Math 114 - The 10 AM Edition Worksheet 3 The Method of Cylindrical Shells Guiding Principle: Any quantity which may be computed as a limit of Riemann sums may be computed as an integral. Useful Formulas: A cylinder of height h and radius r has volume V = πr2 h Good Advice: After class, be sure to re-read this section and work examples of each of the cases in figure 10 on page 388. 1. Suppose that a cylinder has constant height h. Using the formula V = πr2 h find the differential dV . Solution: dV = 2πrh dr. 2. Set up but do not evaluate the integral for the volume of a solid obtained by rotating the area under the graph of f (x) = 4 − x2 on [0, 2] about the y-axis. Use the method of cylindrical shells. Figure 1: Area bounded by y = 4 − x2 on [0, 2] Solution: We have 0 ≤ x ≤ 2. A cylindrical shell at x has radius x and height 4 − x2 . Hence Z 2 V = 2πx(4 − x2 ) dx. 0 3. Set up but do not evaluate the integral for the volume of a solid obtained by rotating the area between the graphs of y = 8 − x3 and y = 8 − 4x for x ≥ 0 about the y-axis. Figure 2: Area bounded by y = 8 − x3 and y = 8 − 4x Solution: First we need to find where the curves intersect. Setting 8 − x3 = 8 − 4x we get x3 − 4x = 0 or, factoring, x(x2 − 4) = 0. The roots are 0 and ±2. Given that x ≥ 0 we conclude that 0 ≤ x ≤ 2. Figure 2 shows the area. A cylindrical shell at x has radius x and height [8 − x3 ] − [8 − 4x] = 4x − x3 . Hence Z 2 2πx(4x − x3 ) dx. V = 0 4. Set up but do not evaluate the integral for the volume of a solid obtained by rotating the area under the graph of f (x) = x3 on [0, 1] about the axis x = −2. Solution: Here the challenge is to get the radius right. Figure 3 shows the situation. Clearly 0 ≤ x ≤ 1. The radius of the cylindrical shell is x − (−2) or x + 2, while the height is x3 . Hence Z 1 2π(x + 2)x3 dx. V = 0 Figure 3: Area bounded by f (x) = x3 on [0, 1] and axis x = −2 5. Set up but do not evaluate the integral for the volume of a solid obtained by rotating the area enclosed by the curves x = y(4 − y) and x = 0 about the x-axis. Figure 4: Area bounded by x = 0 and x = y(4 − y) showing cross section of thin cylinder of “height” 3 at y = 1 Solution: In this example, we need to integrate over y rather than over x. For this area, we have 0 ≤ y ≤ 4. The radius of a cylindrical shell at y is exactly y, while the “height” (distance parallel to the x axis) is y(4 − y). Hence Z 4 2πy · y(4 − y) dy. V = 0