Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Area of surfaces of revolution. Surface area of a cylindrical surface is going to be the delta X * 2π Y or the change of X * 2π Y. If we think about unwrapping a cylinder, it really is just a rectangle. The frustum of a cone is going to be 2π times the average of the Y times the distance of the diagonal. If we think about that and take just a little piece of it, we're going to see that our length is just the change of X ^2 plus the change of y ^2, the square root of that for the length. So we're going to have 2π. Our Y values are going to change from our F of X sub K -, 1 plus our F of X sub K all over two. And then the length of that diagonal is just going to be the square root of the delta X sub K ^2 plus the delta Y sub K square. So then if we look at that and we think about adding up all of those pieces, we end up with a summation K equal 1 to north of 2π. The average of those YS times the distance of the slant. If the function of F is differentiable by the mean value in theorem, there is a point CKF of CK on the curve between P&Q where the tangent is parallel to the segment PQ. So the first derivative at C sub K has to equal the slope or the change of Y over the change of X. If we cross multiply we can see the change of Y equals the slope times the change of X and then we're going to substitute that into the summation formula. So we're going to have the 2π times the average of the YS, but instead of this delta X sub K ^2 plus delta X sub KF prime CK squared, we're going to take this out and we're going to put that in. We're going to take out the delta YK and put that in. We see that there's a delta X sub K in both of them, so we're going to factor out that. And then the square root of the square makes that just plain old delta X sub K, leaving us sqrt 1 + F prime C of K ^2. So the formula is going to be an integral from A to B of 2π the Y. So instead of the average of the Y, now we're just actually taking the Y height square root 1 + F prime X quantity squared DX. Or we could think of it as A to B of 2π Y instead of F of X, just the height sqrt 1 + d YD X ^2 times DX. If we were going to go around the Y axis, it would be CD 2π X square root 1 + d XD y ^2 d Y. A generic formula that's sometimes thought about is the L is equal to the integral of 2π rho where rho is the radius, DS where DS is the bandwidth, or DS generically could be thought of as square root of DX squared plus dy squared. Let's look at a few examples. If we have the example of Y equal X ^2 from zero less than or equal to X less than or equal to 2, and we want to go around the X axis, we're going to set up the integral for the area of the surface generated. So the area of the surface generated is going to be A to B 2π. Our radius here is going to be Y and then we're going to have sqrt 1 + d YD X ^2 DX because we're going around the X axis. One way that I think about it is if I'm going around the X axis, it's got to be a DX, so 2π. If I have to do it in terms of DX instead of Yi need to put in what Y equaled, which was X ^2. The square root 1 plus dy DX up here is just going to be two X SO1 plus the quantity 2X squared DX. When we evaluate that, we get zero to two 2π X squared square root 1 + 4 X squared DX. Now, if we wanted to graph this, we could grab our graphing calculators and we could just graph from zero to two of the graph Y equal X ^2. So we can see that it's going to go zero to 02 to 4:00. If we wanted to find this length, we can't do this currently with any of the skills that we have for our calculus. We can't do AU sub, we can't do anything we know as of yet. So our calculus has our calculators have a feature called the function interval FNINT, and we're just going to put in 2π X squared square root, 1 + 4 X squared, X, 0, two. We're going to get approximately 53.23 units squared. We could also just type this into. If we're using our inspires, we could type this directly into our inspires and get the same answer. Another example here, we're going to have Y equal X / 2 + 1/2 one less than or equal to X less than or equal to three, but around the Y axis. So that means it has to be in terms of DY. So we have A to B 2π X square root 1 + d XD y ^2 d Y. Well, if we're doing it DY, everything has to be in terms of Y. So we're going to take our original equation and we're going to solve for X, multiply everything through by two, subtract our One X = 2 Y -1. We also need to figure out our new bounds. So if X was one, when we stick in 1/2 + 1/2, we get Y being one. If X was 33 halves plus 1/2, four halves or two. So we get one to two for our Y. Taking our DX dy, we get 2. So now literally we're just going to plug it in. So we're going to go from one to two of 2π. The X value in terms of Y is 2 Y -1, the square root 1 plus our DX dy squared DY. This is root 5, and we can pull it out because it's a constant. So 2π root 5 integral, one to two, two y -, 1 dy. So when we evaluate that integral, we get y ^2 -, y from 1:00 to 2:00, or four Pi root 5 units squared. Looking at another example, if we have X equal 1/3 Y to the three halves minus Y to the one half, one less than or equal to Y less than or equal to three for the Y axis, we're going to have DX dy equaling 1/2 Y to the 1/2 -, 1/2 Y to the negative and 1/2 or Y halves over 2 - 1 / 2 Y to the half. So our integral is going to go one to three 2π our X in terms of our Y. So we're going to put in 1/3 Y to the three halves minus Y to the half. Then we have sqrt 1 plus our DX dy squared dy. Now, there's a lot of algebra involved, so we're going to start by squaring that Y to the half over 2 -, 1 / 2 Y to the half. That gives us Y to the 4th minus 1/2 + 1 / 4 Y. Combining our like terms there, we get Y to the 4th y / 4 + 1/2 + 1 / 4 Y, all underneath the square root. That's going to factor into Y to the half over 2 + 1 / 2 Y to the half squared. Then the square root and the square are going to cancel. So then we're going to have a two in both of these denominators. So I'm going to factor out the two so that 2π / 2 would just give us π and we have 1/3 Y to the three halves minus Y to the half times Y to the half plus Y to the negative half dy. We're going to foil that out first, outer, inner last. Remembering when we multiply variables, we add the exponents. So Y did the three halves times Y to the half. Three halves plus 1/2 is 4 halves divided by two is 2. So we get one third y ^2 + 1 third y -, y -, 1, DY the integral from one to three all times π. When we evaluate that, we end up with a -16 ninths π. But because we're doing area, area has to be positive. So we're going to have 16 ninths π unit squared. Thank you and have a wonderful day.