Mathematics
Surface Area of Revolution
The surface area of revolution is the total area of the surface generated by rotating a curve around an axis. This concept is used in calculus to calculate the surface area of three-dimensional objects. It is an important tool in engineering, physics, and other fields that deal with three-dimensional shapes.
Written by Perlego with AI-assistance
Related key terms
1 of 5
7 Key excerpts on "Surface Area of Revolution"
eBook - PDF- Paul A. Calter, Michael A. Calter(Authors)
- 2011(Publication Date)
- Wiley(Publisher)
We showed earlier that a solid of revolution is the figure we get by rotating an area about some axis. A surface of revolution is simply the surface of that solid of revolution. Alternately, we can think of a surface of revolution as being generated by a curve rotating about some axis. Rotation About the x Axis Let us take the curve PQ in Fig. 28–12 and rotate it about the x axis. It sweeps out a surface of revolution, while the small section ds sweeps out a hoop-shaped element of that surface. The area of a hoop-shaped or circular band is equal to the length of the edge times the average circumference of the hoop. Our element ds is at a radius y from the x axis, so the area dS of the hoop is But from Sec. 28–1, we saw that So We then integrate from a to b to sum up the areas of all such elements. 301 We are using the capital letter S for surface area and lowercase s for arc length. Be careful not to confuse the two. Common Error s 2p L b a y B 1 a dy dx b 2 dx Area of Surface of Revolution About x Axis dS 2py B 1 a dy dx b 2 dx ds B 1 a dy dx b 2 dx dS 2py ds 916 Chapter 28 ◆ More Applications of the Integral TI-84 screen for Example 4. Note how the display differs from that of the TI-83. x 0 1 y (2, 4) y = x 2 2 −1 −2 2 4 FIGURE 28–14 ◆◆◆ Example 4: The portion of the parabola between and 16 is rotated about the x axis (Fig. 28–13). Find the area of the surface of revolution generated. Estimate: Let us approximate the given surface by a hemisphere of radius 16. Its area is or about 1600 square units. Solution: We first find the derivative. Substituting into Eq. 301, we obtain Integrating by Calculator: The calculator screen for the integral in this example is shown. ◆◆◆ Rotation About the y Axis The equation for the area of a surface of revolution whose axis of revolution is the y axis can be derived in a similar way. 302 ◆◆◆ Example 5: The portion of the curve lying between the points (0, 0) and (2, 4) is rotated about the y axis (Fig. 28–14).
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].
eBook - PDF- Brian H. Chirgwin, Charles Plumpton(Authors)
- 2014(Publication Date)
- Pergamon(Publisher)
6:2 Volumes and surfaces of revolution If a plane curve y = / (x) is revolved about one of the coordinate axes through a radians (α = 2π if the revolution is complete), the arc of the curve generates a portion of a surface of revolution. The area of this portion can be expressed as a definite integral. Similarly if the plane area enclosed by a loop of a curve, or between the curve, two § 6 : 2 APPLICATIONS OF INTEGRATION 221 ordinates and an axis, is revolved about the axis this area sweeps out a volume which is a sector of a solid of revolution. We consider the case of that portion of the curve y = f(x) lying between x = a and x = b (assumed to be all above the #-axis) revolved through 4 right angles about the #-axis, cf. Fig. 51. We regard the volume as made up of a number of discs, a typical disc being generated by the revolution of a strip of height y = f(x) and width δχ. The volume of this disc is ÒV = ny 2 bx + 0{δχ 2 ). In fact, for Fig.51, ny 2 òx < ÒV
eBook - PDFCalculus
Early Transcendental Single Variable
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
6.5.2 DEFINITION If f is a smooth, nonnegative function on [a, b], then the surface area S of the surface of revolution that is generated by revolving the portion of the curve y = f (x) between x = a and x = b about the x-axis is defined as S = b a 2π f (x) 1 + [ f (x)] 2 dx This result provides both a definition and a formula for computing surface areas. Where convenient, this formula can also be expressed as S = b a 2π f (x) 1 + [ f (x)] 2 dx = b a 2πy 1 + dy dx 2 dx (4) Moreover, if g is nonnegative and x = g(y) is a smooth curve on the interval [c, d], then the area of the surface that is generated by revolving the portion of a curve x = g(y) between y = c and y = d about the y-axis can be expressed as S = d c 2πg(y) 1 + [g (y)] 2 dy = d c 2πx 1 + dx dy 2 dy (5) Example 1 Find the area of the surface that is generated by revolving the portion of the curve y = x 3 between x = 0 and x = 1 about the x-axis. Solution. First sketch the curve; then imagine revolving it about the x-axis (Figure 6.5.6). Since y = x 3 , we have dy /dx = 3x 2 , and hence from (4) the surface area S is Figure 6.5.6 S = 1 0 2πy 1 + dy dx 2 dx = 1 0 2πx 3 1 + (3x 2 ) 2 dx = 2π 1 0 x 3 (1 + 9x 4 ) 1/2 dx = 2π 36 10 1 u 1/2 du u = 1 + 9x 4 du = 36x 3 dx = 2π 36 · 2 3 u 3/2 10 u=1 = π 27 (10 3/2 − 1) ≈ 3.56 Example 2 Find the area of the surface that is generated by revolving the portion of the curve y = x 2 between x = 1 and x = 2 about the y-axis. Figure 6.5.7 Solution. First sketch the curve; then imagine revolving it about the y-axis (Figure 6.5.7). Because the curve is revolved about the y-axis we will apply Formula (5). Toward this end, we rewrite y = x 2 as x = √ y and observe that the y-values corresponding to x = 1 and 6.5 Area of a Surface of Revolution 373 x = 2 are y = 1 and y = 4.
- James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
But what if we rotate about a slanted line, that is, a line that is neither horizontal nor vertical? In this project you are asked to discover formulas for the volume of a solid of revolution and for the area of a surface of revolution when the axis of rotation is a slanted line. Let C be the arc of the curve y - f s xd between the points Ps p, f s pdd and Qsq, f sqdd and let 5 be the region bounded by C, by the line y - mx 1 b (which lies entirely below C), and by the perpendiculars to the line from P and Q. P 0 x y q p C Q y=ƒ y=m x+b Îu 1. Show that the area of 5 is 1 1 1 m 2 y q p f f s xd 2 mx 2 bg f1 1 mf 9 s xdg dx [Hint: This formula can be verified by subtracting areas, but it will be helpful throughout the project to derive it by first approximating the area using rectangles perpendicular to the line, as shown in the following figure. Use the figure to help express Du in terms of Dx.] y=m x+b Îu å tangent to C at {x i , f(x i )} x i ∫ ? Îx ? (continued ) Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 576 CHAPTER 8 Further Applications of Integration Applications to Physics and Engineering Among the many applications of integral calculus to physics and engineering, we con- sider two here: force due to water pressure and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths) and to work, our strategy is to break up the physical quantity into a large number of small parts, approximate each small part, add the results (giving a Riemann sum), take the limit, and then evaluate the result- ing integral.
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)
But what if we rotate about a slanted line, that is, a line that is neither horizontal nor vertical? In this project you are asked to discover formulas for the volume of a solid of revolution and for the area of a surface of revolution when the axis of rotation is a slanted line. Let C be the arc of the curve y - f s xd between the points Ps p, f s pdd and Qsq, f sqdd and let 5 be the region bounded by C, by the line y - mx 1 b (which lies entirely below C), and by the perpendiculars to the line from P and Q. P 0 x y q p C Q y=ƒ y=m x+b Îu 1. Show that the area of 5 is 1 1 1 m 2 y q p f f s xd 2 mx 2 bg f1 1 mf 9 s xdg dx [Hint: This formula can be verified by subtracting areas, but it will be helpful throughout the project to derive it by first approximating the area using rectangles perpendicular to the line, as shown in the following figure. Use the figure to help express Du in terms of Dx.] y=m x+b Îu å tangent to C at {x i , f(x i )} x i ∫ ? Îx ? (continued ) Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 614 CHAPTER 8 Further Applications of Integration Applications to Physics and Engineering Among the many applications of integral calculus to physics and engineering, we con- sider two here: force due to water pressure and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths) and to work, our strategy is to break up the physical quantity into a large number of small parts, approximate each small part, add the results (giving a Riemann sum), take the limit, and then evaluate the result- ing integral.
eBook - PDF- G. M. Fikhtengol'ts, I. N. Sneddon, M. Stark, S. Ulam(Authors)
- 2014(Publication Date)
- Pergamon(Publisher)
365. The area of a surface given by an explicit equation. In view of the fact mentioned in the preceding section, we need to look for another way to justify the concept of the area of a curved surface. We mention one such method, which, although simple, is some-what artificial, where we begin with the case of a surface (S) given by the expUcit equation z=ñx,y). (1) Let x,y vary in a four-sided domain (/)) on the xy plane, and let ζ have continuous partial derivatives dz ^ dz ρ = and q = -^ ^ dx ^ dy in this domain. We divide up the domain (D) by means of a net of curves (with area equal to zero) into the elements § 2. AREA OF CURVED SURFACE 331 and v^e consider one of these elements, (Di), If we construct on the contour of this partial domain (as on a directrix) a cylindrical surface with generators parallel to the z-axis, then it cuts out an element (.S/) on the surface (5). Taking any point Pi (Xi ,yi) within the limits of (A)> we draw, at the corresponding point Mi (Xi ,yi, z^, where Zi =f {pci .yd^ the tangent plane to the surface (5). The cylindrical surface mentioned above also cuts out an elementary figure (Γ^ (Fig. 55) on this plane, FIG. 55. whose area T, serves as an approximation to the area of the element (5,·). Thus the sum of all such areas π i = 1 can be regarded as an approximation to the area of the surface {S). We define the area S of this surface as the limit S = lima = lim ^ Ti i= 1 with the hypothesis that the diameters of all the elements (Si) tend to zero or, equivalently, the diameters of all the plane elements (Z);.) 332 22. SURFACE INTEGRALS cosv dxdy, (2) where, as always, ν denotes the angle made by the normal to the surface with the z-axist.
Index pages curate the most relevant extracts from our library of academic textbooks. They’ve been created using an in-house natural language model (NLM), each adding context and meaning to key research topics.
