# Volume Integration Question !!!! (1 Viewer)

#### flowerp

##### Member
Hi everyone, I would like help with the following question please:

A horn is generated by rotating the curve y = 1 + e^-x about the x-axis between x = 1 and x = 3. Find its volume to three decimal places.

I figured that the volume can be attained by finding the integral of (1+e^-x)^2 from x=1 and x=3 and then multiplying it by pi, however, I do not understand why we do not subtract the volume of the cylinder formed by the asymptote? How exactly does integration work here to find the volume, thanks!

Thank you sooo much !!!

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#### fan96

##### 617 pages
The standard formula for volumes of solids with similar cross-sections is

$\bg_white V = \pi \int_a^b r^2\, dh$

(in this case, $\bg_white r = y, \,\,h=x$)

This can be thought of as approximating the volume of the solid with several cylindrical slices, and then making these slices thinner and thinner.

The volume of each slice is $\bg_white \pi y^2$, as the radius of the slice is exactly the $\bg_white y$ value of the function.

For example, suppose we have some solid formed by revolution of $\bg_white y=f(x)$ across the $\bg_white x$-axis.

By taking $\bg_white n$ slices, the volume of the solid on the unit interval is

$\bg_white V = \lim_{n \to \infty} \frac \pi n f(1/n)^2 + \frac \pi n f(2/n)^2 + ... + \frac \pi n f(3/n)^2$

$\bg_white = \pi \lim_{n \to \infty} \frac{ f(1/n)^2}n+ \frac{ f(2/n)^2 }n + ... + \frac{ f(1)^2 }n$

This is a Riemann sum, (which is how the integral is defined) so it should be clear that this ends up being

$\bg_white V= \pi \int_0^1 f(x)^2\, dx$

Last edited:

#### flowerp

##### Member
The standard formula for volumes of solids with similar cross-sections is

$\bg_white V = \pi \int_a^b r^2\, dh$

(in this case, $\bg_white r = y, \,\,h=x$)

This can be thought of as approximating the volume of the solid with several cylindrical slices, and then making these slices thinner and thinner.

The volume of each slice is $\bg_white \pi y^2$, as the radius of the slice is exactly the $\bg_white y$ value of the function.

For example, suppose we have some solid formed by revolution of $\bg_white y=f(x)$ across the $\bg_white x$-axis.

By taking $\bg_white n$ slices, the volume of the solid on the unit interval is

$\bg_white V = \lim_{n \to \infty} \frac \pi n f(1/n)^2 + \frac \pi n f(2/n)^2 + ... + \frac \pi n f(3/n)^2$

$\bg_white = \pi \lim_{n \to \infty} \frac{ f(1/n)^2}n+ \frac{ f(2/n)^2 }n + ... + \frac{ f(1)^2 }n$

This is a Riemann sum, (which is how the integral is defined) so it should be clear that this ends up being

$\bg_white V= \pi \int_0^1 f(x)^2\, dx$
Thank you very much !!!!