anonymous
  • anonymous
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^4, y=0, x=2, x=6, about y=-4
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I would use the method of cylindrical shells. At any f(x) value between 2 and 6, what is the radius from that y point to y = -4?
anonymous
  • anonymous
Which method u prefer? shell or washer?
anonymous
  • anonymous
I just need help setting it up to integrate then i can take it from there...

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anonymous
  • anonymous
The setup of the integral depends on the method you use....
anonymous
  • anonymous
^ I didn't see your comment.. the washer method :)
anonymous
  • anonymous
Why you choose washersssssss D:
anonymous
  • anonymous
isn't easier? or not? usually for me it is easier..
anonymous
  • anonymous
i agree that washer is appropriate here.... hang on....
anonymous
  • anonymous
You guys have fun with that >_>
anonymous
  • anonymous
|dw:1361159214153:dw|
anonymous
  • anonymous
Can you give me the expression for R and r?
anonymous
  • anonymous
errr.. ;/ crap.
anonymous
  • anonymous
???
anonymous
  • anonymous
no... i don't think it's that.. :)
anonymous
  • anonymous
haha give me a sec
anonymous
  • anonymous
consider this...|dw:1361159501396:dw|
anonymous
  • anonymous
would it be 1/x^4-4
anonymous
  • anonymous
squared
anonymous
  • anonymous
for R, shouldn't it be 1/x^4 + 4 ???
anonymous
  • anonymous
- - + .. err yeah my bad
anonymous
  • anonymous
ok... you also agree that r = 4 (inner radius) ???
anonymous
  • anonymous
yupp it is
anonymous
  • anonymous
so... \(\large dV=\pi(R^2-r^2)dx \) so \(\large V=\pi \int_2^6(R^2-r^2)dx \) you replace the expressions for R and r ....
anonymous
  • anonymous
sorry... i gotta leave... :(
anonymous
  • anonymous
i'm pretty sure what we have is correct....

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