Mathematics
calculate improper integral

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Final Answer

Hello!

y = e^(-x), x = -lny.

The volume is pi*int_0_1_[(lny)^2]dy.

The indefinite integral made by integrating by parts twice:

int[(lny)^2]dy = (du=dy, u=y, v=(lny)^2, dv=(2lny/y)dy) =

y*(lny)^2 - int[y*2*(lny/y)]dy = y*(lny)^2 - 2*int[lny]dy =

(du=dy, u=y, v=lny, dv = dy/y) =

y*(lny)^2 - 2*(y*lny - int[y/y]dy) =

y*(lny)^2 - 2y*lny + 2y (+C).

So the volume is pi*[y*(lny)^2 - 2y*lny + 2y]_0_1.

At y=1 it is clearly 2*pi (ln1=0).
At y=0 we have to determine limits y*lny and y*(lny)^2 as y->+0. They are both 0, I can prove this.

The answer is 2*pi.

Please ask if anything is unclear.

Borys_S (8127)
University of Virginia

Anonymous
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