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Let's do a couple more rotational volume problems, and

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I'm going to make these a little bit more difficult.

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And hopefully after these if you've understood everything

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we've done up to now and the ones I'm about to do, I think

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you're pretty set for most of what you should face

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in most math classes.

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And definitely I think you'll be set for the AP exam,

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and either ab or bc on this concept.

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So let's do another example.

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OK.

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So let's say that I want to-- actually, let me

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do something different.

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Let me draw here.

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So that's my y-axis.

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This is my x-axis.

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Now let me draw the function y is equal to x squared.

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And we know that that could be written as y is equal to x

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squared, or we could write that as x is equal to square root of

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y, depending on what we want to be a function of what.

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This is the y-axis.

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This is the x-axis.

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Let's say I also have the line y equals 2.

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Goes over what I just wrote.

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y equals 2.

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Now this problem is going to be slightly different then

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what we've done so far.

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I'm going to take a rotation, but instead of taking a

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rotation around the y or the x-axis, I'm going to take a

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rotation around another line.

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So let's say I want to take a rotation between

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x is equal to 0.

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Actually let me do something arbitrary.

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Let me say between x is equal to 1, so that's that point,

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and x is equal to 2.

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That's where they intersect.

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This is the point right here, this is 2,2.

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I'm sorry, no, 2,4.

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Because y is equal to x squared.

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So this is 2,4.

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This is the point 4.

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So what point is this.

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This is the point 1, right?

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So our y values go from 4 to 1, our x values go from 1 to 2.

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And that makes sense, because y is x squared.

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And so if we were to kind of take the area that we're going

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to rotate, and I haven't told you what we're going to rotate

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it around yet, and this might prove to be shocking to you.

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So this is the area we're going to rotate.

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Instead of rotating it around the y-axis, I want to rotate

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it around the line y is equal to minus 2.

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So if that's 2, y equals minus 2 should be roughly here.

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So I'm going to rotate this area around this line.

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So what's it going to look like?

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It's going to be a fairly big ring.

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Like if I were to try to draw it-- let me see if I can

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even make an attempt.

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Once again this is always the hardest part, just drawing

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what I'm trying to rotate.

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I'll try to do it from an upward perspective.

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So that's kind of the inner loop, and then there

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will be an outer loop.

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The top is flat right, because it's defined by y is equal

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to 4, so that's the top.

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And the inside is also going to be a hard edge.

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But then the outside is going to curve inward.

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I don't know if you see what I'm saying, because this is the

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outside, it's curving inward.

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So it's going to be a big ring.

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So if I were to draw the axis, this would be the y-axis access

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coming in-- no, no, sorry.

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Whoops.

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The y-axis is actually going to be closer to this hand side.

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The y-axis is going to be in the middle of kind of-- so

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this is going to be the y-axis coming up here.

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And then the x-axis is going to come below that.

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I'm drawing everything at an angle as best as I can.

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The x-axis is going to come a below that.

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And then this line we're rotating it around,

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that's going to be someplace over here.

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That's going to be something like that.

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It's going to go behind there and come back over there.

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Hopefully that makes sense.

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We're just getting a big ring.

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So how are we going to do this?

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Well actually there's a couple of things we can do.

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First we could just use the shell method,

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using the x value.

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So how do we do that?

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The important thing is to always visualize

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the shell or the disk.

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So the shell method we're going to take slivers like this,

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where the width of that sliver is dx.

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I could draw it really big.

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So that's our direct angle, is going to be dx.

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What's the height going to be of the sliver?

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Well it's going to be the top function minus

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the bottom function.

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It's going to be y equals 4 minus x squared.

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So this is going to be 4 minus x squared, the height at

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any point right here.

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And then if I were to do the shell just like we did

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before-- let me see if I can draw a decent shell.

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I think I'm getting better at this.

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This is one edge of the shell, that's the other shell.

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We already figured out that the width of the shell is dx.

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The height is 4 minus x squared, the top function minus

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the bottom function because of the distance between the two.

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And then what's the radius going to be?

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What's the radius of that shell going to be?

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Well, is it going to be just x?

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Is it just going to be x value?

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No, the x value will tell you the distance from the y-axis to

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that shell, and it's going to be from minus 2 to the e value.

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So it's going to be essentially 2 plus x.

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That's going to be the radius at any point.

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And this is where we diverge from what we've done before.

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Before the radius was just x, but now it's 2 plus x.

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So what's the circumference of each shell going to be?

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Well circumference is equal to 2 pi r, and

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our radius is 2 plus x.

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So it's 2 pi times 2 plus x, which equals 4 pi

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plus 4 pi plus 2 pi x.

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That's the circumference.

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And then what's the surface area of this?

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Well it's going to be the circumference times the height.

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So surface area is equal to that.

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The circumference 4 pi plus 2 pi x, all of that times the

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height-- times 4 minus x squared.

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And let's see if we can foil this out or

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distribute this out.

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So 4 pi times 4 is 16 pi.

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4 pi minus x squared minus 4 pi x squared.

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2 pi x times 4 plus 8 pi x, and then 2 pi x times minus x

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squared, so that's minus 2 pi x to the third.

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So that's the surface area of each ring.

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And then if we want the volume of each shell, essentially

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we multiply it times the width, the dx.

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So that's the volume of each shell, and if we want

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the volume of all the shells, we sum them up.

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So we take the integral, that's an integral sign, and I'm

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running out of space like I always do.

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And where did I take the integral from?

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I take the integral from x is equal to 1 to x is equal to 2.

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From 1 to 2.

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That's probably too small for you to read.

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So let's see if we can take the antiderivative of this.

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Let me make some space free just so I don't

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have to write so small.

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So I'll keep this down here, because that's the set

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up of the problem.

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I think I can get rid of a lot of this.

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I think that is pretty good.

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OK.

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And let me switch to another color.

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And we're going to take the antiderivative.

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So what's the antiderivative of this?

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So the antiderivative of 16 pi is 16 pi x.

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And then what's the antiderivative of

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4 pi x squared?

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It's going to be x to the third over 3, so it'll be minus 4

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pi over 3-- sorry-- 4 pi over 3 x to the third.

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181
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Well this will be x squared over 2, so it'll be

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plus 4 pi x squared.

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And then minus, this will be x to the fourth over

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4, so minus pi over 2.

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Just divided by 4.

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x to the fourth.

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I'm going to evaluate that.

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This is a much hairrier problem then what we've been doing.

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2 and at 1.

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So what is it evaluated at 2?

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It is 32 pi minus 4 pi over 3 times 8 plus 4 pi times 4 plus

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16 pi minus 2 to the fourth is 16 divided by 2 minus 8 pi.

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And then I just realized I'm running out time, so I will

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continue this in the next video.

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