Solid of revolution volume
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Disc method: function rotated about x-axis
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Disc method (rotating f(x) about x axis)
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Volume of a sphere
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Disc method with outer and inner function boundaries
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Shell method to rotate around y-axis
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Disk method: rotating x=f(y) around the y-axis
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Shell method around a non-axis line
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Shell method around a non-axis line 2
Shell method to rotate around y-axis Use the "shell method" to rotate about the y-axis
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- We've been doing a lot of rotating around the x-axis, so
- let's start rotating around the y-axis and see what we can do.
- Or at least attempt to.
- Let's me draw my axes.
- That's y-axis.
- That's my x-axis.
- Well let's just do it with an example, but we'll call it f
- of x too because it'll be generalizable.
- Let's just draw y equals x squared.
- Let me just draw the positive because we're going to rotate
- it around the y-axis and it's symmetric anyway, so that's
- y equals x squared.
- This is y-axis.
- This is x-axis.
- Actually, no I'm going to keep it general, then we'll actually
- solve it particularly.
- So we'll call this f of x, but clearly this is
- y equals x squared.
- This is f of x.
- And we know how to take the volume if I were to rotate
- this around the x-axis.
- But what if I wanted to say-- I guess we could call it the area
- between 0 and-- I'm trying to determine how general to be.
- Well let's just say between 0 and 1.
- I think the boundaries might make sense to you.
- Roughly this area, and I'm going to rotate it
- around the y-axis now.
- So what's that final figure going to look like?
- The base of it-- let me see how well I can draw it.
- Nope that's not what I wanted to do.
- The base is going to look something like a
- cylinder like that.
- And then the top of it is also going to be-- no, that's
- not what I wanted to do.
- Let me draw the side lines.
- So it's going to look something like that.
- And then the top of it looks something like that.
- But it's not just going to be cylinder, right?
- If I was doing this entire block it would be a cylinder.
- But the inside of it is going to be kind of hollowed out.
- Let me see how effective I am at drawing that.
- I'll do it in a different color.
- So the inside is going to be hollowed out.
- I don't know if that makes sense to you that.
- It's kind of like on the inside it'll look like a bowl.
- On the outside it'll look like a cylinder or a can.
- Hopefully that makes sense.
- You take this and you rotate this around.
- And the curve that specifies the inside would be y
- is equal to x squared.
- It would rotate all the way around.
- I think that makes sense.
- The drawing is the hardest part.
- So how do we do it?
- Well even the shape might give you an idea.
- We can't use that disk method, what we were doing before when
- we were rotating the x-axis, that was the disk method,
- because we were essentially imagining each of these
- particular disks and then summing them up.
- Now we're going to do something called the shell method.
- So what's the shell method?
- Instead of taking a bunch of disks and figuring out their
- combined volumes, we're going to take a bunch of shells.
- So what's a shell?
- So imagine a rectangle right here.
- Hope you can see it right there.
- Let's say it's at the point x,1.
- What's its height going to be?
- Its height going to be f of x,1.
- That's its height.
- Now imagine taking that sliver and rotating
- it around the y-axis.
- What's it going to look like?
- Well, it's going to look like a shell, it's going to look like
- a cylinder, just like the outside of a cylinder.
- It's going to look not too different then that but I want
- to draw it well because intuition is the most important
- thing that, not getting the problem right.
- Let me see if I can draw this respectably.
- And then we're going to have the bottom of the shell, it'll
- look something like that.
- Let me finish these lines up.
- I think you get the point.
- OK.
- So it's going to look like a shell like that.
- The outside of the shell is going to be solid.
- And it'll have some width, but the inside is hollow.
- Let me do a different color.
- Maybe a darker color to show that that's the inside.
- You know it's like a ring essentially.
- And so what's the height of this ring?
- The height is going to be f of x,1.
- So let me do a brighter color so you know what I'm saying.
- The height of this ring is f of x,1.
- f of x evaluated at that arbitrary point we picked up.
- What is going to be the surface area of this ring?
- You know, this outside.
- Well let's think about it.
- It'll be the circumference of this ring times it's height.
- So what's the circumference of this ring?
- Let's go back to our basic geometry.
- Circumference is equal to 2 pi times the radius.
- So if we know the radius of it, we know the circumference.
- Well what's the radius?
- Well the radius is how far we went from the axis of
- rotation to that point.
- So that's the radius.
- So in our particular example the radius is x,1.
- It's that x point that we're evaluating it at.
- So circumference is going to be equal to 2 pi times that point
- that we're evaluating at.
- And so the surface area-- this magenta thing that I filled
- in-- that's going to be equal to the circumference times this
- height, which we already said is f of x,1.
- Let's call it area surface.
- Surface area is equal to circumference times height,
- which is equal to 2 pi x,1 times f of x,1.
- We figured out the surface are of this.
- Now how do we figure out the volume?
- Well what's the width of it?
- How thick is this ring?
- What's this thickness right here?
- It's a very small thickness.
- But we took this sliver, and this sliver as we learned in
- previous calculus, the width of this little rectangle is dx.
- And you know when we take the integral, it's going to get
- infinitely smaller and smaller and we'll have infinitely
- more and more of them.
- So the width of this is dx.
- Let me draw it big, not so horrible looking.
- So if this is a sliver, it's width is dx.
- It's height is f of x,1.
- x,1 will be right in the center.
- And then it's distance from the center is of course x,1.
- Hopefully that make sense.
- So what's the volume of this shell?
- So the volume of the shell-- this shell, not this one-- the
- volume of the shell is going to be equal to the surface area
- of the shell times how wide that surface is.
- And that width is dx, so it's going to equal this times dx.
- So the volume of that shell is 2 pi x,1 times
- f of x,1 times dx.
- I think you see where I'm going with this now.
- So what would be the volume of the entire rotated
- figure, this thing here?
- Well I'm just going to sum up each of these shells.
- I have one shell there, then here I'll have a slightly less
- high shell, and up here I would have a much bigger shell,
- and I'll add them up.
- Here's one shell that goes around.
- Then they'll be another shell here, and I'll add them all up.
- And that's taking the integral.
- So the total volume of the figure when I rotate it around
- the y-axis is going to be-- and my boundary is from 0 to 1-- 2
- pi-- this one I just told you a particular x,1 but we're going
- to sum them over all of the x's.
- So it's going to be 2 pi x f of x dx.
- This is just a constant, so you could call it
- 2 pi times x f of x.
- So let's take a particular example.
- Let's do it for x squared.
- Let's say the function is x squared.
- So in this case the volume is going to equal-- let's take the
- 2 pi out-- 2 pi integral 0 to 1 x times f of x-- f of x in our
- case is x squared, which I drew earlier-- dx equals 2 pi.
- This is just x to the third, right?
- x to the third.
- So it's going to be 2 pi times the antiderivative
- of x to the third.
- Well that's x to the fourth over 4.
- Evaluate it at 1 minus evaluate it at 0.
- Well that equals 2 pi times 1 to the fourth is 1, so
- 1/4 and then minus 0.
- So it's 2 pi times 1/4.
- So that's pi over 2.
- That's the volume, and we just rotated it around the y-axis.
- I will see you in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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