# Washer vs Disk Method

• Jun 20th 2014, 03:00 PM
Jason76
Washer vs Disk Method
How can I tell which to use (to find volume) without seeing a graph or graphing a bunch of points? Is it possible to look at the x or y intercept?

In another scenario, what if it said the graph should be revolved around y = 10. However, in this case finding a x intercept wouldn't work. How could you tell algebraically if the graph touched y = 10, or not?

$y = 5e^{-x}$

$y = 5$

about y = 10, but we don't know if it touches the y = 10 (part of the x axis).

Area $= 5e^{-x} - 5$
• Jun 20th 2014, 08:32 PM
Prove It
Re: Washer vs Disk Method
In some cases you just have to try both ways and see which way works.

I don't understand your other question.
• Jun 21st 2014, 06:45 AM
HallsofIvy
Re: Washer vs Disk Method
The washer and disk methods are basically the same. Strictly speaking you should only use the "disk" method when you have a full "disk". That is, the area being rotated around the disk goes all the way to the axis of rotation.

Example: find the volume of the solid generated by rotating the area between the curve $y= 4- x^2$ and the x-axis ( $y= 0$) around the x-axis. From any point on the x-axis, between x= -2 and x= 2, the entire line from the x-axis up to $y= 4- x^2$ is rotated and will form a full "disk". Use the disk method: a disk with radius $4- x^2$ will have area $\pi(4- x^2)^2$. If we take "dx" as the thickness, the disk will have volume $\pi(4- x^2)^2 dx$ so the entire volume is the integral $\pi \int_{-2}^2 (4- x^2)^2 dx$.

But suppose the problem is to find the volume of the solid generated by rotating the region between $y= 4- x^2$ and the x-axis around the line x= -1. Now, for each x between -2 and 2, the region rotated does NOT give a "disk". The section from y= 0 to y= -1 is NOT rotated so we have a "hole" in the middle of the disk- a "washer". A washer with "outer radius" R and "inner radius" r has area $\pi (R^2- r^2)$- that is, just the area of the disk out to radius "R", $\pi R^2$ minus the area of the "hole", $\pi r^2$. In this problem the "outer radius", at each x, is the distance from y= -1 to $y= 4- x^2$ so that $R= 4- x^2- (-1)= 5- x^2$ and the radius of the "hole" is the distance from y= -1 to y= 0, $r= 1$. The area of the "washer", the area of the large disk minus the area of the hole, is $\pi ((5- x^2)^2- 1)$ and the volume is given by the integral $\pi \int_{-2}^2 ((5- x^2)^2- 1) dx$.

Of course, just as the area of a washer is the area of the large circle minus the area of the "hole" so we could also find the volume of the solid as the volume of the large region minus the volume of the hole. The volume of the larger region, using "disks" is $\pi\int_{-2}^2 (5- x^2)^2 dx$ and the volume of the hole, again using "disk" is $\pi\int_{-2}^2 1^2 dx= 4\pi$, the volume of a cylinder with radius 1 and height 4.

Obviously those two methods, using "washers" or doing two "disk" calculations and subtracting will give the same result: $\pi\int (R(x)^2- r(x)^2)dx= \pi\int R(x)^2 dx- \pi\int r(x)^2dx$.
• Jun 21st 2014, 07:28 AM
catenary
Re: Washer vs Disk Method
In terms of learning integral calculus it is nice to be able to calculate volume using the washer method and then calculate that same volume using the disk method and then seeing that they both yielded the same result. It is very useful for building confidence in your answers and verifying them, and for understanding better the relationship between the two. Useful when taking exams and you have extra time, it is like reviewing your work.
• Jun 24th 2014, 01:14 PM
Jason76
Re: Washer vs Disk Method
As of now I think the best way is just to graph it. That way I can see if it touches or crosses the axis of revolution. I don't see a shortcut.