5128_Ch07_pp378-433

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Section 7.3 Volumes 411 Quick Quiz for AP* Preparation: Sections 7.1–7.3 You may use a graphing calculator to solve the following problems. 1. Multiple Choice The base of a solid is the region in the first quadrant bounded by the x-axis, the graph of y sin 1 x, and the vertical line x 1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume? C (A) 0.117 (B) 0.285 (C) 0.467 (D) 0.571 (E) 1.571 2. Multiple Choice Let R be the region in the first quadrant bounded by the graph of y 3x x 2 and the x-axis. A solid is generated when R is revolved about the vertical line x 1. Set up, but do not evaluate, the definite integral that gives the volume of this solid. A 3 (A) 2p(x 1)(3x x 2 ) dx (B) 0 3 (C) 1 3 0 3 (D) (E) 0 3 0 2p(x 1)(3x x 2 ) dx 2p(x)(3x x 2 ) dx 2p(3x x 2 ) 2 dx (3x x 2 ) dx 3. Multiple Choice A developing country consumes oil at a rate given by r(t) 20e 0.2t million barrels per year, where t is time measured in years, for 0 t 10. Which of the following expressions gives the amount of oil consumed by the country during the time interval 0 t 10? D (A) r(10) (B) r(10) r(0) 10 (C) r(t) dt 0 10 (D) r(t) dt 0 (E) 10 r(10) 4. Free Response Let R be the region bounded by the graphs of y x, y e x , and the y-axis. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal line y 1. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a semicircle whose diameter runs from the graph of y x to the graph of y e x . Find the volume of this solid. 4. (a) The two graphs intersect where x e –x , which a calculator shows to be x 0.42630275. Store this value as A. The area of R is A (e –x x) dx 0.162. 0 (b) Volume A p((e –x 1) 2 (x 1) 2 ) dx 1.631. 0 (c) Volume A 1 0 2 p x 2 ex 2 dx 0.035.
412 Chapter 7 Applications of Definite Integrals 7.4 What you’ll learn about • A Sine Wave • Length of a Smooth Curve • Vertical Tangents, Corners, and Cusps . . . and why The length of a smooth curve can be found using a definite integral. [0, 2] by [–2, 2] Figure 7.32 One wave of a sine curve has to be longer than 2p. y x y √ 2 k + 2 k Q y k y = sin x Lengths of Curves A Sine Wave How long is a sine wave (Figure 7.32)? The usual meaning of wavelength refers to the fundamental period, which for y sin x is 2p. But how long is the curve itself? If you straightened it out like a piece of string along the positive x-axis with one end at 0, where would the other end be? EXAMPLE 1 The Length of a Sine Wave What is the length of the curve y sin x from x 0 to x 2p? SOLUTION We answer this question with integration, following our usual plan of breaking the whole into measurable parts. We partition 0, 2p into intervals so short that the pieces of curve (call them “arcs”) lying directly above the intervals are nearly straight. That way, each arc is nearly the same as the line segment joining its two ends and we can take the length of the segment as an approximation to the length of the arc. Figure 7.33 shows the segment approximating the arc above the subinterval x k1 , x k . The length of the segment is Δx 2 k . Δy 2 k The sum Δx k 2 Δy 2 k over the entire partition approximates the length of the curve. All we need now is to find the limit of this sum as the norms of the partitions go to zero. That’s the usual plan, but this time there is a problem. Do you see it? The problem is that the sums as written are not Riemann sums. They do not have the form f c k Δx. We can rewrite them as Riemann sums if we multiply and divide each square root by Δx k . O P x k–1 x k Figure 7.33 The line segment approximating the arc PQ of the sine curve above the subinterval x k1 , x k . (Example 1) Group Exploration Later in this section we will use an integral to find the length of the sine wave with great precision. But there are ways to get good approximations without integrating. Take five minutes to come up with a written estimate of the curve’s length. No fair looking ahead. x k x x k 2 y 2 k x k 2 y k 2 x x k y x 1 ( k ) 2 k k This is better, but we still need to write the last square root as a function evaluated at some c k in the kth subinterval. For this, we call on the Mean Value Theorem for differentiable functions (Section 4.2), which says that since sin x is continuous on x k1 , x k and is differentiable on x k1 , x k there is a point c k in x k1 , x k at which Δy k Δx k sin c k (Figure 7.34). That gives us 1 si nc k 2 Δx k , which is a Riemann sum. Now we take the limit as the norms of the subdivisions go to zero and find that the length of one wave of the sine function is 2p 1 si nx 2 dx 2p 1 cos 2 x dx 7.64. Using NINT 0 0 How close was your estimate? Now try Exercise 9. x k
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