Volume of a solid

**redman223** · September 25th 2008, 03:09 PM

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

Rotated about the y-axis.

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?

**skeeter** · September 25th 2008, 04:49 PM

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

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

**redman223** · September 25th 2008, 05:06 PM

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

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

**skeeter** · September 25th 2008, 06:25 PM

are you familiar with using the method of cylindrical shells?

**redman223** · September 26th 2008, 07:46 AM

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?

**skeeter** · September 26th 2008, 05:03 PM

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

I recommend you research the method.

Thread: Volume of a solid

LinkBack

Thread Tools

Display

Volume of a solid

Similar Math Help Forum Discussions

Volume of a solid

Volume of a solid

Volume of a solid

Volume of a solid

Volume of Solid

Search Tags