# Disk Method in Calculus: Formula & Examples

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• 0:03 The Disk Method in Calculus
• 3:05 Disk Method Example
• 4:14 Rotation Around the ''Y''-Axis
• 4:39 The Washer Method
• 5:00 Lesson Summary

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Lesson Transcript
Instructor: David Karsner
The disk method in calculus is a means to finding the volume of a solid that has been created when the graph of a function is revolved about a line, usually the x or y axis. Learn more about it in this lesson.

## The Disk Method in Calculus

Do you remember this toy from your early childhood?

This stack of rings is a great illustration of how the disk method in calculus works. We can find the volume of the cone (the stack of rings) by adding together the volumes of the purple ring to the blue ring to the light blue ring to the green ring and so on.

In other words (less colorful words), the disk method is the process of finding the volume of an object by dividing that object into many small cylinders/disks and then adding the volumes of these small disks together. The radius of the cylinder is given by a function f(x) and the height is the change in x. If we find the limit of the volume as the change in x goes to zero and the number of disk goes towards infinity, then we'll have the actual volume of the object and not just an estimate. This volume is the anti-derivative of the square of the function f(x) from point a to point b, multiplied by 3.14. This lesson will contain an explanation of the process, an example of the process, and a few modifications of the concept.

If you take a section from a to b of the graph of a function f(x) and rotate it around a line, you'll create a three dimensional solid. The volume of this solid can be found using the disk method of integration. The disk method is based on the formula for the volume of a cylinder: V = 3.14hr^2. Imagine a cylinder that is lying on its side. The x-axis is going through its center, the y-axis is up against the left base, the right base is located at x = b, and the top of the cylinder is y = 2.

Notice that the radius in this cylinder goes from the x-axis to the line y = 2, a distance of 2. The height (even though it is laying on its side) is from a to b along the x-axis. The height can be written as b-a or as the change of x. The volume of this cylinder is 3.14(2)^2(change of x). Since this cylinder was created by revolving y = 2 around the x-axis, we can call y = 2 to be our function, f(x). We can then say the volume of this solid is 3.14 times f(x)^2 times the change in x.

Notice that with this solid the function f(x) does not provide a constant value for the radius. Any cylinder that we could impose on this solid would only be an estimate of the volume of the solid. If we divided the f(x) into sections and found the volume of each cylinder, we would get a better estimate when all those disks were added together. The more disks measured the better the estimate would be. If we could measure an infinite number of disks, then we would have the actual volume of the solid. This is the point at which calculus comes into play. Finding the limit of the volume as the height goes to infinitely small is the antiderivative of the square of the function f(x) from a to b times 3.14.

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