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Let's

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continue on with our study of rotation of functions around

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the x, and we'll soon see the y-axis as well.

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So let's do a slightly harder example than what we've been

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doing, but I think it might be obvious how to approach it.

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So there's my y-axis, there's my x-axis, and in a couple of--

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I think it was two problems ago-- we figured out if we had

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the function y is equal to square root of x-- let me try

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to draw it-- so this is y equals square root of x-- if we

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were rotate that around the x-axis, what the volume would

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be between two points, let's say 0 and some other point.

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Now let's just pick an arbitrary point 1.

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I think you know how to do that at this point.

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Now let's make it a slightly more difficult problem.

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Let's say I were to also draw the function y is

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equal to x squared.

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So that looks-- if this is 1, they both meet at 1, right?

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Because square root of 1 is 1, and 1 squared is 1, so that

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would look something like this.

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They should be actually symetric around y

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equals x, but anyway.

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So say y equals x squared looks like that.

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So my question now is, what is the volume if I were to

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take this figure and rotate it around?

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So what do I mean?

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So this area here, if I were to rotate that about the x-axis,

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what would the volume be?

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So now, what we did just with the square root of x, we had

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like a solid cup, right?

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It would look something like this.

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It would be like a cup, and it was solid, we were just trying

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to figure out the volume of it.

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Hopefully that made sense to you.

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Now it's going to be kind of a hollowed-out cup, because we

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have this inside function, and so the inside of the cup

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is going to be empty.

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Hopefully that makes a little sense.

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Remember you're just taking this and then you're rotating

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it around the x-axis.

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Well, the way to think about it-- especially if you having

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trouble visualizing-- actually, the solution might

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help you visualize it.

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The volume of this figure, which I'm having trouble

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drawing, it will be the volume formed by the

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outside rotation of this y equals square root of x.

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We'll do that in the yellow.

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It'll be the volume formed when that is rotated around, and

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then the whole solid volume minus the volume when

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minus this volume.

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So if we took the y equals x squared, y equals x squared

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would look something like that, and then if you rotated it

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around the axis, it would look something like that.

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I don't know if you've ever been to Morocco, but they have

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these tajin plates that the tops look a lot like that.

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Well, you probably haven't.

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Well, anyway.

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It would look something like that.

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So if we subtract out this volume when it's rotated around

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from the volume of y equals square root of x, when that's

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rotated, we'll get this figure that we're trying to figure

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out, this area when it's rotated around.

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And that should be intuitive for you, hopefully, because

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when we just did area under a curve, that's how we would

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figure out the area of this green area.

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We would figure out the area under square root of x, and

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we'd subtract out the area under y equals x squared.

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This time, we're going to say the volume of the revolution of

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y equals square root of x minus the volume of the revolution

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of y equals x squared.

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So let's do the problem.

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So the total volume-- let me do a good color, that looks good--

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total volume is going to be equal to the volume formed when

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we rotate y equals square root of x around the x-axis.

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I said from 0 to 1, and that's because I picked

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where they meet.

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Sometimes in a book, or on an exam, they'll just say, oh, you

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know, the area between y equals x squared and y equals square

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of x, we're going to rotate that around and you have

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to figure out, well they intersected at 1, and you can

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just set the equations equal to each other to figure that out.

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We're going from 0 to 1, because they also

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intersect at 0.

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0 squared is the same thing as square root of 0.

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We're going from 0 to 1, and so what's the

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volume of the larger?

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Or I guess the y equals square root of x rotated around?

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I always forget the formula, that's why I always redraw a

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disk, so if that's the radius of my disk, the disk is going

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to come around like that, so we know that the radius is a

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function of the disk, and that's, of course, the dx

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is the depth of the disk.

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So the radius is the function which, for the outside one, is

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square root of x, and we know area of this disk is pi r

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squared, so we square the radius, take a pi outside, and

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then we multiply that times the width, so that's where we

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get our dx, and of course we sum them all up and that's

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where we get the integral.

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I'm going to do it as two separate integrals.

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Some people will put them both within the same integral, but I

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really want to hit the point home that this is the volume of

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the outside surface, formed by the outside surface or the cup,

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minus the volume formed by the the inside function.

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It's going to be minus pi-- still going to be from 0 to 1--

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I drew fairly huge integral signs, I don't know why-- and

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what's the inside function?

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It's x squared, and that's going to be the radius of its

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own disks, if that's the radius that's the disk, dx

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is the width, so it'll be x squared, squared, times dx.

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So let's see if we can figure that out.

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So, volume equals-- let's take the pis out.

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I think you'll see that that pi is applying to everything,

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so we can take the pi out.

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And then, times the integral, and now we can merge them

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back, because integrals are additive like that.

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You'll see what I'm talking about.

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This integral is the same thing.

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So what's square root of x squared?

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Well that's just x.

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And what's x squared, squared?

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That's x to the fourth, right?

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You multiply the exponents, exponent rules.

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We have a minus sign here.

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Minus x to the fourth, all that times dx, we've got that pi on

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the outside, that equals-- let's keep our pi

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on the outside.

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We're going to have to evaluate the antiderivative at 1 and 0.

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So what's the antiderivative of x?

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Well, that's x squared over 2, minus-- what's the

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antiderivative of x to the fourth?

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Well, it's x to the fifth over 5.

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That's hopefully second nature to you.

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x to the fifth over 5, and we're going to evaluate

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that at 1 and 0.

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1 and 0.

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We're going to subtract them.

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Fundamental theorem of calculus.

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So that equals-- I'm going to switch colors to avoid

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monotony-- that equals pi times, let's evaluate it at 1,

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so it's 1/2 minus 1/5 and when you evaluate it at 0, it's 0

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minus 0, so when you evaluate 0, you get nothing.

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And so what's 1/2 minus 1/5?

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That's pi times 2, get a common denominator of 10, 5/2 is 1/2

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minus 2/10 is 1/5, so this equals, this would be 3,

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so we get 3pi over 10.

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That's the volume formed.

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So it's almost easier to figure out the volume of

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this figure than to draw it.

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Anyway, I think I'll leave you there with this video, and in

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the next video, we're going to start rotating

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around the y-axis.

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See you soon.

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