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Calculus 2

Unit 4: Lesson 10

Volume: disc method (revolving around other axes)

Disc method rotation around horizontal line

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.2 (EK)
Solid of revolution constructing by rotating around line that is not an axis. Created by Sal Khan.

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• if f(x) is the outermost function and g(x) is the innermost function, shouldn't it be (f(x))^2 - (g(x))^2 instead of [(f(x)) - (g(x))]^2 like in this video?
• Good question.

However, he is not subtracting the volume of the solid created if you rotate y=1 around the x-axis from the volume of the solid created if you rotate y=√x around the x-axis, but rather rotating y=√x around y=1. Thus, this is not about outermost and innermost functions, but rather about shifting the function so that we are rotating it around the x-axis but getting the same volume as if we were rotating it around y=1.

I hope this clarifies the problem!
• Isn't the washer method supposed to work here too because we have two functions?
I keep trying to use the washer method and it gives a difference answer.
• No the washer method wouldn't work because there is no 'opening' in the middle of the revolution. Had it been revolved around the y-axis, then you would have to use the washer method.
• If the equation was `y = lnx` instead of `y = √x`, and I was rotating around of `y = -1` instead of `y = 1`, I presume that each washer would equate to `π(lnx + 1)` instead of - 1. Is that presumption accurate? See .
• That's very close. Remember, we are finding the volume, so we will need to find the area of the circle. Since the radius is ln(x)+1, the area is π(ln(x) + 1)^2. Also, to find the volume, we will need to multiply the area of the circle by the width of each washer, which is dx. The dx is important. With dx, you will summing up an infinite number of finite values, which is infinite, but with dx, you will be summing up an infinite number of infinitesimal values, which is finite.

I hope this helps!
• What would the radius be if the function was being rotated around y = 5?
• The expression √x -5 gives your new radius, because the radius is still just the distance between the axis of rotation and the function
The full integral would be v = ∫π(√x - 5)^2dx, on whatever definite interval you want to take the volume
• how do you know whether you're supposed to subract/add the line rotated around to the radius?
• You would subtract the line if it is below the function, or subtract the rotated function if it is below the line. This is because the radius of the solid at any given x value is the distance between the rotated function curve and the line it is being rotated around. Distance technically is found by taking the absolute value of their difference, however assuming the the two never cross inside the section we are taking the volume of, we can just pick the one that is lowest and subtract it from the other.
• At , how do we know that the interval is from 1 to 4?
(1 vote)
• The interval comes from the setup of the problem in the first 45 seconds of the video. We're working with a figure that begins at the point where the curve y = sqrt x crosses the line y = 1, and that occurs when x = 1. The figure ends at x = 4, which is just an arbitrary ending point that Sal states early in the video.
• Hello, Calculus II student here. Some of these easier integration problems take me 5-10 minutes especially when I write out every step. My question is: Is this normal? As in, do most average-above average students write EVERYTHING out? And by everything, I mean how Sal does it.
• Why don't you use U-substitution for the integral
(1 vote)
• If we let u = sqrt(x) - 1, then the derivative of u would be 1/2*sqrt(x), which does not appear in the expression that we're integrating; therefore u substitution is not the best method when we could take a much more feasible option and multiply out the expression that we're integrating.
• What about the x interval [0,1]? The problem asks us to rotate the (whole?) function around the given axis, so i thought there should be something happening in that interval.
Isn't there a sort of a cone with the head pointing to the right? Shouldn't we add the volume of this smaller cone to the one Sal found?
(1 vote)
• Sal didn't say what volume we were computing too explicitly, so we can just assume that the volume he cared about didn't include the part above [0, 1]. But you're correct that there would be another figure there, shaped sort of like the mouth of a trumpet and coming to a point at (1,1).