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Solids of revolution, y axis

  • Thread starter James889
  • Start date
  • #1
192
1
Hi,
The area
[tex]e^x-1[/tex]
Is rotated about the y axis, bounded by y=1, x=0 and x=ln2 find the volume of the solid.

And i am clearly making something wrong, so if anyone could verify my work.

[tex]~ 2\pi\int_0^{ln2}x\cdot(1-(e^x-1)[/tex]
[tex]-2\pi\int_0^{ln2}xe^x-2x[/tex]

Integration:
u=x, du=1
dv=e^x, v=e^x

[tex]\int xe^x -2x= xe^x -e^x -x^2[/tex]
[tex]e^x(x-1)-x^2\bigg|_0^{ln2} = 2\pi \cdot ln2-1 -(ln2)^2[/tex]

The answer is supposed to be [tex]2\pi(ln2-1)^2[/tex]
Thanks!
 

Answers and Replies

  • #2
217
0
where did that 1 come from in your shell method formula?
 
  • #3
699
5
When you integrate you should obtain : 2*pi*(ln22-2*ln2+1)

That then can be the factored to the book's answer.

You lost a 2 with the ln2
 

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