 Hi everyone let's take a look at an example of a washer method problem for finding the volume of a solid of revolution. In this case we're talking about the region bounded by the trig curve y equals cosine of x the x-axis in between x equals zero and pi over two and we are revolving about the horizontal line y equals one and we are asked to find the volume of the resulting solid. So go ahead and graph it on your graphing calculator. I've already gone ahead and done so and taken a snapshot of the graph for you. I did adjust my window a little bit. I did zoom trig the first time this intersection right here where the trig function hits the x-axis of course is pi over two. This is negative pi over two over here. Here is your horizontal line y equals one and which happens to be our axis of revolution also. So if you take a look at the problem again let's make sure we have the correct region that we are revolving. Oftentimes that's the hard part just make sure you have the correct region. So we want the region bounded by the blue cosine curve the x-axis down here on the bottom and between x equals zero which of course is the y-axis over here on the left and y that x-coordinate of pi over two. All right so it's almost this triangular part that you see here in yellow. So imagine taking that yellow region and slipping it upward over that horizontal line y equals one. So with a disc or a washer method problem our representative rectangles of course are perpendicular to the axis of revolution but they are not touching it that is remember the big difference between a washer method and a disc method. So this is a dx problem because of the vertical orientation of our representative rectangle and you can see that depending upon where you drew that rectangle the hole is kind of bigger so like over here on the right you'd have a much bigger hole there than maybe over here if you drew the rectangle over here you have a teeny tiny little hole right in there. So if you think of the formula that we're going to build our definite integral off of it's going to be pi times big r squared minus little r squared. So big r is the distance from the axis of revolution at y equals one to the far side of your representative rectangle. So big r is going to be this distance here. So let's go ahead and get an expression to represent that. So we'll use the idea of top minus bottom so at the top we're hitting that horizontal line y equals one at the bottom we're hitting the x axis at zero so big r simply is one. Now let's do little r. Little r is going to be the distance from the axis of revolution to the close side essentially it's the radius of your hole. So little r again we're going to use top minus bottom so at the top it's hitting that horizontal line again at one minus and at the bottom it's hitting the curve the cosine curve. So one minus cosine x is the expression that we'll use for little r so now we just have to piece it all together in a definite integral. Remember that our limits of integration will have to be x value so in this case we're going to use zero and pi over two. So let's go back to the previous slide and we'll get it all set up. So our volume then is going to be the definite integral from zero to pi over two of pi and then remember we need big r squared so big r was simply one minus little r and we have to square that as we said it's a dx problem. You're most welcome to simply evaluate that in your calculator and the decimal answer you should get is approximately 3.816.