Finding volume though revolution

[tnutty]

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

 

[Billy Bob]

Shells look easier than discs. Do you know shells?

 

[tnutty]

Yes. 2*pi*r*f(x) dx

 

[Billy Bob]

In this case, the shells are "sideways" so it's 2*pi*r*f(y) dy. And the r is y.

 

[tnutty]

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

?

 

[Billy Bob]

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.

 

[tnutty]

2pi*y*f(y)