Surface area obtained by solid of revolution

  • Thread starter lillybeans
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  • #1
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Homework Statement



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

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?
 

Answers and Replies

  • #2
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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:
  • #3
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Deleted.
 
Last edited:
  • #4
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You are correct in that the radius is equal to (x+4).

The volume should be equal to:

∫ from 5 to 7 2π(x+4)(x^2/40-5lnx) dx

Is that what you did?
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?
 
  • #5
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Deleted.
 
Last edited:
  • #6
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OHHH NEVERMIND! I'm stupid, you're looking for surface area, not volume. Ignore everything I've said, sorry!
 

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