Mathematics
Volumes of Revolution
Volumes of revolution refer to the process of finding the volume of a three-dimensional shape formed by rotating a two-dimensional shape around an axis. This concept is commonly used in calculus and is important in understanding the properties of various shapes and objects.
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6 Key excerpts on "Volumes of Revolution"
eBook - PDF- Ron Larson, Bruce Edwards(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
This portion of the volume could have been determined without using calculus. TECHNOLOGY Some graphing utilities have the capability of generating (or have built-in software capable of generating) a solid of revolution. If you have access to such a utility, use it to graph some of the solids of revolution described in this section. For instance, the solid in Example 4 might appear like that shown in Figure 7.22. Generated by Mathematica Figure 7.22 7 .2 Volume: The Disk Method 459 Manufacturing See LarsonCalculus.com for an interactive version of this type of example. A manufacturer drills a hole through the center of a metal sphere of radius 5 centimeters, as shown in Figure 7.23(a). The hole has a radius of 3 centimeters. What is the volume of the resulting metal ring? Solution You can imagine the ring to be generated by a segment of the circle whose equation is x 2 + y 2 = 25, as shown in Figure 7.23(b). Because the radius of the hole is 3 centimeters, you can let y = 3 and solve the equation x 2 + y 2 = 25 to determine that the limits of integration are x = ±4. So, the inner and outer radii are r(x) = 3 and R(x) = radical.alt225 - x 2 , and the volume is V = π integral.alt1 b a ([R(x)] 2 - [r(x)] 2 ) dx = π integral.alt1 4 -4 [(radical.alt225 - x 2 ) 2 - (3) 2 ] dx = π integral.alt1 4 -4 (16 - x 2 ) dx = π bracketleft.alt2 16x - x 3 3 bracketright.alt2 4 -4 = 256π 3 cubic centimeters. Solids with Known Cross Sections With the disk method, you can find the volume of a solid having a circular cross section whose area is A = π R 2 . This method can be generalized to solids of any shape, as long as you know a formula for the area of an arbitrary cross section. Some common cross sections are squares, rectangles, triangles, semicircles, and trapezoids. VOLUMES OF SOLIDS WITH KNOWN CROSS SECTIONS 1. For cross sections of area A(x) taken perpendicular to the x-axis, Volume = integral.alt1 b a A(x) dx.
eBook - PDFCalculus
Single Variable
- Carl V. Lutzer, H. T. Goodwill(Authors)
- 2011(Publication Date)
- Wiley(Publisher)
In the limit, we have an exact calculation of the volume, V = lim n→∞ n X k=1 V k = Z 1 -1 2π(1.5 - x)(3 - x - x 2 ) dx = 52π 3 cm 3 . The technique developed in Examples 3.1–3.4 is summarized below. Suppose R is the region that extends from x = a to x = b, where a < b, whose upper and lower boundaries are the graphs of the continuous functions f and g, respectively. If the line x = c does not intersect R (except perhaps tangentially at the boundary), the solid that’s generated by revolving R about the line x = c has a volume of V = Z b a 2π|x - c| f (x) - g(x) dx. Note: If x ≤ c throughout R, we know |x - c| = x - c. If x ≥ c throughout R, we know |x - c| = c - x. So in either case we can drop the absolute value from the integrand. Try It Yourself: Using cylindrical shells when the axis of revolution is removed from R Example 3.4. Suppose R is the region that extends from x = 1 to x = 3 that is bounded above by the graph of f (x) = 1 + x 2 and below by the graph of g(x) = 2x - 1. If length is measured in microns, determine the volume of the solid that’s generated by revolving R about the line x = -2. Answer: 40π cubic microns. Section 7.3 Volumes by Cylindrical Shells 520 z Horizontal Axes of Revolution In all of the previous examples, the axis of revolution was vertical, and the rectan- gles that we used to make cylindrical shells stretched vertically from one boundary curve to another (see Figures 3.3–3.6). Similarly, we see cylindrical shells when both our rectangles and the axis of revolution are horizontal. Shells with a horizontal axis of revolution Example 3.5. Suppose R is the bounded region between the parabola x = y 2 and the line y = x - 2. Determine the volume of the solid that’s generated by revolving R about the line y = 2.5.
eBook - PDFCalculus
Single Variable
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
Find the volume of the solid of revolution that is generated by revolving the region R about the x-axis. x y a b R y = f ( x) (a) f ( x) x y a b x (b) FIGURE 5.2.9 We can solve this problem by slicing. For this purpose, observe that the cross section of the solid taken perpendicular to the x-axis at the point x is a circular disk of radius f (x) (Figure 5.2.9b). The area of this region is A(x) = π[ f (x)] 2 Thus, from (3) the volume of the solid is V = b a π[ f (x)] 2 dx (5) Because the cross sections are disk shaped, the application of this formula is called the method of disks. x y 1 4 y = √x FIGURE 5.2.10 Example 2 Find the volume of the solid that is obtained when the region under the curve y = √ x over the interval [1, 4] is revolved about the x-axis (Figure 5.2.10). Solution From (5), the volume is V = b a π[ f (x)] 2 dx = 4 1 πx dx = πx 2 2 4 1 = 8π − π 2 = 15π 2 x y −r r x 2 + y 2 = r 2 FIGURE 5.2.11 Example 3 Derive the formula for the volume of a sphere of radius r. Solution As indicated in Figure 5.2.11, a sphere of radius r can be generated by revolving the upper semicircular disk enclosed between the x-axis and x 2 + y 2 = r 2 about the x-axis. Since the upper half of this circle is the graph of y = f (x) = √ r 2 − x 2 , it fol- lows from (5) that the volume of the sphere is V = b a π[ f (x)] 2 dx = r −r π(r 2 − x 2 ) dx = π r 2 x − x 3 3 r −r = 4 3 πr 3 Volumes by Washers Perpendicular to the x-axis Not all solids of revolution have solid interiors; some have holes or channels that create interior surfaces, as in Figure 5.2.8d. So we will also be interested in problems of the following type. 5.2 Volumes by Slicing; Disks and Washers 285 5.2.5 Problem Let f and g be continuous and nonnegative on [a, b], and suppose that f (x) ≥ g(x) for all x in the interval [a, b]. Let R be the region that is bounded above by y = f (x), below by y = g(x), and on the sides by the lines x = a and x = b (Figure 5.2.12a).
eBook - PDF- James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
r at right angles to the axis of the cylinder. Set up, but do not evaluate, an integral for the volume that is cut out. 84. A hole of radius r is bored through the center of a sphere of radius R . r. Find the volume of the remaining portion of the sphere. 85. Some of the pioneers of calculus, such as Kepler and Newton, were inspired by the problem of finding the volumes of wine barrels. (Kepler published a book Stereometria doliorum in 1615 devoted to methods for finding the volumes of barrels.) They often approximated the shape of the sides by parabolas. (a) A barrel with height h and maximum radius R is con- structed by rotating about the x-axis the parabola y - R 2 cx 2 , 2hy2 < x < hy2, where c is a positive constant. Show that the radius of each end of the barrel is r - R 2 d, where d - ch 2 y4. (b) Show that the volume enclosed by the barrel is V - 1 3 h(2R 2 1 r 2 2 2 5 d 2 ) 86. Suppose that a region 5 has area A and lies above the x-axis. When 5 is rotated about the x-axis, it sweeps out a solid with volume V1. When 5 is rotated about the line y - 2k (where k is a positive number), it sweeps out a solid with volume V2. Express V2 in terms of V1, k, and A. 87. A dilation of the plane with scaling factor c is a transforma- tion that maps the point s x, yd to the point scx, cyd. Applying a dilation to a region in the plane produces a geometrically similar shape. A manufacturer wants to produce a 5-liter (5000 cm 3 ) terra-cotta pot whose shape is geometrically similar to the solid obtained by rotating the region 51 shown in the figure about the y-axis. (a) Find the volume V1 of the pot obtained by rotating the region 51. (b) Show that applying a dilation with scaling factor c transforms the region 51 into the region 52. (c) Show that the volume V2 of the pot obtained by rotating the region 52 is c 3 V1. (d) Find the scaling factor c that produces a 5-liter pot. 0 x (cm) x (cm) y=8 y=˛ y=1 0 y=8c y=˛/c@ y=c y (cm) y (cm) T™ T¡ 76.
eBook - PDFSingle Variable Calculus
Concepts and Contexts, Enhanced Edition
- James Stewart(Author)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
(See Figure 3.) We divide the interval into n subintervals of equal width and let be the midpoint of the i th subinterval. If the rectangle with base and height is rotated about the y-axis, then the result is a cylindrical shell with average radius , height , and thickness (see Figure 4), so by Formula 1 its volume is Therefore an approximation to the volume of is given by the sum of the volumes of these shells: V n i 1 V i n i 1 2 x i f x i x S V x y b y=ƒ x i – a b 0 0 x y x i-1 x i y=ƒ x y b y=ƒ V i 2 x i f x i x x f x i x i f x i x i 1 , x i x i x x i 1 , x i a, b FIGURE 3 x y a b 0 y=ƒ a b x y 0 y=ƒ b a 0 x b y 0, x a, f x 0 y f x y S V [circumference][height][thickness] V 2 rh r 1 r 1 2 r 2 r 1 r r 2 r 1 2 r 2 r 1 2 h r 2 r 1 r 2 r 1 r 2 r 1 h r 2 2 h r 2 1 h r 2 2 r 2 1 h V V 2 V 1 V 2 450 CHAPTER 6 APPLICATIONS OF INTEGRATION F I GURE 4 SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS 451 This approximation appears to become better as . But, from the definition of an inte- gral, we know that Thus the following appears plausible: The volume of the solid in Figure 3, obtained by rotating about the y-axis the region under the curve from a to b, is The best way to remember Formula 2 is to think of a typical shell, cut and flattened as in Figure 5, with radius x, circumference , height , and thickness or : This type of reasoning will be helpful in other situations, such as when we rotate about lines other than the y-axis. Using the shell method Find the volume of the solid obtained by rotating about the -axis the region bounded by and . SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x, circumfer- ence , and height . So, by the shell method, the volume is It can be verified that the shell method gives the same answer as slicing.
eBook - PDFCalculus
Late Transcendental
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
continuous 2. b a 1 + [ f (x)] 2 dx 3. (π − 1) 2 + 1 4. (a) 2 0 1 + 4x 2 dx (b) 4 0 1 + 1 4y dy 5.5 AREA OF A SURFACE OF REVOLUTION In this section we will consider the problem of finding the area of a surface that is generated by revolving a plane curve about a line. SURFACE AREA A surface of revolution is a surface that is generated by revolving a plane curve about an axis that lies in the same plane as the curve. For example, the surface of a sphere can be generated by revolving a semicircle about its diameter, and the lateral surface of a right circular cylinder can be generated by revolving a line segment about an axis that is parallel to it (Figure 5.5.1). Figure 5.5.1 In this section we will be concerned with the following problem. 5.5.1 SURFACE AREA PROBLEM Suppose that f is a smooth, nonnegative function on [a, b] and that a surface of revolution is generated by revolving the portion of the curve y = f (x) between x = a and x = b about the x-axis (Figure 5.5.2). Define what is meant by the area S of the surface, and find a formula for computing it. Figure 5.5.2 To motivate an appropriate definition for the area S of a surface of revolution, we will decompose the surface into small sections whose areas can be approximated by elementary formulas, add the approximations of the areas of the sections to form a Riemann sum that approximates S, and then take the limit of the Riemann sums to obtain an integral for the exact value of S. To implement this idea, divide the interval [a, b] into n subintervals by inserting points x 1 , x 2 , . . . , x n−1 between a = x 0 and b = x n . As illustrated in Figure 5.5.3a, the corre- sponding points on the graph of f define a polygonal path that approximates the curve y = f (x) over the interval [a, b].
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