# 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 $\displaystyle y = \frac {1}{x}$, with 1 $\displaystyle \leq$ x $\displaystyle \leq 2$, about the y-axis"

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

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

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

4. Make that:

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

would become:

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

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

Re-write the lot to see clearly

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