If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Calculus, all content (2017 edition)>Unit 6

Lesson 9: Disc method

# Generalizing disc method around x-axis

Generalizing what we did in the last video for f(x) to get the "formula" for using the disc method around the x-axis. Created by Sal Khan.

## Want to join the conversation?

• Why can't you just measure out each disc?
• You have infinite disks, and therefore, it would take a very, very long time...
• i do not understand how does the definite integral comes to know of the exact variation of radius and how to add em up ,
• in technical terms, the definite integral is the Summation of the terms of a function with the first term equal to f(a), the second to f(a+dx), third to f(a+2dx) ..., term j equal to f(a+n.dx)=f(b), when the dx approaches 0, or you could say, when n approaches plus infinity. Each term is multiplied by dx, if you draw that you'll see that this Sum will give you the sum of many areas of rectangles. When the number of rectangles approaches inf. you will have a definite integral. Fortunately, someone proved that all of that I said equals to F(b) - F(a), when F(x) is the antiderivative of f(x)
• When finding the volume of the "disk", why do you use the formula for the area of a circle, and not the volume of a cylinder?
• Take a look at the Riemann sum that is a step before the definite integral. You can see there are physical cylinders that make up the volume for n= positive real number. You can use this formula as a better and better approximation for the volume as n becomes larger and larger.
I think your question involves the limiting case that defines the definite integral as you let n go to infinity. The width of each cross sectional area "shell, washer...any shape" becomes infinitely small (dx). Limit as delta x approaches zero and the number of cross sectional areas you are adding up becomes infinite.
• Can you use this for the y-axis as well?
(dy)
• Yes, you can use any of the methods around either axis. (That is, disks, shells, and washers.) You just have to keep in mind that your variable of integration has changed; so for example, if you're integrating dy and rotating around the x-axis, you'd have to use shells rather than using disks like you would if you were integrating dx. Hope that makes sense.
• What if the x-axis of revolution is moved? How would you deal with it? Any example videos please. Thank you.
• You would still need to figure out the radius of the disks. The radius is the distance between TWO lines: the function f(x) and the axis of rotation. So if you were rotating around the line y=-2 (which is just shifting down two units from how it is now), then the radius of each disk would be x^2 + 2.
• So at is the bottom radius of the taken disk not bigger than the top radius? Is it not more something like a section of a cone rather than a cylinder? Or is it taken like a cylinder because of the limit?
• The radii are different, but as the height becomes infinitely small, so does the difference in radius. It's much like the "problem" with Riemann sums, where the edges of the rectangles stick out to one side or the other of the function graph. As the rectangles get narrower and narrower, the error disappears.
So yes, the limit makes it a cylinder.
• In the examples shown thus far f(x) is greater or equal to 0, but what of we had a cubic that dipped below the x-axis, would we still be able to use that formula?
• Since you square r in the formula, it doesn't matter if the function dips below. So yes!
• Hi Sal. A small question. Is there way to calculate the equation of the figure when y=x^2 is rotated about the x-axis? If there is a video could you please share it?
• Say that the axis that we revolve our function around is also revolving around another axis. How would one calculate that?
• Greetings!

Actually, it could not be an ellipse. You are rotating your y or (f(x)) value around the x axis. The absolute value of f(x) never changes, so the radius is constant as it rotates around the x axis.

Thanks!