# Homework Help: Evaluate the definite integral for the area of the surface.

1. Jun 2, 2010

### lude1

1. The problem statement, all variables and given/known data
Evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

y=(x3/6) + (1/2x), [1,2]

2. Relevant equations
2π∫[r(x)](1+[f'(x)2])

3. The attempt at a solution
First I found the derivative.

f'(x)= (x2/2) + (1/2x2)dx​

And since y is a function of x, r(x) is

r(x)= (x3/6) + (1/2x)​

Then I plug everything in and get

2π∫ [(x3/6) + (1/2x)] * {1 + [(x2/2) + (1/2x2)]2}1/2}dx​

And then I'm stuck. The book tells me that I am suppose to get

2π∫ [(x3/6) + (1/2x)] * [(x2/2) + (1/2x2)]dx​

But I have no idea how they got that. Specifically, I don't know how they got rid of the radical...

2. Jun 2, 2010

### mmmboh

For starters you derived it wrong, the derivative of f(x) is f'(x)=x2/2-1/2x2. Maybe this is the problem. Unless you just wrote your f(x) wrong.

Edit: Yes that is the problem, use the correct f'(x), and then expand f'(x)2 and put everything over a common denominator, and then you can make an equation which is being squared that is equal to that, and then the root cancels out the squares.

Last edited: Jun 2, 2010