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Surface area obtained by solid of revolution

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Find surface area of solid of revolution obtained by rotating the curve:

    y=x2/40-5lnx

    from x=5 to x=7, rotated about x=-4

    3. The attempt at a solution

    The problem is I know how to do this if I rotated it about x-axis/y-axis, but I have no idea how to do it if the axis is shifted. I did it pretending it was rotated about the y-axis (since x=-4 is parallel to the y-axis), and my answer came out wrong.

    I used:

    SA=∫2∏x ds, where ds=√(1+[f'(x)]2) dx

    So I found the derivative of the function, everything was fine. I squared and square-rooted, found the integral to be:

    ∏/10 ∫ x^2+100 dx (evaluated from 5 to 7)

    but i didn't use the piece of information "axis of rotation is x=-4" anywhere in my solution. I don't really know how to go about using it, please help?
     
  2. jcsd
  3. Mar 3, 2012 #2
    Do I just change the formula from SA=∫2∏x ds to SA=∫2∏(x+4) ds since the axis is shifted to the left 4 units, so the distance to the axis of rotation would be 4 units longer?

    Although the answer is still not correct...

    or do I change the function so that it goes from y=x^2/40-5lnx to y=(x-4)^2/40-5ln(x-4)? Or none of the above?:/
     
    Last edited: Mar 3, 2012
  4. Mar 3, 2012 #3
    Deleted.
     
    Last edited: Mar 3, 2012
  5. Mar 3, 2012 #4
    Thanks for answering, but I don't understand the second part of the integral. Why is it (x^2/40-5lnx) dx?

    I thought it should be:

    ∫2∏(x+4) ds

    where ds is

    √(1+[f'(x)]^2) dx, and f'(x) is the derivative of x^2/40-5lnx?
     
  6. Mar 3, 2012 #5
    Deleted.
     
    Last edited: Mar 3, 2012
  7. Mar 3, 2012 #6
    OHHH NEVERMIND! I'm stupid, you're looking for surface area, not volume. Ignore everything I've said, sorry!
     
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