Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. The washer method, The volume of a solid of known integrable cross-sectional area A of X from X equal A to X equal B is the integral of a from A to B or the volume equals the limit as N approaches Infinity of the summation of K equal 1 to north of the area of the cross section or a of X sub K times the thickness or delta XK. This could be thought of as the integral of A to B of a of X DX where a of X is the area of the cross section. Solid generated by rotating a plane or shape about a vertical or horizontal line is called a solid of revolution. So when we look at this, we're going to think about if we're going around the X axis, we're going from our A to our B of our X. Each time. Our cross section is going to be a circle, and the area of a circle is π R-squared. We're going to get the radius by the farthest line out minus the closest line or the height of each of those. So in this case, capital R of X -, 0 quantity squared DX. Now sometimes we actually have a hole or it being hollow inside like with a bead. So now we're going to go from A to B π. But now we're going to have the outer radius squared and we're going to subtract the hollow portion or the inner radius squared. Let's look at some examples. First example is that we are going to go, we're going to have circles that are crossed along the curve zero. Our X values of zero to four of the parabola Y equal the square root of X or X equal y ^2. Now we're not rotating on this example, we're just putting in circular cross sections. So Pi, our radius squared is going to be from our height, our height being Y which is square root of X to the bottom height in this case being 0 so π ^2 of X ^2 or Pi X the integral of zero to four of π X DX. So when we evaluate that, we get π halves X ^2, 0 to 48, which would equal 8 Pi if we did the same concept. But this time we're actually going to put squares. So we have square root of X going to the top and square root of X coming to the bottom. And the area of a square is side squared. So we'd have two square roots of X being one of the sides, 2 square roots of X being another side. So to find that area, we're going to have two square roots of X ^2, which is really just 4X. And we're going to do the volume zero to four of four X DX. And if we evaluate that, we're going to get 2X squared from zero to four or 32 units cubed. Now we're going to do 1 where we're looking at y = X ^2 and Y equal 2 -, X ^2. And we're looking at the cross sections there. So the first thing we need to do is figure out what our bounds are going to be. If I set the two equations equal X ^2 and 2 -, X ^2, if I take all the XS to one side, everything else to the other, I can see X equal positive -1. Another way to do it would be to subtract everything to one side and factor. We'd also see positive -1 So we have π our radius squared. Well, our radius is going to be the upper Y value, the height minus the lower Y value. So we're going to have this 2 -, X ^2 as the upper minus the X ^2 the lower. But that's going to actually be a diameter of the circle, and we need π R-squared. So we're going to take that diameter and divide it by two. When we take the diameter and divide it by two, we're going to get 1 -, X ^2 ^2 from -1 to one times the π in the DX. Distributing that out, we get -1 to one of one minus two X ^2 + X to the 4th DX, all times the Pi. Finding the integration of that, we get Pi X -, 2 thirds X ^3 + 1/5 X to the fifth sticking in our bounds -1 to 116 Pi fifteenths cubic units. Now, all of those examples are not being rotated yet. They're just saying we're adding up a certain area along a cross section. Our next examples are going to look at rotations. So if we have Y equal, X -, X ^2, and y = 0, and we want to rotate it around the X axis, the first thing we need to do is figure out our bounds. So I'm going to set those two YS equal and solve. So X is going to be equal to 0 and X is going to be equal to 1. If we think about going around the X axis, we can see that we're going to have these circles here. So when we have these circles, we're going to have Pi radius squared, our upper upper Y value minus our lower Y value, our upper yx minus X ^2, our lower Y0, and we're going to square that. So we get 0 to one Pi X -, X ^2 -, 0 ^2 DX. If we evaluated that we'd get π, we'd have to foil this out, and we'd have X ^2 minus two X ^3 + X to the 4th DX, which would equal 1 third X ^3 -2 fourths, X to the fourth plus 1/5, X to the fifth from zero to 1, and all that still times π. So we'd have π * 1/3 -, 1/2 + 1/5 subtract 0 and that would end up being Pi 30th cubic units. So if we think about it, this is 10 thirtieths and this is 15 thirtieths and this is 6 thirtieths. If we look at Y equal to Y equal to sine X0 less than or equal to X less than or equal to π halves going around the O and bounded by the Y axis rotated around the line Y equal to. So if we look at this, we have Y equal to Y equal to sine X&Y axis. So we're having a figure that looks something like this with these circles. So the first thing to do is to realize they gave us our bound zero to π halves Pi radius squared. So we're going to think about each time this is our radius. So our upper is going to be AY value of two and our lower is going to be that 2 minus sine X. So 2 - 2 sine X ^2. If we foiled that out, we get π zero to π halves of 4 - 8 sine X + 4 sine squared X DX. So we get π zero to π halves 4 - 8 sine X. We don't know how to evaluate the sine squared X, but we know that sine squared X is really just one minus cosine 2X all over 2. So if we evaluate all this, we should get π zero to π halves 4 - 8 sine X - 4 / 2 is 2 so plus 2 - 2 cosine 2X DX so 4 + 2 6 - 8 sine X - 2 cosine 2X DX from zero to π halves. So Pi six X - 8 cosine X plus sine. Nope, wait minus sine 2X because the derivative of cosine was negative sine. The derivative of sine is cosine. But then we have to take the derivative of the inside doing the chain rule, which is where the two came from. So now if we evaluate this, we get Pi 6 * π halves would be three π + 8. Cosine Pi halves is 0 minus sine of of π, sine of π is 0 -6 * 0. But eight times cosine of 0 is 8 minus sine of 0 is 0. So we should get three π ^2 -, 8 Pi cubic units. Thank you and have a wonderful day.