Section 6.2 Reading Assignment: Volumes Using Cylindrical Shells

Answer Only Exercise 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for.

Make sure to consider how to determine the shell height based on the graph. This type of formula is not universal, so it'd be best to approach this from the point of view of how we can determine this height without already knowing it.

Section 6.2 Reading Assignment: Volumes Using Cylindrical Shells Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages of the textbook. In this section, we find the volume of solids of revolution using a different method. This approach to volume tends to be a bit unintuitive, so it might be worth spending some extra time here to understand the reasoning behind this method. Exercise 1. Read the paragraph starting with "Each shell sits over.." on p. 380 until the end of the page and look at Figure 6.18. Summarize how the formula for the volume of one of these cylindrical shells is obtained. Note that we are looking at the volume of a single shell here, not the total volume of the entire solid. Exercise 2. Read the introduction to the subsection "The Shell Method" (p. 381 - 382). Summarize the Riemann sum argument for the volume of the total solid here, being sure to include where the previous question comes into play for this. Notice that we have two separate quantities to worry about using this method: the shell radius and the shell height. The shell radius will be relatively simple, typically just x or y, but it is worth keeping in mind. Looking at Figure 6.20 (p. 382), notice that the shell height of the shell is parallel to the axis of revolution. This is opposite from the disk/washer method, where the radius lines are perpendicular to the axis of revolution. This means that if we use both methods for the same solid, the resulting integrals should have different variables of integration. This can be a problem for some students when setting up these integrals. Check out Examples 2 and 3 (p. 382 -383) to see how this method works, especially looking at how the graphs of the regions of revolutions (in 2D) help set up the integrals. Example 3 in particular has a wrinkle, which I want you to consider in the next problem. Exercise 3. Read Example 3 (p. 383) and look at Figure 6.21. Explain where the formula for the shell height comes from based on the figure in this example. Be sure to include how the graph is related to the formula in your answer. This last example is an important reason to graph the region of revolution. The graph is an important visual aid to help setup the integral and helps where the algebraic description alone can't.Chapter 6 Applications of Definite Integrals 380 Chapter 6 Applications of Definite Integrals -2 -1 Axis of Axis of revolution -2 revolution (a) FIGURE 6.16 (a) The graph of the region in Example 1. before revolution. (b) The solid formed when the region in part (a) is revolved about the axis of revolution * = =1. another cylindrical slice around the enlarged hole, then another, and so on, obtaining n cylinders. The radii of the cylinders gradually increase, and the heights of the cylinders follow the contour of the parabola: shorter to taller, then back to shorter (Figure 6.16a). The sum of the volumes of the shells is a Riemann sum that approximates the volume of the entire solid. Each shell sits over a subinterval [X-1 X] in the x-axis. The thickness of the shell is Ax = Xx - Xx-1. Because the parabola is rotated around the line x = - 1, the outer radius of the shell is 1 + xx- The height of the shell is the height of the parabola at some point in the interval [X 1, *], or approximately y = f(x) = 3x - M. If we unroll this cylinder and flatten it out, it becomes (approximately) a rectangular slab with thickness An (see FIGURE 6.17 A cylindrical shell of Figure 6.18). The height of the rectangular slab is approximately y = 3.x - x1, and its height y obtained by rotating a verti- length is the circumference of the shell, which is approximately 2n - radius = 2w(1 + .q). cal strip of thickness Ax, about the line Hence the volume of the shell is approximately the volume of the rectangular slab, which is * = -1. The outer radius of the cylinder occurs at x, where the height of the pa- AV = circumference X height X thickness rabola is * - 3x - *,' (Example 1). = 2w(1 + x). (3xx - x] ) . A.x-234 Outer circumference = 24 . radius = 2w(1 + x) Radius = 1 + *4 An, - thickness 1= 2m(1 + 4) FIGURE 6.18 Cutting and unrolling a cylindrical shell gives a nearly rectangular solid (Example 1).Chapter 6 Applications of Definite Integrals 6.2 Volumes Using Cylindrical Shells 381 Summing together the volumes AV of the individual cylindrical shells over the interval [0, 3 ] gives the Riemann sum ZAK = Ezw(x + 1)(3x - XX )AXA. Taking the limit as the thickness Ax -0 and n - co gives the volume integral V = lim > 20(x + 1)(3x - x ) 4. ( 2x(x + 1)(3x - r) dx 2n(3x' + 3x - x - x) dx - 21/ (2x' + 3x - x3) do 73 = 2T 4517 2 We now generalize this procedure to more general solids.The Shell Method Suppose that the region bounded by the graph of a nonnegative continuous function " = /(x) and the x-axis over the finite closed interval [a, b ] lies to the right of the verti- cal line x = L (see Figure 6.19a). We assume a 2: L. so the vertical line may touch the region but cannot pass through it. We generate a solid $ by rotating this region about the vertical line _. Let Pbe a partition of the interval [ a, b ] by the points a = >