Applications of Definite Integral

Transcription

1 Chpter 5 Applitions of Definite Integrl 5.1 Are Between Two Curves In this setion we use integrls to find res of regions tht lie between the grphs of two funtions. Consider the region tht lies between two urves = f() nd = g() nd between the vertil lines = nd = b, where f nd g re ontinuous funtions nd f() g() for ll in [,b]. = f() b The re A of this region is = g() A = b [ f() g() ] d (5.1) Emple 5.1. Find the re of the region enlosed b = 2, = 4 2, nd =

2 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 134 Emple 5.2. Find the re of the region bounded b = nd = 1 for Emple 5.3. Findthereboundedbthegrphsof = 2 +2nd 2 = +2for 1 2.

3 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 135 d = g() = f() Some regions re best treted b regrding s funtion of. If region is bounded b urves with equtions = f(), = g(), =, nd = d, where f nd g re ontinuous nd f() g() for d, then its re is Emple 5.4. A = d [ f() g() ] d (5.2) Find the re of the region bounded b the grphs of = 1 2 nd = +1. Emple 5.5. Find the re of the region bounded b the urves = 2, = + 5, = 2, nd = 1.

4 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn Find the re of the shded region. Eerise 5.1 () = (b) = 2 +2 = 1 2 () = 2 2 = 2. Find the re of the region bounded b the given urves. () = 2 +3, =, 1 1 (b) = 3, = 2 1, 1 3 () = +1, = 9 2, 1 2 (d) = e, = 1, 2 0 (e) = +1, = ( 1) 2, 1 2

5 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 137 (f) = 2 1, = 1, 0 2 (g) = os, = sin2, 0 π/2 (h) = 3 1, = 1, Find the re of the region enlosed b the given urves. () = 2 1, = 7 2 (b) = 2 +1, = 3 1 () = 3, = 3+2 (d) = 3, = 2 (e) =, = 2, = 0 (f) = 3, = 2+ 2 (g) =, =, = 1 (h) =, = 2, = 6, = 0 (i) = 1 2, = 2 1 (j) = os, = 1 2/π (k) = 2, = 2/( 2 +1) (l) = 3, = 1 4 Answer to Eerise () 6 (b) 40 3 () () 20 3 (b) 40 3 () 19.5 (d) e 2 (e) 31 6 (f) 3 (g) 1 2 (h) () 64 3 (b) 1 6 () 27 4 (d) 1 12 (e) 1 (f) 1 6 (g) 1 (h) 8 (i) 1 8 (j) 2 π 2 (k) π 2 3 (l) 8 5

6 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn Volumes b Sliing: Disks nd Wshers In this setion we will use definite integrl to find volumes of solid of revolution. A solid of revolution is solid tht is generted b revolving plne region bout line tht lies in the sme plne s the region; the line is lled the is of revolution. Method of Disks Suppose tht f() 0 nd f is ontinuous on [,b]. Tke the region bounded b the urve = f() nd the -is, for b nd revolve it bout the -is, generting solid. = f() 0 = f() b b We n find the volume of this solid b sliing it perpendiulr to the -is nd reognizing tht eh ross setion is irulr disk of rdius r = f(). We then hve tht the volume of the solid is Emple 5.6. V = b π[f()] 2 }{{} ross-setionl re = πr 2 d Findthevolumeofthesolidobtinedbrottingtheregionboundedb = from = 0 to = 4 bout the -is.

7 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 139 In similr w, suppose tht g() 0 nd g is ontinuous on the intervl [,d]. Then, revolving the region bounded b the urve = g() nd the -is, for d, bout the -is genertes solid. b b = g() = g() One gin, notie from Figure tht the ross setions of the resulting solid of revolution re irulr disks of rdius r = g(). The volume of the solid is then given b Emple 5.7. V = d π[g()] 2 }{{} ross-setionl re = πr 2 d Find the volume of the solid obtined b rotting the region bounded b = from = 0 to = 2 bout the -is.

8 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 140 Method of Wshers There re two omplitions tht n be found in the tpes of volume lultions we hve been studing. The first of these is tht ou m need to ompute the volume of solid tht hve vit or hole in it. The seond of these ours when region is revolved bout line other thn the -is or the -is. Suppose tht f nd g re nonnegtive ontinuous funtion suh tht g() f() for b nd let R be the region enlosed between the grphs of these funtions nd the lines = nd = b. = f() = g() b When this region is revolved bout the -is, it genertes solid hving nnulr or wsher-shped ross setions. Sine the ross setion t hs inner rdius g() nd outer rdius f(), its volume of the solid is V = b } π {[f()] 2 [g()] 2 d Suppose tht u nd v re nonnegtive ontinuous funtion suh tht v() u() for d. If R the region enlosed between the grphs of = u() nd = v() nd the lines = nd = d. d = v() = u() When this region is revolved bout the -is, it lso genertes solid hving nnulr or wsher-shped ross setions. Sine the ross setion t hs inner rdius v() nd outer rdius u(), its volume of the solid is V = d } π {[u()] 2 [v()] 2 d

9 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 141 Emple 5.8. The region R enlosed b the urves = 4 2 nd = 0. Find the volume of the solid obtined b rotting the region R () bout the -is (b) bout the line = 7 () bout the line = 3

10 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 142 Emple 5.9. Find the volume of the solid generted when the region under the urve = 2 over the intervl [0,2] is rotted bout the line = 1. Eerise Find the volume of the solid obtined b rotting the region bounded b = 2, = 0, nd = 0 () bout the -is (b) bout the line = Find the volume of the solid obtined b rotting the region bounded b =, = 2, nd = 0 () bout the -is (b) bout the line = Find the volume of the solid obtined b rotting the region bounded b = e, = 0, = 2, nd = 0 () bout the -is (b) bout the line = Find the volume of the solid obtined b rotting the region bounded b = 3, = 0, nd = 1 () bout the -is (b) bout the -is. 5. The region R enlosed b the urves = 3, the -is, nd the -is. Find the volume of the solid obtined b rotting the region R () bout the -is, (b) bout the -is () bout the line = 3, (d) bout the line = 3, (e) bout the line = 3, nd (f) bout the line = The region R enlosed b the urves = 2, = 0, nd = 1. Find the volume of the solid obtined b rotting the region R () bout the -is, (b) bout the -is, () boutthe line = 1, (d) boutthe line = 1, (e) boutthe line = 1, nd (d) bout the line = 1.

11 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn () 8π 3 4. () 2π 5 (b) 28π 3 (b) π 7 Answer to Eerise () 32π 5 (b) 224π 15 ( e 4 3. () 2πe 2 +2π (b) π 2 +4e2 9 ) 2 5. () 9π (b) 9π () 18π (d) 36π (e) 18π (f) 36π 6. () π 2 (b) π 5 () π 6 (d) 7π 15 (e) 7π 6 (f) 13π Volumes b Clindril Shells A lindril shells is solid enlosed b two onentri right-irulr linders. h r 2 r 1 The volume V of lindril shll hving inner rdius r 1, outer rdius r 2, nd height h n be written s V = [re of ross setion] [height] = (πr2 2 πr1)h 2 = π(r 2 +r 1 )(r 2 r 1 )h ( ) r2 +r 1 = 2π h (r 2 r 1 ) 2 But r 2 +r 1 2 is the verge rdius of the shell nd r 2 r 1 is its thikness, so V = 2π [verge rdius] [height] [thikness] This formul n be used to find the volume of solid of revolution. Let R be plne region bounded bove b ontinuous urve = f(), bounded below b the -is, nd bounded on the left nd right, respetivel, b the line = nd = b. = f() = f() b

12 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 144 The volume of the solid generted b revolving R bout the -is is given b b V = 2π }{{} rdius f() }{{} height }{{} d thikness Let R be plne region bounded bove b ontinuous urve = g(), the -is, nd the line = nd = d. d = g() The volume of the solid generted b revolving R bout the -is is given b Emple d V = 2π }{{} rdius g() }{{} height d }{{} thikness Find the volume of the solid obtined b rotting the region bounded b = nd = 2 in the first qudrnt bout the -is.

13 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 145 Emple Use lindril shells to find the volume of the solid generted when the region R enlosed between = 2, = 1 nd = 0 is revolved bout the -is. Eerise Use the method of lindril shells to find the volume generted b rotting the region bounded b the given urve bout the -is. () = 1, = 0, = 1, = 2 (b) = e 2, = 0, = 0, = 1 () 2 =, = 2 2. Use the method of lindril shells to find the volume of the solid obtined b rotting the region bounded b the given urve bout the -is. () = 1+ 2, = 0, = 1, = 2 (b) = 2, = 9 () =, = 0, + = 2 3. Use the pproprite method to find the volume generted b rotting the region bounded b the given urve bout the speified is.

14 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 146 () =, = 3, = 0; bout -is (b) =, = 5, = 0; bout = 5 () = , = 1, = 1; bout -is (d) = 2, = 1, = 0; bout -is (e) = 2, = 2, = 0; bout = 2 Answer to Eerise () 2π (b) π(1 1/e) () 64π/15 2. () 21π/2 (b) 1944π/5 () 5π/6 3. () π (b) π () 23π 30 (d) π 2 (e) 8π Length of Plne Curve Ar Length Problem. Suppose f is ontinuous on [, b] nd differentible on (, b). Find the r length L of the urve = f() over the intervl [,b]. In order to solve this problem, we begin b prtitioning the intervl [,b] into n equl piees: = 0 < 1 < 2 < < n = b, where i i 1 = = b n, for eh i = 1,2,...,n. Between eh pir if djent points on the urve, ( i 1,f( i 1 )) nd ( i,f( i )) we pproimte the re length l i b the stright-line distne between the two points. f( i )+ l i f( i 1 )+ i 1 i From the usul distne formul, we hve l i ( i i 1 ) 2 +[f( i ) f( i 1 )] 2. Sine f is ontinuous on ll of [,b] nd differentible on (,b), f is lso ontinuous on the subintervl [ i 1, i ] nd is differentible on ( i 1, i ). Rell tht b the Men Vlue Theorem, we hve f( i ) f( i 1 ) = f ( i )( i i 1 ),

15 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 147 for some number i ( i 1, i ). This give us the pproimtion l i ( i i 1 ) 2 +[f( i ) f( i 1 )] 2 = ( i i 1 ) 2 +[f ( i )( i i 1 )] 2 = 1+[f ( i )] 2 ( i i 1 ) }{{} = 1+[f ( i )] 2. Adding together the lengths of these n line segments, we get n pproimtion of the totl r length, n L 1+[f ( i )] 2. i=1 Notie tht s n gets lrger, this pproimtion should pproh the et r length, tht is, n L = lim 1+[f ( i )] 2. n i=1 You should reognize this s the limit of Riemnn sum for 1+[f ()] 2, so tht the r length is given b the definite integrl: whenever the limit eists. Emple L = b 1+[f ()] 2 d, Find the r length of the urve = 3/2 from (1,1) to (4,8).

16 MA111: Prepred b Asst.Prof.Dr. Arhr Pheenburwn 148 Find the length of the urve. 1. = 1 3 (2 +2) 3/2, = , = ln(se), 0 π/4 4. = ln(1 2 ), = osh, 0 1 Eerise 5.4 Answer to Eerise ln( 2+1) 4. ln sinh1