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Calculus Volume 1

6.4 Arc Length of a Curve and Surface Area

Calculus Volume 16.4 Arc Length of a Curve and Surface Area
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 6.4.1. Determine the length of a curve, y=f(x),y=f(x), between two points.
  • 6.4.2. Determine the length of a curve, x=g(y),x=g(y), between two points.
  • 6.4.3. Find the surface area of a solid of revolution.

In this section, we use definite integrals to find the arc length of a curve. We can think of arc length as the distance you would travel if you were walking along the path of the curve. Many real-world applications involve arc length. If a rocket is launched along a parabolic path, we might want to know how far the rocket travels. Or, if a curve on a map represents a road, we might want to know how far we have to drive to reach our destination.

We begin by calculating the arc length of curves defined as functions of x,x, then we examine the same process for curves defined as functions of y.y. (The process is identical, with the roles of xx and yy reversed.) The techniques we use to find arc length can be extended to find the surface area of a surface of revolution, and we close the section with an examination of this concept.

Arc Length of the Curve y = f(x)

In previous applications of integration, we required the function f(x)f(x) to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for f(x).f(x). Here, we require f(x)f(x) to be differentiable, and furthermore we require its derivative, f(x),f(x), to be continuous. Functions like this, which have continuous derivatives, are called smooth. (This property comes up again in later chapters.)

Let f(x)f(x) be a smooth function defined over [a,b].[a,b]. We want to calculate the length of the curve from the point (a,f(a))(a,f(a)) to the point (b,f(b)).(b,f(b)). We start by using line segments to approximate the length of the curve. For i=0,1,2,…,n,i=0,1,2,…,n, let P={xi}P={xi} be a regular partition of [a,b].[a,b]. Then, for i=1,2,…,n,i=1,2,…,n, construct a line segment from the point (xi1,f(xi1))(xi1,f(xi1)) to the point (xi,f(xi)).(xi,f(xi)). Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. Figure 6.37 depicts this construct for n=5.n=5.

This figure is a graph in the first quadrant. The curve increases and decreases. It is divided into parts at the points a=xsub0, xsub1, xsub2, xsub3, xsub4, and xsub5=b. Also, there are line segments between the points on the curve.
Figure 6.37 We can approximate the length of a curve by adding line segments.

To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by Δx.Δx. The change in vertical distance varies from interval to interval, though, so we use Δyi=f(xi)f(xi1)Δyi=f(xi)f(xi1) to represent the change in vertical distance over the interval [xi1,xi],[xi1,xi], as shown in Figure 6.38. Note that some (or all) ΔyiΔyi may be negative.

This figure is a graph. It is a curve above the x-axis beginning at the point f(xsubi-1). The curve ends in the first quadrant at the point f(xsubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.
Figure 6.38 A representative line segment approximates the curve over the interval [xi1,xi].[xi1,xi].

By the Pythagorean theorem, the length of the line segment is (Δx)2+(Δyi)2.(Δx)2+(Δyi)2. We can also write this as Δx1+((Δyi)/(Δx))2.Δx1+((Δyi)/(Δx))2. Now, by the Mean Value Theorem, there is a point xi*[xi1,xi]xi*[xi1,xi] such that f(xi*)=(Δyi)/(Δx).f(xi*)=(Δyi)/(Δx). Then the length of the line segment is given by Δx1+[f(xi*)]2.Δx1+[f(xi*)]2. Adding up the lengths of all the line segments, we get

Arc Lengthi=1n1+[f(xi*)]2Δx.Arc Lengthi=1n1+[f(xi*)]2Δx.

This is a Riemann sum. Taking the limit as n,n, we have

Arc Length=limni=1n1+[f(xi*)]2Δx=ab1+[f(x)]2dx.Arc Length=limni=1n1+[f(xi*)]2Δx=ab1+[f(x)]2dx.

We summarize these findings in the following theorem.

Theorem 6.4

Arc Length for y = f(x)

Let f(x)f(x) be a smooth function over the interval [a,b].[a,b]. Then the arc length of the portion of the graph of f(x)f(x) from the point (a,f(a))(a,f(a)) to the point (b,f(b))(b,f(b)) is given by

Arc Length=ab1+[f(x)]2dx.Arc Length=ab1+[f(x)]2dx.
6.7

Note that we are integrating an expression involving f(x),f(x), so we need to be sure f(x)f(x) is integrable. This is why we require f(x)f(x) to be smooth. The following example shows how to apply the theorem.

Example 6.18

Calculating the Arc Length of a Function of x

Let f(x)=2x3/2.f(x)=2x3/2. Calculate the arc length of the graph of f(x)f(x) over the interval [0,1].[0,1]. Round the answer to three decimal places.

Solution

We have f(x)=3x1/2,f(x)=3x1/2, so [f(x)]2=9x.[f(x)]2=9x. Then, the arc length is

Arc Length=ab1+[f(x)]2dx=011+9xdx.Arc Length=ab1+[f(x)]2dx=011+9xdx.

Substitute u=1+9x.u=1+9x. Then, du=9dx.du=9dx. When x=0,x=0, then u=1,u=1, and when x=1,x=1, then u=10.u=10. Thus,

Arc Length=011+9xdx=19011+9x9dx=19110udu=19·23u3/2|110=227[10101]2.268units.Arc Length=011+9xdx=19011+9x9dx=19110udu=19·23u3/2|110=227[10101]2.268units.
Checkpoint 6.18

Let f(x)=(4/3)x3/2.f(x)=(4/3)x3/2. Calculate the arc length of the graph of f(x)f(x) over the interval [0,1].[0,1]. Round the answer to three decimal places.

Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration. In some cases, we may have to use a computer or calculator to approximate the value of the integral.

Example 6.19

Using a Computer or Calculator to Determine the Arc Length of a Function of x

Let f(x)=x2.f(x)=x2. Calculate the arc length of the graph of f(x)f(x) over the interval [1,3].[1,3].

Solution

We have f(x)=2x,f(x)=2x, so [f(x)]2=4x2.[f(x)]2=4x2. Then the arc length is given by

Arc Length=ab1+[f(x)]2dx=131+4x2dx.Arc Length=ab1+[f(x)]2dx=131+4x2dx.

Using a computer to approximate the value of this integral, we get

131+4x2dx8.26815.131+4x2dx8.26815.
Checkpoint 6.19

Let f(x)=sinx.f(x)=sinx. Calculate the arc length of the graph of f(x)f(x) over the interval [0,π].[0,π]. Use a computer or calculator to approximate the value of the integral.

Arc Length of the Curve x = g(y)

We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of y,y, we can repeat the same process, except we partition the y-axisy-axis instead of the x-axis.x-axis. Figure 6.39 shows a representative line segment.

This figure is a graph. It is a curve to the right of the y-axis beginning at the point g(ysubi-1). The curve ends in the first quadrant at the point g(ysubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.
Figure 6.39 A representative line segment over the interval [yi1,yi].[yi1,yi].

Then the length of the line segment is (Δy)2+(Δxi)2,(Δy)2+(Δxi)2, which can also be written as Δy1+((Δxi)/(Δy))2.Δy1+((Δxi)/(Δy))2. If we now follow the same development we did earlier, we get a formula for arc length of a function x=g(y).x=g(y).

Theorem 6.5

Arc Length for x = g(y)

Let g(y)g(y) be a smooth function over an interval [c,d].[c,d]. Then, the arc length of the graph of g(y)g(y) from the point (c,g(c))(c,g(c)) to the point (d,g(d))(d,g(d)) is given by

Arc Length=cd1+[g(y)]2dy.Arc Length=cd1+[g(y)]2dy.
6.8

Example 6.20

Calculating the Arc Length of a Function of y

Let g(y)=3y3.g(y)=3y3. Calculate the arc length of the graph of g(y)g(y) over the interval [1,2].[1,2].

Solution

We have g(y)=9y2,g(y)=9y2, so [g(y)]2=81y4.[g(y)]2=81y4. Then the arc length is

Arc Length=cd1+[g(y)]2dy=121+81y4dy.Arc Length=cd1+[g(y)]2dy=121+81y4dy.

Using a computer to approximate the value of this integral, we obtain

121+81y4dy21.0277.121+81y4dy21.0277.
Checkpoint 6.20

Let g(y)=1/y.g(y)=1/y. Calculate the arc length of the graph of g(y)g(y) over the interval [1,4].[1,4]. Use a computer or calculator to approximate the value of the integral.

Area of a Surface of Revolution

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let f(x)f(x) be a nonnegative smooth function over the interval [a,b].[a,b]. We wish to find the surface area of the surface of revolution created by revolving the graph of y=f(x)y=f(x) around the x-axisx-axis as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is a curve in the first quadrant beginning at the y-axis. The curve is y=f(x). The second graph is labeled “b” and has the same curve y=f(x). There is also a solid surface formed by rotating the curve about the x-axis.
Figure 6.40 (a) A curve representing the function f(x).f(x). (b) The surface of revolution formed by revolving the graph of f(x)f(x) around the x-axis.x-axis.

As we have done many times before, we are going to partition the interval [a,b][a,b] and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For i=0,1,2,…,n,i=0,1,2,…,n, let P={xi}P={xi} be a regular partition of [a,b].[a,b]. Then, for i=1,2,…,n,i=1,2,…,n, construct a line segment from the point (xi1,f(xi1))(xi1,f(xi1)) to the point (xi,f(xi)).(xi,f(xi)). Now, revolve these line segments around the x-axisx-axis to generate an approximation of the surface of revolution as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is a curve in the first quadrant beginning at the y-axis. The curve is y=f(x). The area under the curve above the x-axis has been divided into regions with vertical lines. The second graph is labeled “b” and has the same curve y=f(x). There is also a solid surface formed by rotating the curve about the x-axis.
Figure 6.41 (a) Approximating f(x)f(x) with line segments. (b) The surface of revolution formed by revolving the line segments around the x-axis.x-axis.

Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.

To find the surface area of the band, we need to find the lateral surface area, S,S, of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let r1r1 and r2r2 be the radii of the wide end and the narrow end of the frustum, respectively, and let ll be the slant height of the frustum as shown in the following figure.

This figure is a graph. It is a frustum of a cone above the x-axis with the y-axis in the center. The radius of the bottom of the frustum is rsub1 and the radius of the top is rsub2. The length of the side is labeled “l”.
Figure 6.42 A frustum of a cone can approximate a small part of surface area.

We know the lateral surface area of a cone is given by

Lateral Surface Area=πrs,Lateral Surface Area=πrs,

where rr is the radius of the base of the cone and ss is the slant height (see the following figure).

This figure is a cone. The cone has radius r, height h, and length of side s.
Figure 6.43 The lateral surface area of the cone is given by πrs.πrs.

Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (see the following figure).

This figure is a graph. It is a frustum of a cone. The radius of the bottom of the frustum is rsub1 and the radius of the top is rsub2. The length of the side is labeled “l”. There is also the top of a cone with broken lines above the frustum. It has length of side s.
Figure 6.44 Calculating the lateral surface area of a frustum of a cone.

The cross-sections of the small cone and the large cone are similar triangles, so we see that

r2r1=sls.r2r1=sls.

Solving for s,s, we get

r2r1=slsr2s=r1(sl)r2s=r1sr1lr1l=r1sr2sr1l=(r1r2)sr1lr1r2=s.r2r1=slsr2s=r1(sl)r2s=r1sr1lr1l=r1sr2sr1l=(r1r2)sr1lr1r2=s.

Then the lateral surface area (SA) of the frustum is

S=(Lateral SA of large cone)(Lateral SA of small cone)=πr1sπr2(sl)=πr1(r1lr1r2)πr2(r1lr1r2l)=πr12lr1r2πr1r2lr1r2+πr2l=πr12lr1r2πr1r2lr1r2+πr2l(r1r2)r1r2=πr12lr1r2πr1r2lr1r2+πr1r2lr1r2πr22lr1r2=π(r12r22)lr1r2=π(r1r2)(r1+r2)lr1r2=π(r1+r2)l.S=(Lateral SA of large cone)(Lateral SA of small cone)=πr1sπr2(sl)=πr1(r1lr1r2)πr2(r1lr1r2l)=πr12lr1r2πr1r2lr1r2+πr2l=πr12lr1r2πr1r2lr1r2+πr2l(r1r2)r1r2=πr12lr1r2πr1r2lr1r2+πr1r2lr1r2πr22lr1r2=π(r12r22)lr1r2=π(r1r2)(r1+r2)lr1r2=π(r1+r2)l.

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the x-axis.x-axis. A representative band is shown in the following figure.

This figure has two graphics. The first is a curve in the first quadrant. Around the x-axis is a frustum of a cone. The edge of the frustum is against the curve. The edge begins at f(xsubi-1) and ends at f(xsubi). The second image is the same curve with the same frustum. the height of the frustum is delta x and the curve is labeled y=f(x).
Figure 6.45 A representative band used for determining surface area.

Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have

S=π(r1+r2)l=π(f(xi1)+f(xi))Δx2+(Δyi)2=π(f(xi1)+f(xi))Δx1+(ΔyiΔx)2.S=π(r1+r2)l=π(f(xi1)+f(xi))Δx2+(Δyi)2=π(f(xi1)+f(xi))Δx1+(ΔyiΔx)2.

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select xi*[xi1,xi]xi*[xi1,xi] such that f(xi*)=(Δyi)/Δx.f(xi*)=(Δyi)/Δx. This gives us

S=π(f(xi1)+f(xi))Δx1+(f(xi*))2.S=π(f(xi1)+f(xi))Δx1+(f(xi*))2.

Furthermore, since f(x)f(x) is continuous, by the Intermediate Value Theorem, there is a point xi**[xi1,xi]xi**[xi1,xi] such that f(xi**)=(1/2)[f(xi1)+f(xi)],f(xi**)=(1/2)[f(xi1)+f(xi)], so we get

S=2πf(xi**)Δx1+(f(xi*))2.S=2πf(xi**)Δx1+(f(xi*))2.

Then the approximate surface area of the whole surface of revolution is given by

Surface Areai=1n2πf(xi**)Δx1+(f(xi*))2.Surface Areai=1n2πf(xi**)Δx1+(f(xi*))2.

This almost looks like a Riemann sum, except we have functions evaluated at two different points, xi*xi* and xi**,xi**, over the interval [xi1,xi].[xi1,xi]. Although we do not examine the details here, it turns out that because f(x)f(x) is smooth, if we let n,n, the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both xi*xi* and xi**xi** are in the interval [xi1,xi],[xi1,xi], so it makes sense that as n,n, both xi*xi* and xi**xi** approach x.x. Those of you who are interested in the details should consult an advanced calculus text.

Taking the limit as n,n, we get

Surface Area=limni=1n2πf(xi**)Δx1+(f(xi*))2=ab(2πf(x)1+(f(x))2)dx.Surface Area=limni=1n2πf(xi**)Δx1+(f(xi*))2=ab(2πf(x)1+(f(x))2)dx.

As with arc length, we can conduct a similar development for functions of yy to get a formula for the surface area of surfaces of revolution about the y-axis.y-axis. These findings are summarized in the following theorem.

Theorem 6.6

Surface Area of a Surface of Revolution

Let f(x)f(x) be a nonnegative smooth function over the interval [a,b].[a,b]. Then, the surface area of the surface of revolution formed by revolving the graph of f(x)f(x) around the x-axis is given by

Surface Area=ab(2πf(x)1+(f(x))2)dx.Surface Area=ab(2πf(x)1+(f(x))2)dx.
6.9

Similarly, let g(y)g(y) be a nonnegative smooth function over the interval [c,d].[c,d]. Then, the surface area of the surface of revolution formed by revolving the graph of g(y)g(y) around the y-axisy-axis is given by

Surface Area=cd(2πg(y)1+(g(y))2)dy.Surface Area=cd(2πg(y)1+(g(y))2)dy.

Example 6.21

Calculating the Surface Area of a Surface of Revolution 1

Let f(x)=xf(x)=x over the interval [1,4].[1,4]. Find the surface area of the surface generated by revolving the graph of f(x)f(x) around the x-axis.x-axis. Round the answer to three decimal places.

Solution

The graph of f(x)f(x) and the surface of rotation are shown in the following figure.

This figure has two graphs. The first is the curve f(x)=squareroot(x). The curve is increasing and begins at the origin. Also on the graph are the vertical lines x=1 and x=4. The second graph is the same function as the first graph. The region between f(x) and the x-axis, bounded by x=1 and x=4 has been rotated around the x-axis to form a surface.
Figure 6.46 (a) The graph of f(x).f(x). (b) The surface of revolution.

We have f(x)=x.f(x)=x. Then, f(x)=1/(2x)f(x)=1/(2x) and (f(x))2=1/(4x).(f(x))2=1/(4x). Then,

Surface Area=ab(2πf(x)1+(f(x))2)dx=14(2πx1+14x)dx=14(2πx+14)dx.Surface Area=ab(2πf(x)1+(f(x))2)dx=14(2πx1+14x)dx=14(2πx+14)dx.

Let u=x+1/4.u=x+1/4. Then, du=dx.du=dx. When x=1,x=1, u=5/4,u=5/4, and when x=4,x=4, u=17/4.u=17/4. This gives us

01(2πx+14)dx=5/417/42πudu=2π[23u3/2]|5/417/4=π6[171755]30.846.01(2πx+14)dx=5/417/42πudu=2π[23u3/2]|5/417/4=π6[171755]30.846.
Checkpoint 6.21

Let f(x)=1xf(x)=1x over the interval [0,1/2].[0,1/2]. Find the surface area of the surface generated by revolving the graph of f(x)f(x) around the x-axis.x-axis. Round the answer to three decimal places.

Example 6.22

Calculating the Surface Area of a Surface of Revolution 2

Let f(x)=y=3x3.f(x)=y=3x3. Consider the portion of the curve where 0y2.0y2. Find the surface area of the surface generated by revolving the graph of f(x)f(x) around the y-axis.y-axis.

Solution

Notice that we are revolving the curve around the y-axis,y-axis, and the interval is in terms of y,y, so we want to rewrite the function as a function of y. We get x=g(y)=(1/3)y3.x=g(y)=(1/3)y3. The graph of g(y)g(y) and the surface of rotation are shown in the following figure.

This figure has two graphs. The first is the curve g(y)=1/3y^3. The curve is increasing and begins at the origin. Also on the graph are the horizontal lines y=0 and y=2. The second graph is the same function as the first graph. The region between g(y) and the y-axis, bounded by y=0 and y=2 has been rotated around the y-axis to form a surface.
Figure 6.47 (a) The graph of g(y).g(y). (b) The surface of revolution.

We have g(y)=(1/3)y3,g(y)=(1/3)y3, so g(y)=y2g(y)=y2 and (g(y))2=y4.(g(y))2=y4. Then

Surface Area=cd(2πg(y)1+(g(y))2)dy=02(2π(13y3)1+y4)dy=2π302(y31+y4)dy.Surface Area=cd(2πg(y)1+(g(y))2)dy=02(2π(13y3)1+y4)dy=2π302(y31+y4)dy.

Let u=y4+1.u=y4+1. Then du=4y3dy.du=4y3dy. When y=0,y=0, u=1,u=1, and when y=2,y=2, u=17.u=17. Then

2π302(y31+y4)dy=2π311714udu=π6[23u3/2]|117=π9[(17)3/21]24.118.2π302(y31+y4)dy=2π311714udu=π6[23u3/2]|117=π9[(17)3/21]24.118.
Checkpoint 6.22

Let g(y)=9y2g(y)=9y2 over the interval y[0,2].y[0,2]. Find the surface area of the surface generated by revolving the graph of g(y)g(y) around the y-axis.y-axis.

Section 6.4 Exercises

For the following exercises, find the length of the functions over the given interval.

165.

y=5xfromx=0tox=2y=5xfromx=0tox=2

166.

y=12x+25fromx=1tox=4y=12x+25fromx=1tox=4

167.

x=4yfromy=−1toy=1x=4yfromy=−1toy=1

168.

Pick an arbitrary linear function x=g(y)x=g(y) over any interval of your choice (y1,y2).(y1,y2). Determine the length of the function and then prove the length is correct by using geometry.

169.

Find the surface area of the volume generated when the curve y=xy=x revolves around the x-axisx-axis from (1,1)(1,1) to (4,2),(4,2), as seen here.

This figure is a surface. It has been formed by rotating the curve y=squareroot(x) about the x-axis. The surface is inside of a cube to show 3-dimensions.
170.

Find the surface area of the volume generated when the curve y=x2y=x2 revolves around the y-axisy-axis from (1,1)(1,1) to (3,9).(3,9).

This figure is a surface. It has an elliptical shape to the top, forming a “bowl”.

For the following exercises, find the lengths of the functions of xx over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

171.

y=x3/2y=x3/2 from (0,0)to(1,1)(0,0)to(1,1)

172.

y=x2/3y=x2/3 from (1,1)to(8,4)(1,1)to(8,4)

173.

y=13(x2+2)3/2y=13(x2+2)3/2 from x=0tox=1x=0tox=1

174.

y=13(x22)3/2y=13(x22)3/2 from x=2x=2 to x=4x=4

175.

[T] y=exy=ex on x=0x=0 to x=1x=1

176.

y=x33+14xy=x33+14x from x=1tox=3x=1tox=3

177.

y=x44+18x2y=x44+18x2 from x=1tox=2x=1tox=2

178.

y=2x3/23x1/22y=2x3/23x1/22 from x=1tox=4x=1tox=4

179.

y=127(9x2+6)3/2y=127(9x2+6)3/2 from x=0tox=2x=0tox=2

180.

[T] y=sinxy=sinx on x=0tox=πx=0tox=π

For the following exercises, find the lengths of the functions of yy over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

181.

y=53x4y=53x4 from y=0y=0 to y=4y=4

182.

x=12(ey+ey)x=12(ey+ey) from y=−1toy=1y=−1toy=1

183.

x=5y3/2x=5y3/2 from y=0y=0 to y=1y=1

184.

[T] x=y2x=y2 from y=0y=0 to y=1y=1

185.

x=yx=y from y=0toy=1y=0toy=1

186.

x=23(y2+1)3/2x=23(y2+1)3/2 from y=1y=1 to y=3y=3

187.

[T] x=tanyx=tany from y=0y=0 to y=34y=34

188.

[T] x=cos2yx=cos2y from y=π2y=π2 to y=π2y=π2

189.

[T] x=4yx=4y from y=0toy=2y=0toy=2

190.

[T] x=ln(y)x=ln(y) on y=1ey=1e to y=ey=e

For the following exercises, find the surface area of the volume generated when the following curves revolve around the x-axis.x-axis. If you cannot evaluate the integral exactly, use your calculator to approximate it.

191.

y=xy=x from x=2x=2 to x=6x=6

192.

y=x3y=x3 from x=0x=0 to x=1x=1

193.

y=7xy=7x from x=−1tox=1x=−1tox=1

194.

[T] y=1x2y=1x2 from x=1tox=3x=1tox=3

195.

y=4x2y=4x2 from x=0tox=2x=0tox=2

196.

y=4x2y=4x2 from x=−1tox=1x=−1tox=1

197.

y=5xy=5x from x=1tox=5x=1tox=5

198.

[T] y=tanxy=tanx from x=π4tox=π4x=π4tox=π4

For the following exercises, find the surface area of the volume generated when the following curves revolve around the y-axis.y-axis. If you cannot evaluate the integral exactly, use your calculator to approximate it.

199.

y=x2y=x2 from x=0tox=2x=0tox=2

200.

y=12x2+12y=12x2+12 from x=0tox=1x=0tox=1

201.

y=x+1y=x+1 from x=0tox=3x=0tox=3

202.

[T] y=1xy=1x from x=12x=12 to x=1x=1

203.

y=x3y=x3 from x=1tox=27x=1tox=27

204.

[T] y=3x4y=3x4 from x=0x=0 to x=1x=1

205.

[T] y=1xy=1x from x=1x=1 to x=3x=3

206.

[T] y=cosxy=cosx from x=0x=0 to x=π2x=π2

207.

The base of a lamp is constructed by revolving a quarter circle y=2xx2y=2xx2 around the y-axisy-axis from x=1x=1 to x=2,x=2, as seen here. Create an integral for the surface area of this curve and compute it.

This figure is a surface. It is half of a torus created by rotating the curve y=squareroot(2x-x^2) about the x-axis.
208.

A light bulb is a sphere with radius 1/21/2 in. with the bottom sliced off to fit exactly onto a cylinder of radius 1/41/4 in. and length 1/31/3 in., as seen here. The sphere is cut off at the bottom to fit exactly onto the cylinder, so the radius of the cut is 1/41/4 in. Find the surface area (not including the top or bottom of the cylinder).

This figure has two images. The first is a sphere on top of a cylinder. The second is a lightbulb.
209.

[T] A lampshade is constructed by rotating y=1/xy=1/x around the x-axisx-axis from y=1y=1 to y=2,y=2, as seen here. Determine how much material you would need to construct this lampshade—that is, the surface area—accurate to four decimal places.

This figure has two images. The first is similar to a frustum of a cone with edges bending inwards. The second is a lamp shade.
210.

[T] An anchor drags behind a boat according to the function y=24ex/224,y=24ex/224, where yy represents the depth beneath the boat and xx is the horizontal distance of the anchor from the back of the boat. If the anchor is 2323 ft below the boat, how much rope do you have to pull to reach the anchor? Round your answer to three decimal places.

211.

[T] You are building a bridge that will span 1010 ft. You intend to add decorative rope in the shape of y=5|sin((xπ)/5)|,y=5|sin((xπ)/5)|, where xx is the distance in feet from one end of the bridge. Find out how much rope you need to buy, rounded to the nearest foot.

For the following exercises, find the exact arc length for the following problems over the given interval.

212.

y=ln(sinx)y=ln(sinx) from x=π/4x=π/4 to x=(3π)/4.x=(3π)/4. (Hint: Recall trigonometric identities.)

213.

Draw graphs of y=x2,y=x2, y=x6,y=x6, and y=x10.y=x10. For y=xn,y=xn, as nn increases, formulate a prediction on the arc length from (0,0)(0,0) to (1,1).(1,1). Now, compute the lengths of these three functions and determine whether your prediction is correct.

214.

Compare the lengths of the parabola x=y2x=y2 and the line x=byx=by from (0,0)to(b2,b)(0,0)to(b2,b) as bb increases. What do you notice?

215.

Solve for the length of x=y2x=y2 from (0,0)to(1,1).(0,0)to(1,1). Show that x=(1/2)y2x=(1/2)y2 from (0,0)(0,0) to (2,2)(2,2) is twice as long. Graph both functions and explain why this is so.

216.

[T] Which is longer between (1,1)(1,1) and (2,1/2):(2,1/2): the hyperbola y=1/xy=1/x or the graph of x+2y=3?x+2y=3?

217.

Explain why the surface area is infinite when y=1/xy=1/x is rotated around the x-axisx-axis for 1x<,1x<, but the volume is finite.

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