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 be talking about finding volumes by the method of cylindrical shells.
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Let us jump right on in.
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How can we find the volume of the following region?
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Let me work in blue, I think.
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How to find the volume of the following region?
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We can go ahead and say, we have the function y = -3x³ + 4x²
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and we want x to be greater than or equal to 0.
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We want y to be greater than or equal to 0.
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We want to rotate this around the y axis.
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Let us see what region we are actually looking at.
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When we draw out this region, this function right here, when we draw it out,
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it is going to look something like this.
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It is going to end up hitting it at 1.33.
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x greater than or equal to 0, everything over here, everything in here are.
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We are looking at this region here.
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We are going to take this region and we are going to rotate it around the y axis.
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We are going to rotate this way.
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The solid of revolution that we are going to generate is going to be something symmetrical.
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This is -1.33.
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We are looking for the volume of this region, how can we do that?
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We have already dealt with washers, it is possible to do something like this.
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It is possible to basically take a washer, we will take a slice perpendicular to the axis of rotation.
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And then, we will add all the washers up along the y axis.
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We are going to integrate along y.
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Let us write this down.
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We could use washers, we try washers, I should say.
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We could try washers which mean we are going to integrate, in this case, along the y axis,
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or I will just say along the y direction.
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Since we are integrating along y, we need x equal to some function of y.
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In other words, we need to express this function as a function of y.
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For this particular function -3x³ + 4x², that is not going to be a very easy thing to do.
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Solving this function for x, in terms of y explicitly, it is not going to be very easy to do at all.
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If in fact if it is possible, I do not even know if it is, in this particular case.
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Here is where we run into a little bit of problem.
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It is doable, theoretically, but expressing this in terms of y is going to be hard.
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Because it is going to be hard, we ask ourselves is there another way of doing this?
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There is, let me actually write out.
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We are going to integrate along y so we need x = f(y).
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For this function, expressing x as a function of y is difficult.
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Is there another way, the answer is yes, there is.
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The answer is using things called cylindrical shells.
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Let us go ahead and redraw our little thing here.
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We have that, we have this, that is one region, that is another region.
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Let me go ahead and make this.
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Here we have 1.33, we have -1.33.
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Here is what I'm going to do.
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Instead of taking a washer which means slicing this solid horizontally
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perpendicular to the axis of rotation which we said is the y axis,
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that is the axis of rotation, I'm actually going to take a piece of it and go perpendicular to the axis of rotation.
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I’m going to pick this little slice right here.
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Not really a slice though, you will see why in a minute.
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Now I'm going to come over here.
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What I’m going to do, I have this solid.
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Essentially what I'm going to do, I’m going to take this little region.
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Now from your perspective, that region, I'm going to end up turning it towards you.
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I’m going to take this solid and you are going to end up taking this little, and actually turn it this way.
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When we rotate around the y axis, we are going to get a circular object.
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I’m going to basically bore into the object from the top, I’m going to turnaround this way.
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I’m going to get a cylindrical shell.
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When I take this down, what I end up getting is something like this, this region right here.
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What I actually done is I have taken this solid.
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When I view the solid from the top, I have actually bore into it and pulled out a cylindrical shell.
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All of this is the solid part, that is what this is, this is from the side.
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It is going to be of some radius here.
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Of course this is going to be some differential length dx.
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What this actually looks like is the following.
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This and this, when I turn it this way, it looks like this.
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The perspective drawing is actually this.
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This is h, this is h, it is the height of the cylindrical shell.
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This is r, that is this right here, that is r.
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The thickness is dx.
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Instead of taking washers and adding them up along the y axis,
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I'm going to take concentric circular cylindrical shells working my way out, from this point to this point.
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Shell, shell, shell, these are all a bunch of concentric cylindrical shells.
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Now I can integrate along the x axis.
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My function was a function of x.
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All I have done is instead of dealing with washers or disks, now I’m dealing with cylindrical shells.
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We need the volume of this one cylindrical shell.
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When I have the volume of this one cylindrical shell, I integrate all of the volumes.
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We need an expression for the volume of the cylindrical shell.
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What is that?
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What is the volume of this shell?
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Cut it and unroll it, in other words, slice it here and unroll it.
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What you end up having is a rectangle, what you end up having is a slab.
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That is what you are going to get.
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When you unroll this shell, you cut it, you unroll it, you are going to get a slab.
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This is c, that is going to be the circumference.
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It is going to be the circumference of the shell.
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This is h, that is the height.
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This right here, this, that is the depth.
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We know that the volume of this is equal to the circumference.
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This is length × width × height, which is circumference × height × depth.
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The circumference × the height × the depth.
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The circumference is equal to, this is r, the circumference is 2π r, 2π × the radius.
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The height is just the height of the shell and the depth is your dx.
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Our volume is going to equal 2π r h dx, circumference × the height × the dx.
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That is the volume of our one little shell.
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Let me draw real quickly again, let draw out in red here.
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We have this and we have this.
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We have this little shell right there.
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This is our shell right, that is our height, this is our radius.
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The radius is just a function of x.
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r is just the x value, the height is f(x).
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Our volume element of our little shell is equal to volume element as a function of x is equal to 2π × x × f(x) × dx.
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I just add up all the shells, as we are working out.
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As we are looking for the top, add up all the shells, working out in concentric circles.
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Now just add up all of the volumes of these individual shells, these differential shells.
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Our integration is going to go from 0, we are adding from here to here.
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We are adding up all of these shells, we are looking it sideways.
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We are going out from the center.
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Integrate from 0 to 1.33.
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Looking at this from the top, that is the solid looked at from the top.
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You are taking concentric cylindrical shells.
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You are integrating, integrating, you are adding them all up.
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Cylindrical shell from 0 to 1.33, that is all we are doing here.
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In general, let us go back to blue.
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Volumes by cylindrical shells, the volume = the integral from a to b of 2π x f(x) dx,
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if we are integrating along the x axis.
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This is for integrating along x and we have v = the integral from a to b of 2π f(y) dy.
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This is if we are integrating along y.
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Again, very important measure, the radius of the shell from the axis of rotation.
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Now because we are measuring from the axis of rotation, it is not always going to be just x and f(x), y f(y).
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This is not just, they give you an equation, plug it into the formula.
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These come from the actual physical situation that we described.
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Finding the shell, finding what the radius is from the axis of rotation.
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There are better formulas than these, more general, that you should actually concentrate on.
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These are just thrown out there because it is what you are going to see in your book.
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You have to let the situation decide what the radius is going to be and what the function is going to be.
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It is better if we write the following.
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Let me do this in blue.
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Better formulas are, volume = the integral from a to b of 2π × the radius of the shell × the height of the shell × dx.
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Or the integral from a to b of 2π radius of the shell, the height of the shell × dy,
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depending on whether we are integrating with respect to x or respect to y.
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You have to find r as a function of x and h as a function of x.
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r as a function of y, h as a function of y.
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I will write where r and h are functions of x.
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Here where r and h are functions of y.
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Probably, we want to avoid those.
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They are accurate for a given situation, when you are revolving around the x axis or y axis.
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But we are always going to be revolving around those axis.
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We might be revolving around any other line x = 5, y = -6.
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These are the general equations.
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Once you have decided based on the situation that you are going to be integrating along x,
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your r and your h, that your going to get from the picture are going to be functions of x.
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If you decided that you are going to integrate along y, then your r and your h,
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the radius of the shell and the height of the shell have to be functions of y.
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These are the equations that you want to use.
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Let us do examples because I think that makes everything clear.
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Use the method of cylindrical shells, specifically cylindrical shells.
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Find the volumes of the solid generated by rotating the region bounded by the following expressions about the given line.
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x² – 5x + 8, 0 to 4, we want to rotate this around the y axis.
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We need to know what this thing looks like, that is the whole idea.
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We need to see what that looks like.
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I’m going to go ahead and complete the square, turn it into a form where I know where the vertex is,
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and then find f(0), f(4), and then, we will rotate that region.
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Let us start off by going, I have got y is equal to x² - 5x + 8.
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I think I’m going to go over here.
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I’m going to take y - 8 = x² - 5x.
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I’m going to complete squaring this, I’m going to take half of the 5/2 and I’m going to square it and add it.
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I'm going to add 25/4 over here which means I'm going to add 25/4 over on the left hand side to retain the equality.
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I get y – 7/4 is equal to x – 5/2² which implies that my vertex is at 2.5 and 1.75.
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When I do y(0), I'm going to get y(0) is equal to 8.
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I'm going to get y(4) is equal to 4.
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Now I have a picture that I can work with.
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Now I got this, now my vertex is at 2.5, 1.75.
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Let us go 1, 2, let us go 1, 2.
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I will keep this as 1, 2, but I will make my 1, 2, 3, 4, 5, 6, 7, 8.
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I have different scales on my x and y axis.
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My vertex is 2.5 and 1.75.
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0 and 8, f(3) to 4.
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I’m at 1, 2, 3, 4.
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That is my region, this is the region that I’m talking about.
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I need to be able to draw it.
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Now I’m rotating this around the y axis.
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It is going to look something like that.
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This is my region.
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Cylindrical shells, the axis of the shell, the axis of rotation is the center of the shell.
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The little sliver that you draw is going to be parallel to the axis of rotation.
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In this particular case, my shell is going to be like this.
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I rotate that, this is going to be the other side of the shell.
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When I take this and I turn it towards me, now I'm going to see my circle like that.
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This thing rotated in perspective drawing, it is going to look like,
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This is the height, that is this right there.
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Let me work in black.
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This is my radius, that is x, the x value.
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This is my height, that is my f(x), that is what is going on here.
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The radius is equal to x and the height is equal to y which is equal to x² - 5x + 8.
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I have gotten the picture telling me what is going on, I’m not just putting it in.
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I know my general formula is volume = the integral from a to b of 2π r h dx.
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In this particular case, r is x, I put that in there.
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h is y which is the x² – 5x + 8.
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I put it in there and then I integrate but it is based on this.
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Once again, the axis of rotation which in this case was the y axis, is the center of the shell.
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It is the axis of the cylindrical shell.
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Side view of the shell looks like that.
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This is the top view of the shell, this is the perspective view of the shell.
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I’m integrating from 0 to 4.
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I’m taking shells concentrically out.
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Let us write it all out, let us go back to blue.
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I have got the volume = the integral from a to b 2π r h dx.
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The integral from 0 to 4, 2π x x² - 5x + 8 dx, this is what is important.
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The rest is just integration.
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This is equal to 2π × the integral from 0 to 4, x³ - 5x² + 8x dx
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= 2π × x⁴/ 4 - 5x³/ 3 + 8x²/ 2 from 0 to 4.
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When I do this, I get 21.33, that is all.
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Again, this is what is important, being able to form the integral.
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The rest is just integration problem.
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Let us do another example.
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Using the method of cylindrical shells, find the volume of the solid generated by rotating the region
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bounded by the following expressions about the given line.
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Let us see what we have got.
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We got y = 1/8 x³.
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Let us see what we have got.
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I have got this and we are going to rotate about the x axis this time.
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Let me make this a little bit smaller, actually.
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I will use a little bit more room.
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Let me go here.
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This is my y axis, this is my x axis, y = 1/8 x³.
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Something like that.
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y = 10, it is going to be up here.
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Let us just say that is the line y = 10 and x = 0.
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This is the region that I'm interested in.
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I’m going to rotate this around the x axis this way.
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This is going to be like this, then, I’m going to get a region like that.
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This is my region.
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They are saying specifically, use cylindrical shells.
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The shells are going to be the length.
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The sides of the shell are going to be parallel to the axis of rotation.
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In other words, the axis of rotation is going to be the center of the shell.
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The axis of rotation is the x axis.
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Our shell is going to be parallel to that, that is this way.
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If I were to take this and turn it that way, looking at it straight on, I will be looking at something like this.
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I hope that make sense.
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The radius of this shell is that value, it is y, that is the radius.
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The height of the shell is that, that is x.
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Let us see what we have got.
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We are going to be integrating along the y axis.
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We integrate along y.
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Because we are integrating along y, we need our functions to be functions of y.
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This is a function of x.
00:28:51.600 --> 00:28:57.100
Let us go ahead and see what we can do, radius = y.
00:28:57.100 --> 00:29:01.500
In this particular case, the radius we said is equal to our y value.
00:29:01.500 --> 00:29:06.500
We have radius is equal to y.
00:29:06.500 --> 00:29:10.500
We need our height, the height is x.
00:29:10.500 --> 00:29:13.100
We need x, in terms of y.
00:29:13.100 --> 00:29:28.800
We have got y = 1/8 x³.
00:29:28.800 --> 00:29:34.900
We have 8y is equal to x³.
00:29:34.900 --> 00:29:40.700
We have x is equal to 2y¹/3.
00:29:40.700 --> 00:29:48.300
x is equal to h, it is the height.
00:29:48.300 --> 00:29:54.600
Therefore, the height is equal to 2y¹/3.
00:29:54.600 --> 00:30:01.900
Now that we have our radius and we have our height, we can go ahead and do this problem.
00:30:01.900 --> 00:30:13.800
Volume = the integral from a to b, 2 × π × the radius × the height dy.
00:30:13.800 --> 00:30:24.000
We are integrating from 0 to 10 shells.
00:30:24.000 --> 00:30:36.700
Looking at it from the top, I see these concentric shells going outward from a radius of 0 all the way up to a radius of 10.
00:30:36.700 --> 00:30:42.800
Shell, shell, shell, until I have covered all of them, that is what is happening.
00:30:42.800 --> 00:30:56.200
We have got 2 × π × y × 2y¹/3 dy.
00:30:56.200 --> 00:31:06.300
This is going to equal 4π × the integral from 0 to 10 of y.
00:31:06.300 --> 00:31:15.900
y × y¹/3 is going to be y⁴/3 dy.
00:31:15.900 --> 00:31:26.600
It is going to be y⁷/3 / 7/3 from 0 to 10.
00:31:26.600 --> 00:31:39.700
Our final answer is going to be 3/7, 10⁷/3, that is all.
00:31:39.700 --> 00:31:43.500
Example number 3, let us go back to blue here.
00:31:43.500 --> 00:31:52.400
Cylindrical shells, we have y = √2x, we have y = 0, and we have a line x = 2.
00:31:52.400 --> 00:31:56.200
Let us go ahead and draw this out.
00:31:56.200 --> 00:32:07.000
Let us do x = 2, we want to rotate along x = -2.
00:32:07.000 --> 00:32:10.400
Let us go ahead and draw this out this way.
00:32:10.400 --> 00:32:15.100
We have got this.
00:32:15.100 --> 00:32:18.300
This is our y axis.
00:32:18.300 --> 00:32:26.700
I do not need to make it this big and this one I might need to.
00:32:26.700 --> 00:32:31.900
This is our 0,0, this is our x, the y = √2x.
00:32:31.900 --> 00:32:36.000
Some functions is going to look like that.
00:32:36.000 --> 00:32:42.700
y = 0 that is this line, x = 2.
00:32:42.700 --> 00:32:49.000
Let us go over here.
00:32:49.000 --> 00:32:52.900
This is the region that we are interested in, that is our region.
00:32:52.900 --> 00:32:57.600
We are going to rotate about the line x = -2.
00:32:57.600 --> 00:32:59.400
Here is -1 and -2.
00:32:59.400 --> 00:33:02.300
This is our axis of rotation.
00:33:02.300 --> 00:33:12.700
When we rotate about that lines, that means we are going to go two more over here.
00:33:12.700 --> 00:33:16.500
We are going to have that.
00:33:16.500 --> 00:33:23.700
We are going to have this region right here.
00:33:23.700 --> 00:33:31.500
This, this, this, this is our region, this is the y axis, this is the x axis.
00:33:31.500 --> 00:33:37.900
We are going to take this region, we are rotating it around the line x = -2.
00:33:37.900 --> 00:33:46.600
This is our solid of revolution, we are going to use cylindrical shells.
00:33:46.600 --> 00:33:51.300
The axis of rotation is the center of the shell.
00:33:51.300 --> 00:33:58.200
The sides, the walls of the shell are parallel to the axis of rotation.
00:33:58.200 --> 00:34:05.200
It is going to be here and here.
00:34:05.200 --> 00:34:09.300
Looking at it from the side, cylindrical shell is opening out.
00:34:09.300 --> 00:34:29.900
Now we are integrating along the x and we are going to integrate from 0 to 2.
00:34:29.900 --> 00:34:32.500
We have taken care of the lower and upper limit of integration.
00:34:32.500 --> 00:34:37.200
Now we need to find the radius, we need to find the height.
00:34:37.200 --> 00:34:41.200
We measure the radius from the axis of rotation.
00:34:41.200 --> 00:34:43.300
The axis of rotation is over here.
00:34:43.300 --> 00:34:54.600
It is at the point -2, distance is what we are worried about.
00:34:54.600 --> 00:34:55.800
Let me work in black.
00:34:55.800 --> 00:35:05.700
The radius from the center, from the axis of rotation which is the center of the shell to the wall of the shell,
00:35:05.700 --> 00:35:13.000
that is going to be this distance which is 2 + this distance which is x.
00:35:13.000 --> 00:35:16.200
Our radius is equal to 2 + x.
00:35:16.200 --> 00:35:21.400
Notice, it is not just x, our axis of rotation has changed.
00:35:21.400 --> 00:35:25.500
It is the radius, the length, that matters.
00:35:25.500 --> 00:35:36.500
Our height that is going to equal this which is f(x).
00:35:36.500 --> 00:35:43.900
Our height is equal to √2x, now we have everything that we need.
00:35:43.900 --> 00:35:46.600
Let me go ahead and erase this.
00:35:46.600 --> 00:35:48.600
This is our y = -2.
00:35:48.600 --> 00:35:52.100
Notice, distance is what we want.
00:35:52.100 --> 00:35:55.900
Even though this is -2, this is not negative, this is 2.
00:35:55.900 --> 00:36:08.400
2 + the x, in order to take me from the center of rotation of the shell to the actual wall of the shell, that is the radius.
00:36:08.400 --> 00:36:12.200
I think I will do it in red, I love changing the colors.
00:36:12.200 --> 00:36:22.900
Volume = the integral from a to b of 2π r h dx, that is our general formula,
00:36:22.900 --> 00:36:38.500
= the integral from 0 to 2 of 2 × π × 2 + x × h which is √2x × dx.
00:36:38.500 --> 00:36:43.500
This is what we want, that is the integral.
00:36:43.500 --> 00:36:48.700
The rest is just integration.
00:36:48.700 --> 00:36:57.800
Let us go ahead and go over there.
00:36:57.800 --> 00:37:07.300
We have got 4 √2π from 0 to 2 of x ^ ½.
00:37:07.300 --> 00:37:11.400
I pulled out the √2 and I just left the √x in there.
00:37:11.400 --> 00:37:30.500
dx + 2 √2π, the integral 0 to 2, x³/2.
00:37:30.500 --> 00:37:40.700
I distributed and just separated the integral.
00:37:40.700 --> 00:38:07.700
I get 4 √2 × π × x³/2 / 3/2, from 0 to 2 + 2 √2 × π x⁵/2 / 5/2 from 0 to 2.
00:38:07.700 --> 00:38:25.100
When I work all of this out, I end up getting 64π/ 3 + 128π/ 5, that is my total volume.
00:38:25.100 --> 00:38:28.500
Again, the rest is just arithmetic which I will leave to you.
00:38:28.500 --> 00:38:32.200
Finding the integral was the important part.
00:38:32.200 --> 00:38:35.800
Integration is important, it is actually where a lot of the problems happen
00:38:35.800 --> 00:38:40.700
because you are dealing with arithmetic issues, +, -, √, this and that.
00:38:40.700 --> 00:38:45.100
In any case, that is the nature of the game.
00:38:45.100 --> 00:38:51.200
Let us take a look at example 4, let us see what we have got here.
00:38:51.200 --> 00:38:57.600
Cylindrical shells, y = x⁴, x = y⁴, we want to rotate about x = 1.
00:38:57.600 --> 00:39:01.200
Let us go back to blue here.
00:39:01.200 --> 00:39:12.200
Let us draw this out.
00:39:12.200 --> 00:39:14.200
y = x⁴ is going to look something like that.
00:39:14.200 --> 00:39:18.600
x = y⁴ is going to look something like that.
00:39:18.600 --> 00:39:26.400
They are going to meet at 1, rotate about the line x = 1 which is this line.
00:39:26.400 --> 00:39:31.600
Now we have got that and we have got that.
00:39:31.600 --> 00:39:43.200
This is our solid that we want to find the volume of, rotated about x = 1, that is the center of the shell.
00:39:43.200 --> 00:39:52.200
The walls of our shell, our representative shell is going to be something like this.
00:39:52.200 --> 00:40:00.600
We are going to be integrating along x.
00:40:00.600 --> 00:40:12.500
We are going to be integrating from 0 to 1, lower and upper limits of integration.
00:40:12.500 --> 00:40:23.300
Let us see, what do we want to do next?
00:40:23.300 --> 00:40:27.100
We need to find r and h.
00:40:27.100 --> 00:40:55.300
Our radius from the axis of rotation is going to be, my radius is going to be that.
00:40:55.300 --> 00:41:04.600
This is my radius and my height is going to be this.
00:41:04.600 --> 00:41:17.100
My radius is going to be this distance - that distance.
00:41:17.100 --> 00:41:26.400
1 - the x value, our radius is 1 – x.
00:41:26.400 --> 00:41:44.900
My height that is going to equal the top function of x - the bottom function of x, that is my height.
00:41:44.900 --> 00:41:59.700
It is going to equal the top function of x which is x¹/4.
00:41:59.700 --> 00:42:03.200
I need functions of x here because I’m integrating along x.
00:42:03.200 --> 00:42:08.400
This one is fine, I need to convert that to a function of x.
00:42:08.400 --> 00:42:17.100
x = y⁴, this is the same as y = x¹/4, that is my top function.
00:42:17.100 --> 00:42:25.800
It is x¹/4 - my bottom which is x⁴.
00:42:25.800 --> 00:42:32.200
Now that I have my r and my h, volume is simple.
00:42:32.200 --> 00:42:58.600
Volume = the integral from a to b of 2π r h dx which = the integral from 0 to 1 2π 1 - x × x¹/4 – x⁴ dx.
00:42:58.600 --> 00:43:01.200
That is what I want.
00:43:01.200 --> 00:43:28.100
When we solve this, we end up with 2π, the integral from 0 to 1 of x¹/4 – x⁵/4 – x⁴ + x⁵ dx
00:43:28.100 --> 00:43:51.600
= 2π x⁵/4 / 5/4 – x⁹/4 / 9/4 – x⁵/ 5 + x⁶/ 6, from 0 to 1.
00:43:51.600 --> 00:44:04.600
When I work that out, I get 0.322 or whatever numbers you happen to put in there, when you put the 1 and 0 in.
00:44:04.600 --> 00:44:08.700
Let us go ahead and try one more example here.
00:44:08.700 --> 00:44:17.400
Cylindrical shells, we have the function y = x + 3/x, y = 10.
00:44:17.400 --> 00:44:22.300
We want to rotate about the line x = 12.
00:44:22.300 --> 00:44:41.500
Let us see what we have got here.
00:44:41.500 --> 00:44:44.100
As far as the graphing is concerned, you can use your graphing calculator,
00:44:44.100 --> 00:44:47.900
you can use any graphical tool that can find on the internet.
00:44:47.900 --> 00:44:53.400
You can go ahead and express this as a rational function and use the techniques of differential calculus to graph it.
00:44:53.400 --> 00:45:00.100
Suffice it to say that y = 10, we are looking at that.
00:45:00.100 --> 00:45:07.900
The graph itself actually looks like this.
00:45:07.900 --> 00:45:15.400
It goes down and it comes up like that, whatever technique that you need to use in order to graph it,
00:45:15.400 --> 00:45:20.400
that is what the graph of this looks like.
00:45:20.400 --> 00:45:22.700
y = 10, that is this line right here.
00:45:22.700 --> 00:45:27.000
That is y = 10, this is the region that we are concerned about.
00:45:27.000 --> 00:45:29.400
It is this region that we are going to be rotating.
00:45:29.400 --> 00:45:32.900
They say rotate around the line x = 12.
00:45:32.900 --> 00:45:37.800
It turns out that x = 12 is actually right about there.
00:45:37.800 --> 00:45:44.500
This is 12, therefore, our region is going to be something like that.
00:45:44.500 --> 00:45:50.000
I probably draw a little bit better than that.
00:45:50.000 --> 00:45:56.400
It is going to come down to about right there.
00:45:56.400 --> 00:45:58.100
Something like that.
00:45:58.100 --> 00:46:01.600
This is a solid that we are dealing with.
00:46:01.600 --> 00:46:13.200
Our axis of rotation is 12 which mean that the sides of the shell are going to be parallel.
00:46:13.200 --> 00:46:30.000
When we measure r and h, r again, we are measuring from the axis of rotation to the original function, that is r, that is h.
00:46:30.000 --> 00:46:43.400
r is equal to this length 12 - this length which is x.
00:46:43.400 --> 00:46:45.500
It is 12 – x.
00:46:45.500 --> 00:47:00.600
The height is 10 – f(x), it is this height - that height.
00:47:00.600 --> 00:47:14.500
It is 10 – f(x) which = 10 - x - 3/x.
00:47:14.500 --> 00:47:18.000
Now we need the a and b.
00:47:18.000 --> 00:47:25.300
We have our r, we have our h which is this thing.
00:47:25.300 --> 00:47:31.600
The question is what are a and b?
00:47:31.600 --> 00:47:37.000
I'm going to integrate from where the two graphs meet, the x value.
00:47:37.000 --> 00:47:43.500
This one and this one.
00:47:43.500 --> 00:47:51.900
Where do the two graphs meet?
00:47:51.900 --> 00:48:07.000
Just set them equal to each other, x + 3/x = 10, and solve.
00:48:07.000 --> 00:48:27.000
Let us do x + 3/x + 10, x + 3/x = 10 which gives us x + 3/x -,
00:48:27.000 --> 00:48:32.200
Let us get this right, 3/ x - 10 is equal to 0.
00:48:32.200 --> 00:48:39.900
We are going to get x² + 3 - 10x = 0.
00:48:39.900 --> 00:48:44.900
We have x² - 10x + 3 = 0, this is a quadratic.
00:48:44.900 --> 00:48:53.600
When I solve this, I get x = 0.31 and I get x = 9.69.
00:48:53.600 --> 00:48:59.700
I'm going to be integrating from 0.31 to 9.69.
00:48:59.700 --> 00:49:08.400
My volume is equal to the integral from a to b 2π r h dx.
00:49:08.400 --> 00:49:21.400
The integral from 0.31 to 9.69 2π, my radius we said was 12 – x.
00:49:21.400 --> 00:49:31.300
Our function, our h, our height was 10 – x - 3/x.
00:49:31.300 --> 00:49:33.600
We are integrating along dx.
00:49:33.600 --> 00:49:38.800
Plug this into your calculator, this is going to be one of those situations where you definitely going to need a calculator.
00:49:38.800 --> 00:49:50.700
My final answer for volume is going to be 2π × 301.29, that is all.
00:49:50.700 --> 00:49:59.100
The important thing is coming up with this.
00:49:59.100 --> 00:50:01.500
Thank you so much for joining us here at www.educator.com.
00:50:01.500 --> 00:50:02.000
We will see you next time, bye.