WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to talk about taking volumes of solids by the method of washers.
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Let us jump right on in.
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In the last lesson, we introduced this notion of taking a function, rotating that function around a certain axis.
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Generating a solid, a solid of revolution, and then taking slices.
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And then, taking the cross sectional area, finding that little differential volume.
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And then, integrating, adding up all of those volumes.
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Now let us continue that, let me go ahead and work in blue here.
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Sometimes when we rotate a region around a given axis, we do not get a complete solid of revolution.
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In other words, we do not get a full… Like if I take a cup and fill it full of water, that whole cup full of water is a solid.
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Sometimes there are some hollows, there are things that are not quite complete.
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Let us take a look at an example, I think it will make this clear.
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We have the region bounded by the functions y = x³ and y = x.
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What we are going to do is we are going to rotate this region that is bounded by those two curves around the x axis.
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Let us see what kind of solid we get.
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Our question is what is the volume of the resulting solid?
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Let us go ahead and draw.
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We have the y = x³ is going to look something like that.
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The y = x line is going to look something like this.
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This is our y = x, this is our y = x³, and the region that we are concerned about is this region right here, the one that bounds those.
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This, we are going to rotate around the x axis.
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We are going to spin it this way.
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The thing that we actually end up generating is the following.
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I’m going to make the axis just a little bit higher.
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The solid that we ended up generating is going to be the following.
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It is going to have this one, it is going to be this way.
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This is a rough drawing.
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There is this solid, notice this is not a complete solid, this is a hollow inside.
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What we have done is take this region, spit it out, generate this.
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If I were to take this thing and turn it a little bit this way, you would see something like that.
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This is all hollow, this sort of triangular part, it is the other volume.
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How are we going to do this, how are we going to find this volume?
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This is how we do it, let us go ahead and draw another picture of it to make it a little bit smaller.
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Once again, we have our region that looks something like this.
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I will go ahead and draw this, something like that.
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Again, we do the same thing, we are still going to slice.
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Around the axis that we rotate, we are going to take a slice perpendicular to that axis.
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We are going to take a little bit of a slice, it is going to look like this.
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The slice is not going to be a single piece.
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It is going to be a single piece, in a sense that it is one slice.
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Again, there is this little gap right in between here.
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I take this slice and I’m going to rotate it out, and see what it is that I look.
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It is going look like a washer, instead of a solid disk, it is just going to be a disk with a hole in it, what we call a washer.
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Let us go ahead and go out here.
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This is going to be an outer disk, it is going to be that one.
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This right here, it is going to occupy an inner circle.
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That is what you get when you splice it.
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When you turn it, you are going to actually get a washer.
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This is a washer of thickness dx.
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This is the thickness from here to here.
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That is the thickness of that washer, that is the dx.
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What we are going to be doing is we are going to be integrating along the x axis.
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In other words, we are going to be integrating along the axis of rotation, along the x axis.
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Exactly the same as before, the only thing that is different now is we want the area of this disk.
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The area of the disk is the area the outer circle - the area of the inner circle.
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Because the area that we are looking for is that one.
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We are going to multiply by its thickness which is dx.
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That is going to give us a differential volume element.
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And then, we are going to add all of the volumes, all of the washers along the x axis.
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Our differential volume element is nothing more than the area of this × dx.
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The area of this is π × the outside radius² - π × the inside radius².
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This is the outside radius, this is the inside radius × dx which is the thickness of the washer.
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That is the volume of that washer, we want to add them all up.
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Let me we write this, dv = π × just pull out the π, outside radius² - inside radius² dx.
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When we integrate both sides, the integral of dv is going to be just v.
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The volume = the integral from a to b of π × the outside radius² - the inside radius² dx.
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That is the general formula for finding the volume of a solid,
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that has gotten a solid of revolution generated by rotating a particular region around a given axis.
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When that particular slice that you took is a washer, it is not quite complete.
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It is this solid disk but it is a disk with a hole in it.
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In this particular case, we have this function which is y = x.
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This function y = x³, and of course, this is the x axis, that is the axis of rotation.
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What are a and b?
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The lower and upper limits of integration, they are just going to be just
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from here we are integrating along the x axis to here.
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These two graphs, we know that they meet at 1.
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a is equal to 0 and b is equal to 1.
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Let us go ahead and fill this in.
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We have the volume = the integral from 0 to 1 of π × the outside radius.
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The outside radius is this one, it is from 0 to x, it is the y value.
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However far we go along the x axis, it is the y value.
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It is just going to be x, the outside radius is x and we square it.
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The inside radius, it is just this point right here.
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We go along x and it goes up to the x³ part.
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This is going to be x³² dx.
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That is going to equal π × the integral from 0 to 1 of x² - x⁶ dx.
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This is really what we want.
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The most important thing in all of these problems is actually finding the integral, establishing the integral.
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Solving it is important but it is the secondary concern.
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We need to take a situation, convert it into an integral.
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My students, when I teach this course, I generally have them stop here.
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I do not want them to solve the integral.
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This is what I'm interested in, I'm interested that they actually can form the integral.
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The rest is just techniques of integration, or software, if you need.
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Let us go ahead and solve this though.
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The volume is equal to π × this is going to be x³/ 3 – x⁷/ 7 from 0 to 1.
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The final answer is 1/3 - 1/7, that is it.
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That is all we are doing here.
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Same exact thing that we did before, except now we are just looking for the cross sectional area
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which happens to be outer circle - inner circle, because we are dealing with a washer.
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All of these problems are dealt with the same way, in some form or another.
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That is all we are doing.
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We are finding some area, we are multiplying it by a differential length dx.
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That is our differential volume.
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We are going to add up all those volumes, that is all it is.
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We just need to make sure that we see what is happening pictorially.
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That is going to be the hardest part in all of these problems.
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What is it look like, how does one visualize this?
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Let us go ahead and, a little bit further before we start our examples.
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When you are faced with a region that generate washers upon slicing,
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the volume is equal to the integral from a to b of π × the outside radius² - the inside radius² dx.
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If you are going to be integrating along the dy, then this is just outside radius - inside radius dy.
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The problem itself will tell you whether you are generating along x or generating along y.
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I will do this in red, this really is the most important part about all of these things.
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Not just this particular section but all of these, generating volumes and things like that.
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The most important thing you always measure the radii or any length for that matter from the axis of rotation.
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Whatever the axis of rotation is, x = 5, y = -1, it is from that point that you are measuring.
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From there, the distance from that to the particular functions that we are dealing with.
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This is the generic, once you have a picture, you are going to decide what our outside is, what our inside is.
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Again, you always measure radii from the axis of rotation.
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Let me go back to blue.
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Examples will make this clear.
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Our general process, number 1, always draw a picture.
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Two, identify your solid, identify the solid.
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In other words, once you actually perform the rotation, make sure you have an idea of what the solid looks like.
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It does not have to be a 100% full graph but you should at least be able to see this side view, what it actually looks like.
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Three, identify the washer.
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In other words, take a slice, draw it out, identify the washer.
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This identification of the washer, whether it is this way or this way,
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this will decide whether you are integrating with respect to x or y.
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Instead of saying with respect to, integrating along x or y.
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The final part, number 4, measure radii from the axis of rotation to the original graph.
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Once you generate your solid, you are going to have that other half.
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From the axis of rotation, measure to the original graph, that is going to give you your length.
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Again, it will make more sense when we start our example problems.
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With that, let us go ahead and start our example problems here.
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Find the volume of the solid obtained by rotating the region bounded by the following functions around the given axis.
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We have the function y = x³, y = x, x greater than or equal to 0.
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We want you to rotate it around y = 3.
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Same functions as before, different axis of rotation.
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Let us draw this out, see what it looks like.
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Let us go down here.
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We are rotating this around y = 3.
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I’m going to go this way, there we go.
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y = x is that line, y = x³ is going to be that thing.
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What we are going to do is going to rotate it around this line.
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This is our y = 3 line.
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When we rotate that way, now what we are going to do is we are going to get something that looks, it is going to be over here.
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Let me extend this out, it is not a problem.
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We are going to get something like that.
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We are going to rotate that way.
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What we are interested in is this region right here.
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That solid, whatever it is, is going to have a volume.
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We want to find that volume.
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We draw our region, let me see, 1, process, we drew our region.
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Two, we identify the solid.
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Three, we identify the washer.
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We are going to take a slice of this, we are not going to go this way, nothing there.
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Let me go to red actually.
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We are going to slice it this way.
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We are just going to take a slice that way.
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When we take this and turn it around, what we are going to end up getting is right over here.
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Our view is slightly to the side, not fully turned so you see it this way.
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But just a little bit so you can see what is actually going on.
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Here is this washer of a given thickness.
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The region that we are interested in is that region right there.
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There we go, we know the washer is a vertical washer.
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We are going to be adding all of the washers this way.
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We are going to be integrating along the x axis, integrate in the x direction.
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We will just say integrate along x.
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This is going to be from 0 to 1 again, because from 0 and 1.
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From here, we are going to integrate that way.
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We are going to integrate that way, along the axis.
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Part 4, measure the radii from the axis of rotation to the original graph.
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Outside radius, inside radius, outside radius, inside radius.
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Let me go back to blue here.
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Let me go to black.
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Outside radius is that radius.
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Inside radius is that radius.
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We said to measure the radii to the original graph.
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The original graph is here.
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Let me erase that.
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That is our outside radius, that is our inside radius.
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Measure it to the original graph because that is actually where the function is.
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The function is here.
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Our outside radius, distance is what we are worried about.
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This distance is 3, this distance here is x³, because if this is x, this height here is x³.
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Our outside radius is 3 - x³.
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The whole distance - this little distance, that gives me this distance.
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The inside radius, it is the whole distance 3 - this distance, from here to here that is x.
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Now that we have our outside and inside radius, the rest is really simple.
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We just have volume = the integral, let me write the formula one more time.
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a to b π × the outside radius² - the inside radius² dx.
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It is always a nice idea to write the equation over and over again.
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That is a way of memorizing it, before you start the problem.
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In this case, it is going to be the integral from 0 to 1 of π × outside radius 3 - x³² - inside radius 3 - x² dx.
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That is the integral that we want to solve.
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Volume = π × integral from 0 to 1.
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In other words, this is technically where you can stop.
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You have converted the problem to the integral.
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Now it is just an integration problem.
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0 to 1, we got 9 – 6x³ + x⁶ - 9 - 6x + x², all of that dx.
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9 – 9, the - - this function².
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The minus is going to distribute over everything.
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We are going to get = π × integral from 0 to 1.
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We are left with -6 x³ + x⁶ + 6x - x² dx,
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is equal to the π × -6x⁴/ 4 + x⁷/ 7 + 6x²/ 2 - x³/ 3 evaluated from 0 to 1.
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When I add it all together, I get 1.31 π.
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If I have done my arithmetic correctly, which is always a tossup.
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In any case, that is it, getting the integral is the most important part.
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Changing the actual problem itself, the physical thing that we are looking at, into some sort of integral.
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The rest is integration.
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Let us take a look at example 2, find the volume of a solid obtained by rotating the region
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bounded by the following functions around the given axis.
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Again, we have y = x³, this time we have y = √x, x is greater than or equal to 0.
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We want to rotate this around x = 2.
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Let us go ahead and draw our region.
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I probably do not want to make it quite so big.
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Let me spare myself a little bit of room here.
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I have got y = x³, it is going to be like that.
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y = √x, it is going to look like that.
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They happen to meet at 1.
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Basically, y = x³ is going to go off that way.
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y = √x is going to go off that way.
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They are going to meet here.
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The region that I'm interested in is this region right here.
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Of course, x is greater than or equal to 0, I’m interested in this region.
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I’m rotating it around x = 2.
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If this is 1, this is 2, I’m rotating it that way which means I'm going to have that and I’m going to have that.
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This is the region than I'm interested in, this thing.
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I’m looking at it straight on, that is my solid.
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Draw the region, we have drawn the region.
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Identify the solid, we have identified the solid.
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Region is this way, rotating around this axis.
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The slice that we take is going to be perpendicular to the axis.
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If you are ever not sure how you are going to slice, you are going to slice perpendicular to the axis.
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This direction is not perpendicular to the axis.
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This direction is perpendicular the axis.
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Our washer is going to be, that is a characteristic washer.
00:27:41.100 --> 00:27:49.500
Since that is a characteristic washer, we are going to be adding washers in the vertical direction from 0 to 1.
00:27:49.500 --> 00:27:52.100
This is 0, this is 1.
00:27:52.100 --> 00:27:54.700
Our lower and upper limit of integration are 0 and 1.
00:27:54.700 --> 00:27:59.400
We are integrating along the y axis.
00:27:59.400 --> 00:28:03.400
When we are integrating along the y axis, we need functions of y.
00:28:03.400 --> 00:28:16.200
Identify the washer which means we integrate along y.
00:28:16.200 --> 00:28:30.400
You know what, I think I should probably write up my words a little bit better than this.
00:28:30.400 --> 00:28:39.500
Let us go ahead and identify our outside radius and our inside radius.
00:28:39.500 --> 00:28:47.300
Let us integrate along y and we are going to be going from 0 to 1.
00:28:47.300 --> 00:29:00.100
We know that the volume is equal to the integral from a to b of π × the outside radius² - the inside radius² dy.
00:29:00.100 --> 00:29:01.800
That is going to be our general equation.
00:29:01.800 --> 00:29:10.800
dy, we need functions of y not functions of x.
00:29:10.800 --> 00:29:24.500
We express these functions, that function, and that function.
00:29:24.500 --> 00:29:34.100
We express these functions as functions of y.
00:29:34.100 --> 00:29:45.400
y = x³ that is going to become x = y¹/3.
00:29:45.400 --> 00:29:54.400
y = √x that is going to turn into x = y².
00:29:54.400 --> 00:30:00.500
These are the functions that I’m going to be dealing with.
00:30:00.500 --> 00:30:06.000
I’m integrating along y which means my formula is going to have a dy.
00:30:06.000 --> 00:30:08.600
The differential is in the y direction.
00:30:08.600 --> 00:30:13.000
Therefore, my functions have to be in terms of y, that is what is going on.
00:30:13.000 --> 00:30:16.200
Let us go ahead and find our outside and inside radius.
00:30:16.200 --> 00:30:24.900
Outside radius is, we measure from the axis of rotation to the original function which is this one right here.
00:30:24.900 --> 00:30:30.700
Our outside radius is that one.
00:30:30.700 --> 00:30:35.200
Our inside radius is this one.
00:30:35.200 --> 00:30:39.700
Our outside radius is 2.
00:30:39.700 --> 00:30:44.000
This distance 2 - the distance from here to here.
00:30:44.000 --> 00:30:49.200
The distance from here to here is this, it is y².
00:30:49.200 --> 00:30:55.600
This x distance, whatever y is, our distance is y², that is the whole idea.
00:30:55.600 --> 00:31:00.300
We have 2 - y² is our outside radius.
00:31:00.300 --> 00:31:04.300
Our inside radius is the distance, it is from here to here.
00:31:04.300 --> 00:31:18.500
This distance is from here to here which is 2 - this distance which is the y¹/3, 2 – y¹/3.
00:31:18.500 --> 00:31:25.400
There we go, now we have got it.
00:31:25.400 --> 00:31:32.700
Our volume is equal to the integral from 0 to 1 of π × the outside radius
00:31:32.700 --> 00:31:48.000
which is 2 - y²² - the inside radius which is 2 – y¹/3² dy.
00:31:48.000 --> 00:31:51.000
That is the most important part, coming up with this integral.
00:31:51.000 --> 00:31:54.500
The rest is just integration.
00:31:54.500 --> 00:31:59.400
We have got π × the integral from 0 to 1.
00:31:59.400 --> 00:32:00.900
Let us go ahead and multiply all of these out.
00:32:00.900 --> 00:32:25.800
We have got 4 - 4y² + y⁴ - 2 × 2 is 4 - 4 × y¹/3 + y²/3 dy.
00:32:25.800 --> 00:32:33.800
Our 4’s go away, the negative sign distributes.
00:32:33.800 --> 00:32:38.100
I have not done the integral yet, be very careful.
00:32:38.100 --> 00:33:01.000
The integral from 0 to 1 of -4y² + y⁴ + 4y¹/3 – y²/3, all of that dy.
00:33:01.000 --> 00:33:06.000
In your AP exam as you know, there are going to be some sections where you are going to have to do the integration by hand.
00:33:06.000 --> 00:33:08.700
There are going to be some sections where you are going to do the integration,
00:33:08.700 --> 00:33:12.200
where you are going to be allowed to use your calculator.
00:33:12.200 --> 00:33:14.800
In this particular case, it depends.
00:33:14.800 --> 00:33:16.300
Sometimes, it is nice to use a calculator.
00:33:16.300 --> 00:33:17.800
Sometimes, it is nice to do it by hand.
00:33:17.800 --> 00:33:19.800
You will have both options.
00:33:19.800 --> 00:33:23.900
The idea is to be able to integrate this by hand, if you need to.
00:33:23.900 --> 00:33:29.100
But again, on the calculator section you are just going to be plugging this in and getting an answer.
00:33:29.100 --> 00:33:34.100
You are not going to be doing the integration like we are doing here.
00:33:34.100 --> 00:33:43.300
We have got π × this is going to be -4y³/ 3.
00:33:43.300 --> 00:33:46.500
This is going to be + y⁵/ 5.
00:33:46.500 --> 00:33:52.000
This is going to be + 4y⁴/3 / 4/3 - y⁵/3 / 5/3.
00:33:52.000 --> 00:34:08.300
Again, it is just my particular habit of not simplifying these and making this 3y⁵/3/ 5.
00:34:08.300 --> 00:34:11.300
It is up to you, the rest is just arithmetic.
00:34:11.300 --> 00:34:20.900
I ended up with 1.27 π, when I actually put this into my calculator.
00:34:20.900 --> 00:34:24.000
Let us try our third example.
00:34:24.000 --> 00:34:29.600
Find the volume of a solid obtained by rotating the region bounded by the following functions around the given axis.
00:34:29.600 --> 00:34:33.600
We have y = 2 sin x, y = 2 cos x.
00:34:33.600 --> 00:34:36.500
The x from π/4 to 5π/ 4.
00:34:36.500 --> 00:34:41.700
We want to rotate this around y = -3.
00:34:41.700 --> 00:34:45.800
Let us see what the heck is going to be going on here.
00:34:45.800 --> 00:35:00.200
Let me work in blue for my graph.
00:35:00.200 --> 00:35:01.600
I will go this way.
00:35:01.600 --> 00:35:04.500
This is my x and y axis.
00:35:04.500 --> 00:35:16.100
y = 2 sin x, it is a normal sin function, except the height is 2.
00:35:16.100 --> 00:35:20.800
Down here this is -2.
00:35:20.800 --> 00:35:31.800
Cos function starts here, it passes here, here, here, and here.
00:35:31.800 --> 00:35:37.900
The cos function looks like this.
00:35:37.900 --> 00:35:43.100
Not the best drawing in the world but I think we understand what is happening.
00:35:43.100 --> 00:35:48.100
π/4 to 5π/ 4, that is this point and this point.
00:35:48.100 --> 00:35:57.600
Our integration, if we decide to go along the x axis, it is probably going to be from π/4 to 5π/ 4.
00:35:57.600 --> 00:35:59.700
Let us generate the solid and see what we have.
00:35:59.700 --> 00:36:01.400
We have taken care of this.
00:36:01.400 --> 00:36:04.000
Let us rotate around the line y = -3.
00:36:04.000 --> 00:36:07.800
y = -3, let us go ahead and put it like right there.
00:36:07.800 --> 00:36:21.700
This is -3, we are going to rotate this region right here around this axis right here.
00:36:21.700 --> 00:36:22.900
Let us see what we get.
00:36:22.900 --> 00:36:29.300
We end up with something that looks like this.
00:36:29.300 --> 00:36:34.500
Again, I do not promise the best graph but here is something.
00:36:34.500 --> 00:36:36.500
Something like that.
00:36:36.500 --> 00:36:46.400
We have this nice region, this thing that we have rotated around there, rotated around there.
00:36:46.400 --> 00:36:49.900
We have this solid of revolution.
00:36:49.900 --> 00:36:52.000
Now we are going to take a slice.
00:36:52.000 --> 00:36:55.700
We take a slice perpendicular to the axis of rotation.
00:36:55.700 --> 00:37:00.100
The axis of rotation of this way, we are going to slice it this way.
00:37:00.100 --> 00:37:03.000
Our slice is going to be vertical.
00:37:03.000 --> 00:37:12.000
Our washer is that.
00:37:12.000 --> 00:37:25.300
When I take this, rotated it out a little bit like that, what you are going to get is a side view of your washer.
00:37:25.300 --> 00:37:27.100
That is your washer.
00:37:27.100 --> 00:37:31.700
It is going to have a certain thickness, that is going to be our dx.
00:37:31.700 --> 00:37:37.800
In this particular case, we are going to be integrating along the x axis.
00:37:37.800 --> 00:37:48.600
We are going to integrate along the x axis and we are going to go from π/4 to 5π/ 4.
00:37:48.600 --> 00:37:55.200
Lower and upper limit of integration, respectively.
00:37:55.200 --> 00:38:04.500
A little bit of an issue here with sins and cos and other functions that tend to occupy both above the axis and below the axis.
00:38:04.500 --> 00:38:07.500
We said we are going to measure our radius.
00:38:07.500 --> 00:38:08.900
We are going to have an outside radius.
00:38:08.900 --> 00:38:13.900
We are going to have an inside radius.
00:38:13.900 --> 00:38:21.100
Let me go to black, our outside radius, we are going to measure from the axis of rotation to the original function.
00:38:21.100 --> 00:38:24.800
Our outside radius is that one.
00:38:24.800 --> 00:38:30.900
Our inside radius is this one.
00:38:30.900 --> 00:38:34.700
Distance, what is this distance?
00:38:34.700 --> 00:38:45.700
It is the distance from here to here which is 3, because from -3 to 0 is 3, + this distance here.
00:38:45.700 --> 00:38:51.800
That is the function, that is the 2 sin x function.
00:38:51.800 --> 00:39:01.100
Our outside radius is going to be 3 + 2 sin x.
00:39:01.100 --> 00:39:04.000
Now inside radius, this is where it gets a little interesting.
00:39:04.000 --> 00:39:15.100
Again, it is going to be this distance 3 - this distance.
00:39:15.100 --> 00:39:19.700
3, it is actually going to be +, here is why.
00:39:19.700 --> 00:39:31.000
Wherever you are from π/4 to 5π/ 4, the cos is going to take on positive values and it is going to take on negative values.
00:39:31.000 --> 00:39:35.700
The function itself accounts for the negative sin here.
00:39:35.700 --> 00:39:46.500
Because sometimes it is going to be this + that, sometimes it is going to be this – that.
00:39:46.500 --> 00:39:48.200
The function itself accounts for it.
00:39:48.200 --> 00:39:53.100
It is just 3 + 2 cos x.
00:39:53.100 --> 00:39:57.800
In this particular case, this particular slice, the cos x function is below the axis.
00:39:57.800 --> 00:40:04.200
It is going to be 3 - 2 cos x.
00:40:04.200 --> 00:40:08.300
I will leave it as + because the 2 cos x is going to be negative itself.
00:40:08.300 --> 00:40:11.200
It is going to be 3 – a number.
00:40:11.200 --> 00:40:14.400
I hope that make sense.
00:40:14.400 --> 00:40:17.600
There we go, we have our outside radius, we have our inside radius.
00:40:17.600 --> 00:40:20.200
Now our volume is really simple.
00:40:20.200 --> 00:40:27.200
Our volume is going to be the integral from π/4 to 5π/ 4.
00:40:27.200 --> 00:40:33.800
We are going to integrate from π/4 to 5π/ 4.
00:40:33.800 --> 00:40:59.300
It is going to be π × outside radius 3 + 2 sin x² - the inside radius which is 3 + 2 cos x².
00:40:59.300 --> 00:41:03.400
We are integrating along x.
00:41:03.400 --> 00:41:08.300
This is the integral that we want.
00:41:08.300 --> 00:41:11.500
Sometimes that will be enough.
00:41:11.500 --> 00:41:14.400
Let us actually integrate it.
00:41:14.400 --> 00:41:35.100
We have volume = π × the integral from π/4 to 5 π/4 of 9 + 12 sin x
00:41:35.100 --> 00:41:52.500
+ 4 sin² x - 9 + 12 cos x + 4 cos² x dx.
00:41:52.500 --> 00:42:03.600
This is going to equal, the 9’s cancel, I always have a thing with 0, I do not know where it is.
00:42:03.600 --> 00:42:26.800
π/4 to 5π/ 4 of 12 sin x + 4 sin² x - 12 cos x - 4 cos² x.
00:42:26.800 --> 00:42:34.100
There we go, this is dx.
00:42:34.100 --> 00:42:43.700
This one, I just decided to go ahead and use a calculator to solve this, as opposed to doing the actual integration.
00:42:43.700 --> 00:42:50.400
Part of the reason is, we can handle this one and this one, we have already learned techniques for that.
00:42:50.400 --> 00:42:55.300
But in the lessons that follow, we are going to be discussing techniques of integration
00:42:55.300 --> 00:43:00.000
for other types of functions that we have to integrate.
00:43:00.000 --> 00:43:03.200
One of them is going to be the sin² function and the cos² function.
00:43:03.200 --> 00:43:11.000
Technically, as far as the progress of this course is concerned, we have not learned how to deal with these integrals formally.
00:43:11.000 --> 00:43:14.500
Just go ahead and use your calculator, in this case.
00:43:14.500 --> 00:43:21.700
But make sure that your calculator is in radian mode.
00:43:21.700 --> 00:43:24.400
Otherwise, you will get the wrong answer.
00:43:24.400 --> 00:43:33.400
Make sure your calculator is in radian mode, when you do any of the integrals.
00:43:33.400 --> 00:43:36.400
Definitely check that before you go in the AP exam or any other exam,
00:43:36.400 --> 00:43:40.500
when you are doing these and you are going to be using your calculator, radian mode.
00:43:40.500 --> 00:43:47.500
Everything is in radian mode in calculus.
00:43:47.500 --> 00:43:56.800
When I actually plug this in and do the integral, I get a volume equal to 33.94.
00:43:56.800 --> 00:44:00.500
Again, finding the integral is the important part.
00:44:00.500 --> 00:44:07.400
Let us do one more example.
00:44:07.400 --> 00:44:11.500
Find the volume of a solid obtained by rotating the region around the given axis.
00:44:11.500 --> 00:44:14.400
This time we have x² + y² = 9.
00:44:14.400 --> 00:44:17.900
We are given an equation in implicit form not explicitly.
00:44:17.900 --> 00:44:21.500
x is greater than 0, y is greater than 0, rotate around x = 3.
00:44:21.500 --> 00:44:26.400
We have to decide what we are going to do.
00:44:26.400 --> 00:44:29.600
Let us go ahead and turn this into an equation that we actually recognize.
00:44:29.600 --> 00:44:40.900
This is going to be x²/ 3² + y²/ 1² = 1.
00:44:40.900 --> 00:44:47.000
This is the equation of an ellipse.
00:44:47.000 --> 00:44:51.300
We are going to take the x is greater than or equal to 0 and the y greater than or equal to 0.
00:44:51.300 --> 00:44:56.800
What we are looking at is here.
00:44:56.800 --> 00:45:01.700
What we are looking at is an ellipse.
00:45:01.700 --> 00:45:05.300
The x direction, we are going to go 1, 2, 3.
00:45:05.300 --> 00:45:07.900
y direction we are going to go 1.
00:45:07.900 --> 00:45:16.000
But we want only the region that is above the x, axis above the y axis.
00:45:16.000 --> 00:45:19.200
It is just this, region right here.
00:45:19.200 --> 00:45:27.100
It continues on this way but the region that we are interested in is this region right here.
00:45:27.100 --> 00:45:30.600
They say rotate around x = 3.
00:45:30.600 --> 00:45:34.600
That is this axis right here.
00:45:34.600 --> 00:45:47.700
When we rotate around that axis, we are going to end up getting that region.
00:45:47.700 --> 00:45:48.800
We are going to take a slice of that region.
00:45:48.800 --> 00:45:52.600
We are going to take a slice perpendicular to the axis of rotation.
00:45:52.600 --> 00:45:57.300
The axis of rotation is vertical, the slice we are going to take is horizontal.
00:45:57.300 --> 00:46:07.400
Our representative washer, when we slice this solid, our solid is this.
00:46:07.400 --> 00:46:11.500
When we take a slice of it, we are going to end up getting a washer.
00:46:11.500 --> 00:46:17.300
When I take that washer and turn it this way, it is going to look exactly like you think it is going to look.
00:46:17.300 --> 00:46:21.700
It is going to look like this.
00:46:21.700 --> 00:46:22.800
It looks like that.
00:46:22.800 --> 00:46:29.200
It is going to have a thickness which is going to be dy.
00:46:29.200 --> 00:46:31.500
When we add up all these washers, we are adding them vertically.
00:46:31.500 --> 00:46:40.800
We are going to be integrating along the y axis.
00:46:40.800 --> 00:46:48.700
We are going to be integrating from 0 to 1, nice perfect.
00:46:48.700 --> 00:46:52.700
So far so good, now we are integrating along y.
00:46:52.700 --> 00:46:55.400
We need to express these functions as functions of y.
00:46:55.400 --> 00:47:11.100
We are going to turn this into a function of y.
00:47:11.100 --> 00:47:12.800
Let me just write that out.
00:47:12.800 --> 00:47:32.800
Integrating along y means expressing functions as functions of y.
00:47:32.800 --> 00:47:40.300
Now I have got x² + 9y² = 9.
00:47:40.300 --> 00:47:45.900
I have got x² = 9 – 9y².
00:47:45.900 --> 00:47:52.800
I have x = √9 - 9y².
00:47:52.800 --> 00:47:55.700
Do not worry about simplification, just leave it as is.
00:47:55.700 --> 00:48:00.400
Pulling out 9 and 3, it is not even worth it.
00:48:00.400 --> 00:48:04.100
Let us go ahead and talk about our inside radius and our outside radius.
00:48:04.100 --> 00:48:06.500
Let us talk about our outside radius first.
00:48:06.500 --> 00:48:07.600
I will go to black.
00:48:07.600 --> 00:48:14.000
We are going to measure from the axis of rotation to the original function.
00:48:14.000 --> 00:48:24.200
The outside radius is that one, the inside radius is this one.
00:48:24.200 --> 00:48:28.500
The outside radius is just 3.
00:48:28.500 --> 00:48:34.300
Nice because it stays always 3.
00:48:34.300 --> 00:48:39.000
The inside radius that is the radius that is going to be short or long.
00:48:39.000 --> 00:48:41.300
That is the one that changes.
00:48:41.300 --> 00:48:45.600
Inside radius, it is going to be 3.
00:48:45.600 --> 00:48:54.600
It is going to be this distance - this distance because we want this distance right here.
00:48:54.600 --> 00:48:57.500
This whole thing - this thing.
00:48:57.500 --> 00:49:00.200
This thing is that.
00:49:00.200 --> 00:49:09.500
It is 3 - √9 - 9y².
00:49:09.500 --> 00:49:13.300
There you go, now that we have that, very simple.
00:49:13.300 --> 00:49:23.400
We have our volume is equal to the integral from 0 to 1 of π × the outside radius
00:49:23.400 --> 00:49:42.600
which is 3² - the inside radius 3 - √9 - 9y²² dy.
00:49:42.600 --> 00:49:48.400
This is the integral that we were looking for.
00:49:48.400 --> 00:49:52.500
Let us go ahead and see if we can solve this.
00:49:52.500 --> 00:50:06.000
We have volume is equal to π × the integral from 0 to 1 of 9 - 9 -
00:50:06.000 --> 00:50:26.500
6 × 9 - √9y² + 9 - 9y² dy, which is equal to π × the integral from 0 to 1.
00:50:26.500 --> 00:50:46.200
This 9 and this 9 cancel, negative sign becomes 6 × √9 - 9y², this becomes -9 and this becomes +9y² dy.
00:50:46.200 --> 00:50:49.400
I just go ahead and plug this into my calculator.
00:50:49.400 --> 00:50:58.100
I get myself 8.14, that is my volume.
00:50:58.100 --> 00:51:15.500
Once again, you should note that some of these integrals, you have not learned how to deal with yet.
00:51:15.500 --> 00:51:17.800
Do not worry about it, that is absolutely fine.
00:51:17.800 --> 00:51:25.400
Some of these integrals, you have not dealt with formally.
00:51:25.400 --> 00:51:27.000
We have not dealt with formally.
00:51:27.000 --> 00:51:36.900
We will, very soon.
00:51:36.900 --> 00:51:40.100
That is it, that is volumes by the method of washers.
00:51:40.100 --> 00:51:41.800
Thank you so much for joining us here at www.educator.com.
00:51:41.800 --> 00:51:43.000
We will see you next time, bye.