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?