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### Course: AP®︎/College Calculus BC>Unit 8

Lesson 12: Volume with washer method: revolving around other axes

# Washer method rotating around vertical line (not y-axis), part 1

Setting up the definite integral for the volume of a solid of revolution around a vertical line using the "washer" or "ring" method. Created by Sal Khan.

## Want to join the conversation?

• How does one determine whether it's best to integrate with respect to x vs. with respect to y? It isn't always obvious, at least not to me...
• If rotating around x-axis or line parallel to x-axis: Integrate with respect to x.
If rotating around y-axis or line parallel to y-axis: Integrate with respect to y.
• what if the rotating axis is not parallel to either the x axis or y axis? For example, x+y=2 or something.
• I dont think there is a quick solution to that. You would have to transform your function such that the axis of rotation becomes the x-Axis.
• At , the outer radius is defines as 2-y^2.
Can anybody explain why ones takes 2 - the function. My first guess was the function + 2 so that i would reach the line where x =2. This is wrong, but i can´t say i really understand why.

And, taking 2-y^2 would that be the gap between the undefined point on the x-axes and x=2? Where is this on our shape?
Thanks.
• At this stage in the process we're trying to find the radius of the outer disk. The radius is the distance from the center of the disk to the perimeter. At the beginning of the video we're told that the figure is created by rotation around the line x=2, so we know the center of the disk is at x=2. The outer edge of the larger disk is at the point x=y^2, so the distance between the two points is 2-y^2.
• how did you come up with the values of the limits of integration right towards the end of the video?
• those are the y values where both functions intersect.
• at how do you get your limits of integration? I understand that you did by inspection, but how would you do it by making the two equations equal to each other ?
• In that example, Sal is integrating with respect to y. To solve for zero and one analytically, we could make the equations equivalent to each other and solve. Here's the example:
sqrt(x) = x^2
x = x^4

This last equation is true only when x = 0 or x = 1.
• What would the solid look like of the sine function around the y-axis? A series of co-joined balls with a radius of pi/2 perhaps?
• Be careful: what you were visualizing is a rotation about the x-axis (not y-axis).

In order to find out how it would look like about the y-axis, you'd have to specify how to construct the area whose rotation will produce the solid. That's because "sin(x)" by itself doesn't produce any area, it's just a continuous line. If, for example, you were to specify that it's "the area between the curve sin(x) and the x-axis", then the rotation about the y-axis would yield a "ripples on a pond" sort of figure.
(1 vote)
• I just want to say thank you for doing this. It is helping me a ton, but I still have one question. How do you know when you don't have to integrate the outer volume but have to integrate the inner volume (and then subtract it) or vice versa? I hope that makes sense...it just seems that sometimes there's only one volume you need to integrate while the other volume is staying constant. I'm having a hard time figuring out when that is. Thank you so much!
• I recommend to watch the exercises and think them through on your own.

As a quick guide,
1. Look at the rotational axis, is it parallel to the x or y-axis.
2.Check the offset ( distance of your axis of rotation)
3.Determine the boundaries.
Integrate and calculate the result.
(Practice makes perfect )
• Do i ever HAVE to use the shell method, or can i always use disc/washer if i want?