# Volume, washer method and shell method

1. Dec 2, 2003

### JKLM

I don't understand how to set up the washer and shell equations. When you are given the function and the line to rotate it around, or two functions and a line.

2. Dec 3, 2003

### HallsofIvy

Suppose we are given an area bounded by certain curves in the xy-plane. That area is rotated around an axis to form a solid. Each point of the figure is rotated in a circle. If you slice the solid perpendicular to the axis you will see a "cross section" (the area revealed by the slice) that is either a full disk or a "washer" (the area between two circles. It's easy to calculate the area of a circle and you can imagine the solid as made of a lot of very thin circles.

For example, suppose the part of the parabola y= x2 for 0< x< 1 is rotated around the y-axis. You get a parabolic solid and if you slice through it perpendicular to the y-axis, at any y, the "cut end" is a circle.

In fact, imagine that this figure is a potato! Slice it into very thin slices to make potato chips. Each slice is a circle (disk, more properly). Its area is &pi;r2 and, taking "h" to be its thickness, its volume is &pi;r2h.

If we put the slices back together we could reform the potato and the volume of the potato is the sum of the volumes of the slices.

Of course, "r", the radius of the circle, varies from slice to slice. If our potato were really shaped like y= x2 rotated around the y-axis, then, since we are slicing perpendicular to the y-axis, "r" is equal to the x value: x= &radic;(y).

The area of the slice is then &pi;r2= &pi;y and, setting this up as a "Riemann sum", the thickness, because it is measured along the y-axis, is dy: the volume of each slice is &pi;y dy and the sum of all the slices (as we imagine the slices becoming infinitesmally thin) becomes the integral of &pi;y dy from y=0 to 1:
(1/2)&pi;y2 (evaluated between 0 and 1)= &pi;/2.

3. Dec 3, 2003

### JKLM

i understand the theory behind the washer method, but what i dont get is the application of the theory, i dont understand how to set up the equations

4. Dec 3, 2003

### AngelofMusic

With the washer/disc method, if you're given a function f(x) and you want to rotate it, say around the x-axis - then you simply take a representative rectangle with its base on the x-axis (and its length going up to the function).

Then the width/base of the representative rectangle will be dx, while the length will be f(x).

When you rotate the function about the x-axis, that rectangle will essentially become the radius of the circular disc. So,

$$V=\pi\int_{a}^{b}f(x)^2dx$$

For something with two functions, just take:

$$V=\pi\int_{a}^{b}f(x)^2-g(x)^2dx$$

Where f(x) is a function above g(x).

These are the steps I use to set up the equations for the disc method. Does that help?

5. Apr 13, 2010

### the2112

explains everything

Last edited by a moderator: Sep 25, 2014
6. Apr 19, 2010

### phys-lexic

this helps me when writing the integral... hey vex... if the drawn washer/disk is vertical (goes up and down) integrate over x (use dx), if it is horizontal (goes left and right) integrate over y (use dy)