CHAPTER 6 Applications of Integration
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1 PART II CHAPTER Applications of Integration Section. Area of a Region Between Two Curves Section. Volume: The Disk Method Section. Volume: The Shell Method Section. Arc Length and Surfaces of Revolution Section. Work Section. Moments, Centers of Mass, and Centroids Section.7 Fluid Pressure and Fluid Force Review Eercises Problem Solving
2 CHAPTER Applications of Integration Section. Solutions to Even-Numbered Eercises. A Area of a Region Between Two Curves d d. A d. d.. A d d. sec cos d 7 ( ), (, ) (, ) 7 π ( ) (, ) π, ( ) π, π ( ) π, π. f. g A Matches (a) (, ) (, ) A d 7 d 7 7 (, ) ( ), (, ) (, )
3 Section. Area of a Region Between Two Curves. The points of intersection are given b when, A d d 7. The points of intersection are given b: when, A f g d d d (, ) (, ) (, ) (, ). A d. (, ).... (,.). The points of intersection are given b (, ) (, ),, A d (, )
4 Chapter Applications of Integration. The points of intersection are given b: when, A f g d d d (, ) (, ). A f g d d d 7. (, ) 7. A d (, ) ln ln.7 (, ). The point of intersection is given b: Numerical Approimation:. (, ) when A f g d d d. The points of intersection are given b: when, ± A d d Numerical Approimation:. (, ) (, ) (, ) (, ) (, )
5 Section. Area of a Region Between Two Curves 7. f, g The points of intersection are given b: g.. when,,, A d d 7 Numerical Approimation: A ln ln. Numerical Approimation:. sin cos π (, ) d. ( )., (, ) A cos sin d. f d A d. A sec tan d sec.77 π π ( ) π, (, ). From the graph we see that f and g intersect twice at and. A g f d (, ) d ln. ln (, ). A (, ) cos sin sin cos d (π, )
6 Chapter Applications of Integration. ln A d ln (,.) (, ) ln.. (a) e,,, (b) A e d. No, it cannot be evaluated b hand. (c). = e. F t dt t t (a) F (b). F e d e e e (a) F F (b) F e.7 (c) (c) F F e e.. A d d 7 7 d 7 d = (, ) (, ) = + (, ) =
7 Section. Area of a Region Between Two Curves. f f,, At f. Tangent line: or (, ) = + f () = + ( ), The tangent line intersects f at. A d arctan. Answers will var. See page 7... on, on, Both functions smmetric to origin d d. Thus, d. A d (, ) (, ) (, ). Proposal is better, since the cummulative deficit (the area under the curve) is less.. A d b b d b b d b b b b ( ( b), b ) ( b, b) b b.
8 7 Chapter Applications of Integration 7. lim n i i where and i i is the same as n n d. f () = (, ) (, ) t.t 7..t.t dt.t.t dt.t.t billion $.7 billion 7. % : P, e.t %: P, e.t Difference in profits over ears:,e.t,e.t dt, e.t e.t.. Note: Using a graphing utilit ou obtain $,.,...7,. $, 7. The curves intersect at the point where the slope of equals that of,.. k (a) The value of k is given b... k k..... (b). Area d... d (a) (b) A.. V A.. m (c) V., pounds. True
9 Section. Volume: The Disk Method 7 Section.. V.. Volume: The Disk Method d d V d d. V d. d ± V V d. d d d.,, (a) R, r V d (b) R, r V d (c) R, r V d d d (d) R, r V d d
10 7 Chapter Applications of Integration., intersect at, and,. (a) R, r V d d (b) R, r V d d. R, r. V 7 d (, ) d 7 R, r sec V sec d ln sec tan tan ln ln ln 7. sec sec d π π π π π. R, r V d
11 Section. Volume: The Disk Method 7. R r,. V d d ln ln ln R, r V d d ln. ln. R r,. V d d R e, r V e d e d e e.. V d d d d (, )
12 7 Chapter Applications of Integration..,,, V d d V 7 (, ) 7 cos d.7. V ln d.. V arctan. d.. A. or V d Matches (b) b V a A d c A d. (a) z (b) z (c) z a < c < b.. R r r h, h V r h d r h h r h h r h = r h r ( h, r) h
13 . r, r V r d h r r d r r h r r r h h r r h h h r r h h r R r, r. (a) (b) Set Section. Volume: The Disk Method 7 V d Let < c < and set c d c c. c c Thus, when, the solid is divided into two parts of equal volume. c c d, To find the other value, set of the volume). Then d (one third of the volume). Then c, c. d d (two thirds h d, d, d. The -values that divide the solid into three parts of equal volume are and......,.,. V.. <.... d.. d.... cubic centimeters..... (a) First find where b intersects the parabola: b b b b b V b d b d b b b b d b d z b b b b b b b b CONTINUED
14 7 Chapter Applications of Integration. CONTINUED (b) graph of Vb b b (c) Vb b b Vb > b is a relative minimum. Minimum Volume is 7.7 for b.7. (a) V f d Simpson s Rule: b a, V cm n (b) f..... (c) V f d. cm. V m. (b) A bh V d Base of Cross Section (a) A b V d CONTINUED
15 Section. Volume: The Disk Method 77. CONTINUED (c) A r (d) V d A bh V d. The cross sections are squares. B smmetr, we can set up an integral for an eighth of the volume and multipl b. A b r r V r d r r r. R V r R r d R r R r R r d R r Rr R r R r R r R r R R r a :. (a) When represents a square. (b) When A a : a a represents a circle. a a d a a d a = a = To approimate the volume of the solid, form n slices, each of whose area is approimated b the integral above. Then sum the volumes of these n slices.
16 7 Chapter Applications of Integration Section. Volume: The Shell Method. p. p h V d d h V d d. p. p h V d h V d. p h V d d. p h sin V sin d sin d cos π π π π. p p on,. p h V d d h V d d
17 Section. Volume: The Shell Method 7. p h V d d. p h V d d. (a) Disk (b) Shell R, r V d d R, r V d d ln ln (c) Disk R, r V d. (a) Disk (b) Same as part a b smmetr R a r a V a d a a a a a d ( a, ) (, a) ( a, ) (, a) (, a) ( a, ) a a 7 a 7 a a a 7 a a a
18 Chapter Applications of Integration. (a) z (b) z (c) z a < c < b. represents the volume of the solid generated b revolving the region bounded b,, and about the -ais b using the Shell Method. d d represents this same volume b using the Disk Method. d. (a) (b) V d. Disk Method. (a). tan,,, Volume Matches (e) (b) V e d. π π. Total volume of the hemisphere is r. B the Shell Method, p, h. Find such that d d.. Diameter:.
19 Section. Volume: The Shell Method. r V R r d r r r R r d r d r r R r r r r r R. (a) (b) b Area region ab n a n d lim R n lim n n lim n abn b ab n a n n b ab n a bn n ab n n ab n n R n n ab n b n n n n abn n n (c) Disk Method: (d) b V ab n a n d b a lim R n lim n a b n a b n b n n d n n b bn n ab n n R n b ab n n n ab n n n n n n n lim b ab n n (e) As n, the graph approaches the line.. (a) V f d,7 cubic feet (b) Top line: Bottom line: V d d d,,,7 cubic feet (Note that Simpson s Rule is eact for this problem.) d
20 Chapter Applications of Integration Section. Arc Length and Surfaces of Revolution.,, 7,. (a) d 7 (b) 7 s d 7,, s d ,, 7 7 s d 7 d d,, b s d a d d 77.. e e. (a), e e,, e e,, s e e d e e d e e e e.7 (b) L d (c) L.. (a), (b) = + (c) L. L d
21 Section. Arc Length and Surfaces of Revolution = cos. (a) cos, (b) sin (c). sin L sin d. (a) ln, (b) (c) L.7 L d. (a),, (b) d d (c) L.! L d d Alternativel, ou can convert to a function of.. d d L d d Although this integral is undefined at, a graphing utilit still gives L.. d d tan d s Matches (e) = tan ( π, ( (, ) π π. f,, (a) (b) d. d. (c) s d.7 (d).7
22 Chapter Applications of Integration. Let ln, e, and d Equivalentl, e,, and d e, Numericall, both integrals ield L. e L d. L e d e d.. e e e e e e e e s e e d e e d e e e 7 ft e Thus, there are 7 7 square feet of roofing on the barn cosh..7 sinh.. s.7 sinh. d. (Use Simpson s Rule with n or a graphing utilit.).. s d d arcsin arcsin arcsin 7. s 7.,, S d d 7.
23 Section. Arc Length and Surfaces of Revolution.,, S d.,, S d d ln,, e e S d e d.. The precalculus formula is the distance formula between two points. The representative element is i i i i i.. The surface of revolution given b f will be larger. r is larger for f.. r r r r r S r r r r d r r r d r r r r. From Eercise 7 we have: a r S r d r a d r rr a r rr a r r r a rh where h is the height of the zone h r r h + = r
24 Chapter Applications of Integration. (a) We approimate the volume b summing disks of thickness and circumference C i equal to the average of the given circumferences: V i (b) The lateral surface area of a frustum of a right circular cone is sr r. For the first frustum, S.. Adding the si frustums together, S r i i cubic inches C i C i i (c) r (d) V r d 7. cubic inches S r r d 7. square inches h r R s. Individual project, see Eercise,.. (a) (b) Ellipse:, + = L d (c) You cannot evaluate this definite integral, since the integrand is not defined at. Simpson s Rule will not work for the same reason. Also, the integrand does not have an elementar antiderivative.. Essa
25 Section. Work 7 Section. Work. W Fd, ft lb. W Fd 7,, ft lb b W. F d is the work done b a force F moving an object along a straight line from a to b. a. (a) (b) W 7 W W (c) (d) W ft lbs d ft lbs. d d 7 7 d ft lbs d 7 ft lbs W d in lb. ft lb.. F k. F k k7 k 7 7 F d 7 W n cm Nm d W 7. k d k k 7 W d 7 k..7 ft lbs k k W d ft lb h,, W d,, h,, h lim W, miton. h ft lb,. Weight on surface of moon: tons Weight varies inversel as the square of distance from the center of the moon. Therefore, F k k k. W.. d.. mi ton. ft lb
26 Chapter Applications of Integration. The bottom half had to be pumped a greater distance then the top half.. Volume of disk: Weight of disk: Distance the disk of water is moved: W d,,, newton meters. Volume of disk: 7 Weight of disk:. Distance: (a) W. d.. ft lb (b) W. d. 7.7 ft lb. Volume of each laer: Weight of each laer:. Distance: W. d. d. (, ) =. ft lb. Volume of laer: V Weight of laer: W Distance:. W. d... d. d The second integral is zero since the integrand is odd and the limits of integration are smmetric to the origin. The first integral represents the area of a semicircle of radius. Thus, the work is W, ft lb,. ft lb. Ground level
27 Section. Work. The lower feet of chain are raised feet with a constant force. W ft lb The top feet will be raised with variable force. Weight of section: Distance: W d W W W 7 7 ft lb 7 ft lb. The work required to lift the chain is 7. ft lb (from Eercise ). The work required to lift the -pound load is W 7. The work required to lift the chain with a -pound load attached is W ft lbs. W d ft lb. Work to pull up the ball: W, ft lb. Work to pull up the cable: force is variable Weight per section: Distance: W d ft lb p k V k W k V dv ln V ln 7. ft lb W W W,, ft lb. (a) (b) W FD, ft lbs W,,,,,. ft lb (c) F,.,. 7,7.,..7, (d) F when. feet. F is a maimum when. feet. (e) W F d,. ft lbs. W e d, ft lb. W sinh d 7. ft lb
28 Chapter Applications of Integration Section. Moments, Centers of Mass, and Centroids The center of mass is translated k units as well.. (Person on left) feet.,, m (, ). m (, ) m (, ). 7. 7, 7, 7 m (, ) m (, ) m (, ) m (, ). m M M m M M m d d d d d ( ),,,
29 Section. Moments, Centers of Mass, and Centroids. m M M m M M m,, d d d d d (, ). m d 7 d M M d d d d d d M m ; M m,, (, ) ( ), (, )
30 Chapter Applications of Integration. m d B smmetr, M and. M d (, ),, 7. m M M M m d d d (, ). M m,, m M M M m d d d d d (, ) M m,,. A M M d ln ln d d
31 . A M d d M b smmetr. d d Section. Moments, Centers of Mass, and Centroids. m M M e e d.7 e d e d.7 e d.7 (, ) M m. M m.. Therefore, the centroid is.,.. m M M m d. d. b smmetr. Therefore, the centroid is,.. d.. A bh ac A ac ac ac c ab c a d ac ab c c a ac c b c a b c d c c c, b a, c c b c a b c d ac abc a c b a This is the point of intersection of the diagonals. = c b (, ) ( b, c) (, ) ( a + b, c) ( a, ) = c b ( a)
32 Chapter Applications of Integration. b smmetr. A r A r r r,, r r r d r r r r r r r A d A,, 7 d d d d d 7 r r. (a) M b smmetr (c) M n b because b n is an odd function. M n b n b b n n b n b b n bnn n n n bnn A n b n b b n d b n b n n b d n b b n d n b b n n b b n bnn n n b n b n n bnn M A n bnn n n n b nn n n b b n d (b) > b because there is more area above b than below. (d) n n (e) lim lim n n n b b (f) As n, the figure gets narrower. n b b b = n n b 7 b = b 7b
33 Section. Moments, Centers of Mass, and Centroids. Let f be the top curve, given b l d. The bottom curve is d..... f...7. (a) d.... Area f d d M.7... f.7.77 M A.77.,,.7 f d f d d f d d.7 (b) f...7 (c) d..7. M A,,. d.... Centroids of the given regions: and Area: A ,,,, 7 7 7, m m m,,. m 7 7, P, 7 m 7 7 B smmetr,. 7, P, m , m,,,.
34 Chapter Applications of Integration. Centroids of the given regions:, and,. V ra Mass:,,.,. A d M d d (, ) Let u, u, du d: u u du M M A 7 7 u u du u u (, ) r V ra.. A planar lamina is a thin flat plate of constant densit. The center of mass, is the balancing point on the lamina.. Let R be a region in a plane and let L be a line such that L does not intersect the interior of R. If r is the distance between the centroid of R and L, then the volume V of the solid of revolution formed b revolving R about L is V ra where A is the area of R.. The centroid of the circle is,. The distance traveled b the centroid is. The arc length of the circle is also. Therefore, S. C d
35 Section.7 Fluid Pressure and Fluid Force 7 Section.7 Fluid Pressure and Fluid Force. F PA. lb. F.h.h.,. lb. h. L F. d.. Force is one-third that of Eercise. d 7. lb h L F. d.. lb. h L F. d. d. 7. lb. h L lower part L upper part F d,, d +, Newtons
36 Chapter Applications of Integration. h. L F d 7, Newtons h L F.7 d.7 d.7 7. lb. h L F.7 d.7 d (, ).7.7. lb. h L F d The second integral is zero since it is an odd function and the limits of integration are smmetric to the origin. The first integral is twice the area of a semicircle of radius. Thus, the force is.7. lb. d d. (a) F wkr lbs. (a) (b) F wkr. lbs F wkhb. lbs (b) F wkhb. 7, lbs
37 Review Eercises for Chapter. From Eercise : F 7. lb. h. Solving for, ou obtain. L F. d. d. lb h L 7 7 F. 7 d.7 d 7.7 lb. Fluid pressure is the force per unit of area eerted b a fluid over the surface of a bod.. The left window eperiences the greater fluid force because its centroid is lower. Review Eercises for Chapter. A d. (, ) (, ) A d d d (, ) (, )
38 Chapter Applications of Integration. A d. d (, ) A csc d ln csc cot ln ln. π ( ) ( ) π,, (, ) π π π π π. A 7 cos d cos sin sin 7 ( ( ) 7π, ) π, d. Point of intersection is given b:.7..7 A d.7. π, ( ) (.7,.). A d. d. (, ) (, ) (, ) A d d (, ) (, )
39 . A d d A d (, ) (, ) Review Eercises for Chapter. (a) R t..7 t. e.t (b) R t. e.t Difference R t R t dt 7. billion dollars. (a) Shell V d (b) Shell V d d (c) Disk V d (d) Disk V d d 7
40 Chapter Applications of Integration. (a) Shell a V b a a d b a a a d b a a a a b (b) Disk V a b a a a ab b a a d (, b) a + b = (, b) a + b = ( a, ) ( a, ). Disk V d arctan. Disk V e d e d e e e. (a) Disk V d d (b) Shell u u d u du V d u u du u u u du 7 u7 u u
41 Review Eercises for Chapter. A bh a a. a a V a d a a a a a Since a, we have a. Thus, s d a. meters. a a a. Since f tan has f sec, this integral represents the length of the graph of tan from to. This length is a little over unit. Answers (b).. S d d. F k k k F W d in lb.7 ft lb. We know that dv galmin galmin dt 7. galft 7. ft min V r h h dv dt dh dt dv dt 7.. ftmin. Depth of water:.t Time to drain well: t minutes. gallons pumped Volume of water pumped in Eercise :.7 gallons.7 dh dt.7 7 Work 7 ft ton
42 Chapter Applications of Integration. (a) Weight of section of cable: Distance: W d, ft lb ft ton (b) Work to move pounds feet verticall: Total work work for drawing up the cable work of lifting the load ft ton ft ton 7 ft ton, ft lb ft ton.. b W F d a F,, W d d 7,, 7 ft lbs A d A d d d d (, ). A d A d d (, ) 7 7,,
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