Volume of revolution. Could someone check this please?

[03125]

Homework Statement

The region bounded by y=e^{-x^{2}}, y=0, x=0, and x=1 is revolved about the y-axis. Find the volume of the resulting solid.

Homework Equations

integral from a to b of pi*f(y)^2

The Attempt at a Solution

If y=e^-x^2 and I am revolving about the y-axis then I need to rewrite f(x)
to f(y)

y=e^-x^2

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

ln(y)=-x^2

-ln(y)=x^2

sqrt(-ln(y))=x

f(y)=sqrt(-ln(y))

Then I integrate f(y) as my radius such that:

integral of(pi*r^2) from 0 to 1

pi is a constant so it gets pulled out, the integral of
-ln(y)=-(y*ln(y)-y+c)

so pi*-(1*0-1+c) => pi (because the c's will cancel)

plugging in 0, ln(0)*0-0+c = 0 (and the c's cancel)

so I am believe that the answer to this should be pi, but the program that I use to submit my homework answers is telling me that the answer is not pi.

I also checked this on my TI89 and got pi, but perhapse I made a mistake on both of my methods.

Any feedback is appreciated, thank you!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Dick]

It's not pi. If you are integrating dy the lower limit for integrating -ln(y) isn't 0. The upper limit for x changes when the curve hits x=1. You need to split it into two parts. You can also do it in one go if you integrate dx using shells. You'll get the same answer.

 

[03125]

I was able to get it, thank you very much for your help =)