Setting up an Integral for the area of a surface of revolution

  • Thread starter darkblue
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  • #1
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Homework Statement



Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve y=xe-x 1=<x=<3 about the y-axis.

Homework Equations



S=integral from a to b x 2pix ds where ds=sqrt(1+(dy/dx)2)dx

The Attempt at a Solution



The first thing I tried to do is solve for the equation in terms of x, and then use the equation above. I figured it makes sense to solve for x since we are rotating the curve about the y-axis. I wasn't able to solve for x, so then I tried to use this method in my textbook where you leave x as it is, and then substitute u for whatever is within the square root sign in such a way that you can eliminate x. I tried to do that, but its turning into a mess since you get 1+(e-x-xe-x)2 underneath the square root and I don't really see how substitution could be used here...any ideas?
 

Answers and Replies

  • #2
35,287
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All you need to do is set up the integral. Don't worry about trying to evaluate this integral.
 
  • #3
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So does this mean that the way I have set it up is correct? I had a feeling it wasn't right because I couldn't see what steps I'd take next in the event that I had to solve it.
 
  • #4
35,287
7,140
Seems to be OK, but I'm a little rusty on these surface area integrals. You have an extra x in what you typed, though, right after b. Did you mean for that to be there?
darkblue said:
S=integral from a to b x 2pix ds where ds=sqrt(1+(dy/dx)2)dx
 
  • #5
30
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oops, i meant to put a "*" for multiplication.

Thanks for your help!
 

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