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lude1
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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...