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

Lesson 10: Volume with disc method: revolving around other axes

# Disc method rotating around vertical line

Volume of solid created by rotating around vertical line that is not the y-axis using the disc method. Created by Sal Khan.

## Want to join the conversation?

• What is a 'principle root?' Is that just another name for a square root?
• Principle root is the positive square root. For example, √25 asks the question "what number when multiplied together gives 25?" This question has two answers: 5 and -5. The principle root of 25 however is unambiguously 5. Unless you see ± before the radical symbol, it usually means the principle root.
• The diagram is confusing to me because the parabola at -2 would be at y = +3 and revolve from the left piece to a cup or from the right piece to an inverted rim. This diagram looks like the parabola was transformed so that it minimum occurs at -2,-1.
• You aren't the first person to find this example confusing, and I suspect Sal would spend a little more time explaining the setup, or maybe change his diagram a bit, if he were to redo it. The main thing you need to understand for proper visualization is that we/re using only the part of the parabola from x = 0 to x = 2 (which translates into the right branch of the parabola from y = -1 to y = 3).

If that clue isn't enough to give you a clear visualization, step away from this example for a minute and imagine a different one, where we have the line segment traced by x = 2 from y = -1 to y = 3, and we rotate it around the vertical line at x = -2. This would simply be a cylinder centered at x = -2. Now imagine that we change the vertical line segment that we rotated by bending in the bottom of it, not all the way to the center of the cylinder (at x = -2) but only to x = 0, and we rotate that curve around the same vertical at x = -2. The bottom of the figure would still have a flat region, just as the cylinder did, but now the flat region is smaller because its radius extends only from x = -2 to x = 0. That's the "gumdrop-shaped" figure Sal drew in his diagram.

To give it 3 dimensions he drew it at an angle instead of a straight-on cross section. Drawing it this way called for the bottom of the figure to be shown as curved instead of straight as it would be in a cross section, and that makes it look as if the parabola continues downward to the left from x = 0, when in reality the parabola ends at x = 0 and the curved line to the left of that point is a perspective view of the round, but flat, bottom of this figure.
• Why is this evaluated using the disk method and not the washer method? If you're evaluating it from -1 to 3 on the y-axis, wouldn't that leave a hollowed out section in the middle around the axis of rotation?
• Yeah I kinda agree with you in this, it wasn't really clear for me too till I watched it couple times, but Sal did mention that he is integrating about the "rotational axis" which is in this case x=-2, so it's a solid face shape and there is no hole so he used the disk method not the washer method! hope I didn't confused you more! :/
• considering y=x^2 - 1 is a parabola with a local minimum at x = -1, wouldn't the solid created by rotating around x=-2 require the shell method?
• You could do this with the shell method (integrating in x)--or you could solve for x = sqrt(1+y) and then use the disk/washer method (integrating in y).

The shell method is only required when it is not possible to solve for x in terms of y (though sometimes it is easier anyway).
• How do we get -1 and 3?
• Hi!

We rotated y = x^2-1 around the line x = -2. x = -2 is parallel to our y-axis. Therefore, Sal picked 2 points on the y-axis as our interval (y = -1, y = 3).

Think about it, the sum of all the discs that are being created when we let y = x^2-1 rotate around the line x = -2 is the total volume of the figure created when y = x^2-1 rotated around x = -2. This is basically the same thing as letting y = x^2-1 rotate around the y-axis but instead, we rotated it around a line that is parallel to it, in this case x = -2. The difference is that the radius becomes different but the same sort of discs are created inside of the figure.

To answer your question: By picking -1 and 3 we will calculate the volumes of all the discs between y = -1 and y = 3, which gives us the volume of the figure between y = -1 and y = 3.

Hope I could be of any help!

// Kris
(1 vote)
• Why use a principle root only and not include the negative root?
• The equation y=x^2-1 is that of a shifted parabola with point of contact with the line x= (-2) at (-2,3). Then on rotating the figure around the axis x= -2 shouldn't we get a cup like figure ( as shown in the video) along with a conical figure in it ( not shown in the video ) ?
• Earlier, you subtracted a function by a constant (x=2 or something) but now, you are subtracting a constant (in this case -2) from your function. How do you know if the given function should be subtracted by the given constant or vice versa?
(1 vote)
• Due to the squaring in the integral, it does not matter if i do ( 2-f(x) )^2 or ( f(x)-2 )^2 because they are mathematically equivalent.
The difference is like saying the distance from the function to the line y=2 and the distance from the line y=2 to the function. They are both correct.