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## AP®︎/College Calculus AB

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

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

# Disc method rotation around horizontal line

Solid of revolution constructing by rotating around line that is not an axis. Created by Sal Khan.

## Want to join the conversation?

• 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. • 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? • 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) • 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) • 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) • For the radius of the first circle discussed, would it make sense to think of the radius as [(x)^(1/2)]/2? Since the radius of a regular disc seems to be equivalent to the diameter of this disc. I want to know why that is wrong(since I got a different answer) considering my line of thought is logical. And another thing that perplexed me is how when I just manipulated the original function with phase shifts to form [(x-1)^(1/2)]+1 and used as as the radius to integrate in the interval of (1,4), I still didn't get similar results as Sal. I will really appreciate it if someone out there answers my questions. I hope you understand me well. If I seem to be incomprehensible please let me know what I should elucidate. Best
(1 vote) • These are much more difficult to answer with words than pictures but I'll do my best. To put simply though, the bit of the function that Sal integrated and the 2 that you did in your 2 questions are 3 different shapes.
1: Perhaps it seems that way because of the place he chose to draw his disc. I assume by "regular disc," you mean one that has been rotated around y=0 (the x-axis.) The diameter of the disc whose "circle part" is centered on y=1 is only the same as the radius of a disc centered on y=0 at x=4 (in this example.) Think about moving to a different spot along the function. I'll arbitrarily choose x=25. The function y=√(x) will have a y value of 5. So the radius when it's rotated around y=1 will be 4 (5-1) making the diameter 8. The radius if the function is rotated around the x-axis is 5 (5-0). Of course, 4 ≠ 5.
2: You shifted the whole function up 1 and to the right 1. We already went over going up 1 in the first answer. Unless I'm drunk, with just the shift up 1, rotating around y=1 should get you the same answer as rotating the un-shifted function around y=0 so the reasoning from before still stands. With the shift to the right, think of it as adding on the little bit at the beginning ( x=0 to x=1) that Sal didn't include as part of his problem but removing the (larger) bit at the other end (x=3 to x=4.) The shape moved but the boundaries of integration didn't so you're integrating a different part of the overall shape. The problems caused by each shift are unrelated and so they don't cancel each other out if that's what you were thinking.
(1 vote)