^(3/2) with the limits from 0 to sqrt(2)
Once I substitute the sqrt(2) I get something way out of the park.
calc says: 46.832.....

Did I do my integration right?

2. Nov 16, 2009

### clamtrox

You can check if it's right by differentiating. After you have checked and seen that it's not, the easiest way forward is to switch variables to polar coordinates.

3. Nov 16, 2009

### darkwolfe5

We havent gotten to polar coordinates yet...that's what we're supposed to start on Monday :-(

4. Nov 16, 2009

### clamtrox

Oops, I mean hyperbolic coordinates (that probably doesn't help, but still :)

5. Nov 16, 2009

### darkwolfe5

well I've had someone on another site tell me:

<quote>
∫ √( 2x² + 1 ) dx

= √2* ∫ √( x² + 0.5 ) dx

= √2* { (x/2)*√( x² + .5) + (0.5/2)*ln [ x + √( x² + 0.5) ] } between x = 0 and x = √2.
</quote>

But that doesnt seem to calculate right to me either.