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Finding volume though revolution

  • Thread starter tnutty
  • Start date
  • #1
327
1
Write a definite integral that gives the volume of the solid of revolution formed by revolving the region
bounded by the graph of x = e^(−y^2) and the y-axis between y=0 and y=1, about the x-axis.

Not sure how to set up the problem
 

Answers and Replies

  • #2
392
0
Shells look easier than discs. Do you know shells?
 
  • #3
327
1
Yes. 2*pi*r*f(x) dx
 
  • #4
392
0
In this case, the shells are "sideways" so it's 2*pi*r*f(y) dy. And the r is y.
 
  • #5
327
1
Then if f(x) = e^(-y^2)

then let y = e^(-y^2)

ln(y) = ln( e^(-y^2))

ln(y) = -x^2

-sqrt( ln(y) ) = x

- ln(y)^.5 = x

so its

V = 2*pi*y*-ln(y^.5)

1
INT [V]
0

?????
 
  • #6
392
0
No, much easier than that. (Did you draw the graph? I hope so!) Just f(y)=e^(-y^2).

No solving for anything required. This is a very quick problem.
 
  • #7
327
1
2pi*y*f(y)
 

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