# Disk Method about the x axis

• Apr 10th 2012, 04:26 PM
crzyberzu123
Disk Method about the x axis
Does anyone know how to solve x(5-x)^(1/2) about the x axis using the disk method?
• Apr 10th 2012, 04:47 PM
skeeter
Re: Disk Method about the x axis
Quote:

Originally Posted by crzyberzu123
Does anyone know how to solve x(5-x)^(1/2) about the x axis using the disk method?

I assume the area formed between the graph of the function and the x-axis in quadrant I ... ?

$V = \pi \int_0^5 \left(x \sqrt{5-x} \right)^2 \, dx$
• Apr 10th 2012, 05:06 PM
crzyberzu123
Re: Disk Method about the x axis
Would you set the equation equal to x so that you could make everything in terms of y?
• Apr 11th 2012, 12:35 AM
earboth
Re: Disk Method about the x axis
Quote:

Originally Posted by crzyberzu123
Does anyone know how to solve x(5-x)^(1/2) about the x axis using the disk method?

May I re-write the question as I understand it?:

Determine the volume when the area, enclosed by the x-axis and the graph of $y = x \cdot \sqrt{5-x}$, rotates about the x-axis.

Quote:

Originally Posted by crzyberzu123
Would you set the equation equal to x so that you could make everything in terms of y?

The volume is composed by cylindrical discs whose radius is y and whose height is dx. The volume of one of these discs is calculated by:

$V_{single\ disc} = \pi \cdot y^2 \cdot dx$

The complete volume of the solid is the sum of all those discs:

$V=\int (\pi \cdot y^2)dx$

Plug in the term of y and determine the borders of the area and you'll get the equation skeeter has posted.