Surface of recolution of a curve integrating wrong :-(

[darkwolfe5]

Homework Statement

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.

Homework Equations

A = 2π ∫ f (x) * sqrt[1+f ' (x)] dx

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?

 

[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.

 

[darkwolfe5]

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.

We haven't gotten to polar coordinates yet...that's what we're supposed to start on Monday :-(

 

[clamtrox]

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

 

[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 doesn't seem to calculate right to me either.