# Thread: Volume of a solid

1. ## Volume of a solid

$
y= \frac {1}{x^5}~
y=0~
x=4~
x=6~
$

I keep coming up with:
$
\frac {1}{9*6^9} - \frac {1}{9*4^9}
$

I don't know what I am doing wrong. pi*r^2 right (which would be pi*(1/x^5)^2)? So it should just be the integral of:

$
\int_4^6 \frac {1}{x^{10}}~dx
$

Right?

2. rotation is about the y-axis, so, using cylindrical shells ...

$V = 2\pi \int_4^6 \frac{1}{x^4} \, dx$

3. Could you explain those steps? I have trouble when the y-axis is involved.

Where did the 2pi and the x^4 come from?

4. are you familiar with using the method of cylindrical shells?

5. I am, but I am used to rotating it about the x-axis or y=something other than zero. So rotating it about x=something other than zero or the y-axis throws a wrench into my methods.

Normally I would just use the curve of the line as the radius and in the pi*r^2 equation right? Then subtract the smaller volume from the larger one and take the integral of that area right?

6. you are thinking of washers ... not cylindrical shells.

I recommend you research the method.