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Washer method rotating around horizontal line (not x-axis), part 1

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.4 (EK)
Washer method when rotating around a horizontal line that is not the x-axis. Created by Sal Khan.

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• why in the last video is the function minus the horizontal line and in this video is the horizontal line minus the function?
• Because in this video the original function is below the horizontal line.
• In my perspective all of these rotations around perfectly vertical or horizontal axes are simply translating all those functions up/down so essentially that axis you are rotating along is the new "x/y axis", in a relative sense. Is this concept correct?
Also (this is probably another separate question) so what if we were trying to rotate along, say, y=2x? And does rotation along an axis which is not a straight line exist (or is it just that I can't imagine it at the moment)?
• First question: Yes, that is a way of thinking about it! :)
Second questions: No, you can do what you described but it would not be a rotation. A rotation must not alter the shape of the "object" that we are rotating. Imagine rotating something in real life, you never alter the shape of the object, just its position and orientation.
"rotating" something around a curved line in math would be like bending something.
• At , why is the outer radius 4- x^2 +2x instead of x^2 - 2x -4? Or is it the same thing because even though it's the negative version of the terms, it's going to get squared anyway?
• No, it's because the horizontal line is above the function, instead of being below it like in the last video, so we subtract the function from the horizontal line. It's also because 4 is the larger value in the interval from 0 to 3, so it keeps the result positive.
Your intuition is correct that it doesn't matter since squaring the result will make it positive anyway, but it keeps the math easier without dealing with confusing sign changes.
(x^2 - 2x - 4)^2 = (- x^2 + 2x + 4)^2 = (4 - x^2 + 2x)^2
x^4 - 4x^3 - 4x^2 +16x + 16
• at around , why does the height equal to 4 - (x^2-2x)? Why do we subtract by the function?
• How does "rotating a function around a line and obtaining a solid of revolution" help us in our life? Where is this concept actually applied?
(1 vote)
• Realistically it's unlikely you'll ever have to actually apply this technique. Many of the problems we tackle in calculus are designed to improve (or test) our overall understanding of the concepts and ability to apply them in a variety of contexts. There happens to be a story about this type of calculation, though. Thomas Edison hired scientists to work in his laboratory and one day he saw one of them doing a complicated calculation to figure out the volume of a glass bulb of some kind. Chances are the scientist was using this kind of calculus technique. Edison walked over, took the glass bulb, filled it with water, and poured the water into a graduated beaker to find the volume in about five seconds.
• I get the part about the inner radius. But I am still confused about how the outer radius is not 4+ x^2 -2x? so from x-axis to horizontal line the distance is 4 but then don't we have to add x^2+4 to get the radius??
• The outer radius is defined in a later video as the distance from the axis of rotation to the outer function. To get this, you would take the axis of rotation (in this case: 4) and subtract it by the outer function (x²-2x). Ultimately, as in before Sal simplifies it, the outer radius would be: 4-(x²-2x). Sal just simplified it right away to 4-x²+2x by distributing the negative. I hope that makes more sense...
• can you find the volume if you rotate the function around a slanted axis like y=x for example?
• Yes, but it would have to be done with triple integrals, which is something you learn in multivariable calculus.
• So even if the region was touching the horizontal axis, you would take the 4-f(x) in this particular problem?
• Yes, because you are rotating around y=4 instead of the horizontal axis. All distances have to be relative to the axis that the figure is being rotated around.
(1 vote)
• if the rotating line is negative or below the x-axis and the function is positive or over the x-axis, how would you calculate the height between the rotating horizontal line and the function ?
• Imagine just a simple line, like `y = 3`. It'll lie parallel to the x axis. If you want to shift it up by one, you'd add one, right?: `y = 3 + 1`. If you want to shift it down by one, you'd subtract one: `y = 3 - 1`.

The shifting of weird curves works in the same way:

1) Take the rotating line to be your new x axis.
2) If your function `f(x)` is below that line, then it was shifted by an amount `C` downwards from that line. These are examples:

`y = 3 - 1` ← A concrete, simple one.
`y = f(x) - C` ← One more general.

3) If your function `f(x)` is above that line, then it was shifted by an amount `C` upwards. These are examples:

`y = 3 + 1` ← A concrete, simple one.
`y = f(x) + C` ← One more general.

Don't get confused by the weirdness of your function or whether the new line is above or below the original x axis. You can use this equation: `y = f(x) ± C`, to figure out the value of `C` in the first place and, at the end, replace `f(x)` by the function you're working with.
(1 vote)
• At , when Sal is finding the outer radius, he writes 4 - x^2 + 2x. Shouldn't it be 4 - x^2 - 2x as in the original function?