# Volume of a rotated region?

How can I find the volume of the solid generated when the region enclosed by $y=0, x=0, x=1$ and $(1+e^-2x)^0.5$ is rotated through 360 degrees about the x axis?

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Sorry, fairly big change to the question there... I've replaced the 3 with an e. –  user8078 Mar 10 '11 at 17:56
That makes no difference with the method you'll use to find the volume. –  Matthew Conroy Mar 10 '11 at 18:05

I'm not completely sure that I understand what you mean, but try this: $$\pi\int_0^1 f(x)^2 \text{ d}x$$ where $f(x)$ in this case is your function $(1+e^{-2x})^{\frac{1}{2}}$.

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I see that Matthews answer explains why this is the way to go. –  please delete me Mar 10 '11 at 17:56
why pi not 2 pi? –  user8078 Mar 10 '11 at 18:00
Because the area of a circle is given by $\pi r^2$. –  please delete me Mar 10 '11 at 18:06
Brilliant, I get it know. Cheers. –  user8078 Mar 10 '11 at 18:09

The volume of an object can be found by integrating the cross-sectional area through all cross-sections.

A cross-section of this solid, taken perpendicularly to the $x$-axis at a distance $x$ from the $y$-axis, is a disk with radius $(1+e^{-2x})^{0.5}$. Express the area of this disk as a function of $x$, and then integrate this function from $x=0$ to $x=1$.

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