Solid of revolution volume
-
Disc method: function rotated about x-axis
-
Disc method (rotating f(x) about x axis)
-
Volume of a sphere
-
Disc method with outer and inner function boundaries
-
Shell method to rotate around y-axis
-
Disk method: rotating x=f(y) around the y-axis
-
Shell method around a non-axis line
-
Shell method around a non-axis line 2
Disc method (rotating f(x) about x axis) The volume of y=sqrt(x) between x=0 and x=1 rotated around x-axis
⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.
- Welcome back.
- On the last video we came to the conclusion that we could
- figure out the volume when we rotate a function about the
- x-axis, so let's apply that to an actual exercise.
- I'm going to erase everything because I don't want you to
- memorize this, because frankly I haven't memorized this.
- And if you do, you'll forget it one day and then you
- won't know how to do it.
- But if you understand why it works then you'll never forget.
- As long as you remember basic integration.
- Maybe you want to memorize it if your teacher tends to give
- you a test that don't have much extra time in them, just
- to speed up the process.
- But you should know what's going on.
- So let me draw the axes again.
- That's my y-axis.
- That's my x-axis.
- And so since our first example was y equals square root of
- x, let's stick with that.
- And for reasons that might become apparent that tends to
- be one of the more typical examples when you rotate
- things around axes.
- So let me see if I can draw it as well as I drew it last time.
- Almost.
- OK, so that's y equals square root of x, it's just f of x
- this time, I've defined it.
- This is the x-axis.
- That's the y-axis.
- And I'm going to rotate this around the x-axis again.
- So I'm going to get a sideways looking cup thing.
- And let's say I want to figure out the volume of that cup
- between the points 0-- and to make it simple, let's just
- say the points 0 and 1.
- So essentially we're just going to get a cup, a sideways
- cup, it's going to look something like this.
- It's going to look something like this.
- That's going to be the-- that's a horrible-- the
- opening of the cup.
- Actually why don't I use the circle tool.
- It just dawned on me that I had a circle tool.
- So the opening of the cup will look like that.
- Actually I could draw it right here.
- This would be the opening of the cup.
- Well-- you can see sometimes that my videos are
- a little unplanned.
- There you go.
- So that would be the opening of the cup.
- This is excellent.
- This tool is very well suited for what I'm doing here.
- We're rotating around that way.
- We're turning that function.
- So the cup's going to look like that, so the bottom part of the
- cup's going to look like this.
- And it's solid, so we want the volume of the whole thing.
- In future videos I'm going to show you actually how to figure
- out the surface area of the cup, which I find in some
- ways more interesting.
- So how do we think about that again?
- Let's just rederive it, but this time we'll use
- a specific equation.
- So we just have to figure out what is the volume of one disk
- and then sum up all the disks.
- So let's say this disk right here-- actually let's just take
- this disk at the end point right here that I've already
- drawn something for.
- So what's the radius of this disk?
- The radius of that disk is f of x at that point.
- Well f of x at that point is just square root of x.
- Radius is equal to square root of x.
- And so the area of that disk is going to equal pi r squared.
- Well the radius is square root of x, so it equals pi times
- square root of x squared.
- So it equals pi times x.
- That's the area of each disk.
- And then if we want the volume, you just have to multiply the
- area of that surface times the depth of the disk.
- I'm just trying to show.
- You can imagine that this is kind of like a quarter and this
- is the side of the quarter.
- We saw in the last video that depth, that's just a very
- small change in x, because we want each disk to be
- infinitesimally thin.
- So the width is just dx at any point.
- So the volume of each disk is equal to the area, which we
- just figured out was pi x times the depth, times dx.
- That's the volume of each disk.
- So the total volume is going to be equal to the
- sum of all of these.
- That was one disk I drew, then you're going to have another
- one here, you're going to have another one here,
- another one here.
- You're going to have infinitely many, and you want them to be
- super, super, super thin so that you get an accurate
- measure of the exact volume of this curve.
- Otherwise it would just be an approximation, and that's
- where we use the integral.
- So it will be the integral from.
- And my original boundaries were 0 to 1.
- The disk we used as an example, this is probably you know the
- last disk, so this one will actually have a radius of the
- square root of 1, which is 1.
- Not that you have to know that, I'm just trying to keep
- emphasizing the visualization.
- So what will be the integral?
- Well we're going to go from 0 to 1, and we're going to sum up
- a bunch of these disks, which we've already defined,
- so it's pi x dx.
- This is looking to be a fairly straightforward integral.
- So what's the integral of that?
- Pi is just a constant and the antiderivative of x is x to
- the 1/2 over-- I'm sorry.
- It's x squared over 2.
- I've been a little rusty since I last did some
- antiderivatives.
- So we get x squared, we get pi times x squared over 2.
- That's the antiderivative of that.
- And then we have to evaluate it at 1 and then subtract
- it and evaluate it at 0.
- And so what do we have.
- We get 1/2 pi, so we get pi over 2 minus 0 pi, minus 0.
- So it equals pi over 2.
- There we go.
- We just figured out the volume of this cup from 0 to 1.
- Reasonably interesting.
- Let's see if we can do that again to figure out the-- just
- to give you another example, just hit the point home-- to
- see if we can figure out the volume of a sphere,
- the equation for the volume of a sphere.
- So what's the equation for a circle?
- It's x squared plus y squared is equal to r squared.
- And let's write that in terms of y is a function of x, just
- so we have something that we can work with the
- way we learned it.
- So we get y squared is equal to r squared minus x squared, and
- then we get y is equal to the square root of r squared
- minus x squared.
- Actually now that I realize it, I'm going to not do this,
- because I think I'm going into too complicated a problem.
- I did that on a fly.
- But in the next video I will do slightly more complicated
- without going to this one, because I probably don't have
- time for it I just realized.
- Anyway I'll see you in the next video.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
|
Have something that's not a question about this content? |
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
abuse
- disrespectful or offensive
- an advertisement
not helpful
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
wrong category
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site
Share a tip
Suggest a fix
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.