Solids of revolution, y axis

[James889]

Hi,
The area
$$e^x-1$$
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.

$$~ 2\pi\int_0^{ln2}x\cdot(1-(e^x-1)$$
$$-2\pi\int_0^{ln2}xe^x-2x$$

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

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

The answer is supposed to be $$2\pi(ln2-1)^2$$
Thanks!

 

[emyt]

where did that 1 come from in your shell method formula?

 

[Dustinsfl]

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