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Well we already know that definite integrals can help us

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figure out areas underneath curves or between curves.

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What we'll show in this video is that you can actually use

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pretty much the exact same principles to figure out the

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volumes of rotational solids.

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So what do I mean?

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So let me just draw a couple of examples.

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So let me start with a fairly straightforward function.

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That's my y-axis.

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This is my x-axis.

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Let me draw my function.

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I'm going to draw y equals the square root of x, but I'll keep

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it general right now, just so this applies generally

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to a lot of things.

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So y equals the square root of x looks something like that.

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It keeps going.

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Actually, let me redraw that, because I didn't want it

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to curve down like that at the end.

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It goes up like that, and then it just keeps

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going up like that.

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That's better.

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OK.

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So we'll call that f of x.

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This is our x-axis.

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This is our y-axis.

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And we already know that if we wanted to figure out

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the area under this curve between two points.

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Let's say between the point a-- well, we can do it

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between any two points.

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Let's say this point a and this point b, and we wanted to find

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this area between the two curves.

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I'm drawing everything crooked today.

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Let me do a different color.

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If we wanted this area right here, we would essentially just

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be-- just as a review-- be summing up a bunch of small

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squares, where each square has a bunch of rectangles, has a

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width dx, and its height at that point would be whatever x

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value here is-- it would be f of x.

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And if we take the sum of all of these areas, of all of these

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rectangles, we would get the area to this curve.

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And we learned in the definite integral video that that's just

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equal to the definite integral from a-- that's a lower bound--

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from a to b of f of x times d of x.

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Where each [? rafter ?]

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angle is f of x times d of x.

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And hopefully this makes a little bit of intuitive

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sense to you.

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I think a lot of people go through calculus just learning

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how to do it mechanistically, just learning you know how to

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do it like a robot without really understanding

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necessarily what's going on.

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And if you understand what's going on, you'll never be

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really lost when you see something a little bit

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different than what you might have practiced.

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So with that out of the way, let's think about something.

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What if we took this function and we rotated

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it about the x-axis.

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So this might take a little bit of visualization, but what you

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imagine is-- let me see if I can draw.

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So I take this curve and if I were to rotate it about the

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x-axis, it would look something like this.

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It would look like a sideways copper vase.

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It would look like that, where that would be the opening.

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That would be the opening on the inside.

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I can even shade it.

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Show you my drawing skills.

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OK.

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Hopefully that make sense.

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You know, that would be the y-axis there, and the x-axis

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would pop out the middle.

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That's if you took this and you rotated it around.

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So let me draw an arrow to show that we're rotating it around.

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And if we did that, what would be the volume created by--

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well let's use the same boundaries, between a and b.

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So if we took this piece and we rotated it around, what would

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it look like between a and b.

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It would look something like this.

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Let me see if I can draw it.

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So you would have a-- whoops, I'm not trying that well.

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This is really testing the limits of my ability to use

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this computer to draw things.

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It'd be kind of a circle on one end, and then it would curve

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down a little bit and it would be another circle

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on the other end.

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And if I were to draw the x-axis, the x-axis would

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kind of pop out of the middle right there.

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That right there would be the point b, x equals b.

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If we were to kind of go behind or look into the object we

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would see the other surface of this rotational solid.

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And this point right here, that would be a.

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That would be a.

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And then of course the x-axis would keep going, and then

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that would be the y-axis.

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The visualization really is the hardest part

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about these problems.

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So first just actually imagine what you're doing.

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So I just did this section, if I rotated it about the x-axis.

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But if I were to draw the whole curve, the whole curve would

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look something like this.

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It would look something like that, and we're

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just rotating it around.

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Hopefully that makes sense.

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We're rotating it around that way.

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So how do we do that?

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Well we use the exact same principle.

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When we figured out the area, we would figure out the area of

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each of these small squares, and then we would take the sum

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of an infinite number of infinitely small squares,

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and we got this.

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So to do the volume, what we do is instead of having each

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rectangle, we kind of rotate each of these rectangles

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around the x-axis.

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If that's the rectangle, it has width dx, and

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it has height f of x.

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So this height right here, that's f of x at this point.

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If I were to rotate this rectangle around the x-axis,

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what do I end up with?

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Well I'll end up with a disk.

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Let me see if I can draw that reasonably well.

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I'm trying to show you some perspective when I draw.

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So that would be the top surface of the disk.

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And this would be the side of the disk.

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And so this is the top surface at the disk.

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And what would be the radius of this disk, what would

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be this height right here?

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Well that radius, that's going to be f of x.

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That's this height.

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Imagine if you took this and rotated it around, that's the

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same thing as this height right here, right?

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So that height or the radius of the disk is f of x.

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And then what's the width of the disk?

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Well that's just d of x.

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That's the same thing as this.

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We just rotated it around.

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So what would be the volume of this disk?

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It would be the area of this side.

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It'll be this area right here times this height.

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Well what's the area?

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Well we know the radius, right?

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Area is equal to pi r squared.

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What's this radius?

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My radius is f of x, right?

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So the area of this disk is equal to pi times the radius

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squared, it so it's pi times f of x, the whole thing squared.

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So what would be the volume of this entire disk?

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So it'll just be this area times dx.

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I'm running out of space and colors.

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So the volume of that disk is going to be equal to area of

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that disk, pi f of x squared.

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The whole function, whatever length this is at any point

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squared, that gives us the area, times the depth you

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can say, so that's d of x.

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Now that gives us just the volume of this one disk

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when it's rotated around.

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So if we wanted the volume of this entire object that I drew

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here, we would just sum up a bunch of these disks.

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We would take each of these rectangles, rotate them around,

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figure out the volume of that disk it creates, and

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then sum them up.

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And so essentially we're going to take an infinite sum of a

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bunch of these small little disks so we can

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take the integral.

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So this is the volume of each disk.

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We could call that a volume of a disk.

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So what's the volume of the whole thing?

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Well we just take a sum, an integral sum of

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each of these disks.

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So the volume when you rotate it is going to be equal to the

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definite integral between-- and remember, our boundaries were a

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and b-- between a and b of this quantity right here--

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pi f of x squared dx.

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So hopefully that makes sense to you.

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Just remember, this is the width of each disk.

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This is the radius of the disk, or the radius of the surface,

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so it would be squared, and that makes sense, that's

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the height, f of x.

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And we have pi r squared, so that's where the pi comes from.

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Some people just memorize that.

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I don't recommend you do that, and we'll see that later.

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But I'm out of time.

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In the next video I'll actually apply this

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to an actual problem.

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See you soon.

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