Example 1 Show that the volume of a sphere of radius r is V = Solution If we place the sphere so that its center is at the origin (see Figure 4), then the plane Pr intersects the sphere in a circle whose radius (from the Pythagorean Theorem) is y = vr2 - x2. So the cross-sectional area is A(x) = Ty = 1(2 - 22) Using the definition of volume with a = -r and b = r, we have V = A(x) da = (r? -x2) da = 27 (72 - x2) dac (The integrand is even.) 3 3 =_ Using the Disk Method Find the volume of the solid obtained by rotating about the x-axis the region under the curve 3,! = J5 from o to 1. lustrate the denition of volume by sketching a typical approximating cylinder. Solution The region is shown in gure g). If we rotate about the z-axis, we get the solid shown in Ligurh}. When we slice through the point 3:, we get a disk with radius v5. The area of this cross-section is and the volume of the approximating cylinder (a disk with thickness As) is 11(3) Aa': = 1m Am The solid lies between a = and a = 1, so its volume is 1 1 $2 1 V=f A(e)da:=f md$=1r] = l) 0 2 D 2 Example 3 V Rotating About the y-axis Find the volume of the solid obtained by rotating the region bounded by y = 23, y = 8, and x = 0 about the y-axis. Solution The region is shown in Figure 7(a) and the resulting solid is shown in Figure 7(b). Because the region is rotated about the y-axis, it makes sense to slice the solid perpendicular to the y-axis and therefore to integrate with respect to y. If we slice at height y, we get a circular disk with radius *, where x = yy. So the area of a cross- section through y is A(y) = Tac2 = 7(3/y)2 = my2/3 Animation: Figure: 6.2.007a. C Animation: Figure: 6.2.007b. and the volume of the approximationg cylinder pictured in Figure 7(b) is A(y) Ay = Try2/3 Ay Since the solid lies between y = 0 and y = 8, its volume is 96TT 5 Figure 7 y = 8 Ay y. (x. y) x=0- y=x or x = Jy 0 (a) (b)Example 4 Using the Washer Method The region % enclosed by the curves y = a and y = 12 is rotated about the x-axis. Find the volume of the resulting solid. Solution The curves y = x and y = a intersect at the points (0, 0) and (1, 1). The region between them, the solid of rotation, and a cross-section perpendicular to the x-axis are shown in Figure 8. A cross-section in the plane Pz has the shape of a washer (an annular ring) with inner radius a2 and outer radius a, so we find the cross- sectional area by subtracting the area of the inner circle from the area of the outer circle: A(z) = Ta2 - 7(202) = 71(202 - 24) Therefore we have V = A(x) da = 7(x2 - 24) da = 7 271 3 5 Jo 15 Figure 8 y = x A(x) y= x2 (0, 0) (a) (b) (c)_ Rotating About 3 Horizontal Line Find the volume of the solid obtained by rotating the region in Eran-113M about the line 3,.- = 2. Solution The solid and a cross-section are shown in gure 9. Again the cross-section is a washer, but this time the inner radius is 2 :c and the outer radius is 2 :5\". I Figure 9 The cross-sectional area is 14(3) = (2 _ ff (2 x)\" and so the volume of S is Example 6 Rotating About a Vertical Line Find the volume of the solid obtained by rotating the region in Example 4 about the line * = -1. Solution Figure 11 shows a horizontal cross-section. It is a washer with inner radius 1 + y and outer radius 1 + vy, so the cross-sectional area is A(y) = (outer radius) - "(inner radius) = 1(1+ Vy) -7(1+y)2 The volume is V = [ A(y) dy = * / [(1+ vy)? - (1+ y)?] dy - 7 (2 y- y-37) dy = may3/ 2371 = Figure 11 -1 +vy -Ity -x = Vy -x=y x=-1Example 7 Triangular Cross-Sections Figure 12 shows a solid with a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles. Find the volume of the solid. Figure 12 Computer-generated picture of the solid in Example 7 Solution Go Explore It: A Solid with Triangular Slices Let's take the circle to be a2 + y? = 1. The solid, its base, and a typical cross-section at a distance a from the origin are shown in Figure 13. Figure 13 y = VI-x B(x, y) I V3y 60 B (a) The solid (b) Its base (c) A cross-section Since B lies on the circle, we have y = v1 - x2 and so the base of the triangle ABC is |AB| = 2v1 - x2. Since the triangle is equilateral, we see from Figure 13(c) that its height is v3y = v3 v1 - x2. The cross-sectional area is therefore A(z) = 2 .2V/1-x2 . V3V1-2 = V3(1-22) and the volume of the solid is V = A(2) da = [ V3(1 -27) daExample 8 V Find the volume of a pyramid whose base is a square with side _ and whose height is h. Solution We place the origin O at the vertex of the pyramid and the x-axis along its central axis as in Figure 14. Any plane Pr that passes through a and is perpendicular to the -axis intersects the pyramid in a square with side of length s, say. We can express s in terms of a by observing from the similar triangles in Figure 15 that $/2 S L/2 I and so s = Lx/h. [Another method is to observe that the line OP has L/(2h) slope and so its equation is y = Lx/(2h).] Thus the cross-sectional area is A(a) = 82 = L2 62 22 Figure 14 Figure 15 The pyramid lies between x = 0 and x = h, so its volume is [2 23]h L2 h 12 3 0 3 Note: We didn't need to place the vertex of the pyramid at the origin in Example 8. We did so merely to make the equations simple. If, instead, we had placed the center of the base at the origin and the vertex on the positive y-axis, as in Figure 16, you can verify that we would have obtained the integral V = ha (h - y)2 dy = L' h 3Read this section briey. Then answer the following questions. 0 In this section, we learn to compute volumes of solids using two methods: \"Slicing\" and \"Spinning.\" There are 8 Examples of computing volumes in this section. In which Examples is \"slicing\" a volume used to compute its volume? List the Example numbers. In which Examples is \"spinning\" a piece of area about an axis used to compute the volume? List the Example numbers. If you are given parametric equations of a curve, what is the formula for finding its arclength? If the curve is given as y as a function of x, what is the formula for finding the arclength