# Thread: volume of solid revolution

1. ## volume of solid revolution

Hi I just need a check on if the answer I got is correct.
"Find the volume of the solid obtained by rotating the region between $y = \frac {1}{x}$, with 1 $\leq$ x $\leq 2$, about the y-axis"

y(1) = 1
y(2) = $\frac{1}{2}$
x = $\frac{1}{y}$

$\pi \int$ (from 1 to $\frac {1}{2}$) , $\frac{1}{y^2}$ dy + $\int$(from $\frac{1}{2}$ to 0) $2^2$ dy
= $\pi [-\frac{1}{y}]\frac{1}{\frac{1}{2}}$ + $2^2[y] \frac{\frac{1}{2}}{0}$
= $\pi$ [ -1 + 2 + 2]
= $3\pi$

Any help would be great thanks.

2. Integrate with respect to x and you only have to calculate one integral (for cylindrical shells).

Alternatively, your method is fine if you either now calculate and remove a central cylindar (radius one around the y axis) or go back and adust the integrands so that your Reimann strips aren't coming right down to the wire (the y axis) but only to x = 1, and so that they rotate to form washers instead of disks. (Look again at your sketch.)

3. .. + $2^2[y] \frac{\frac{1}{2}}{0}$
So that would become:
+ $1^2[y] \frac{\frac{1}{2}}{0}$
because of 2-1=1 right?

4. Make that:

... $+ [4y] _0^{\frac{1}{2}}$

would become:

$+ [3y] _0^{\frac{1}{2}}$

because of $2^2 -1^2 = 3$

Re-write the lot to see clearly

And see http://en.wikipedia.org/wiki/Disk_integration