Evaluate the definite integral for the area of the surface.

[lude1]

Homework Statement

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]

Homework Equations

2π∫[r(x)](1+[f'(x)²])

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...

 

[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.