8.2 Areas in the Plane

Transcription

1 39 Chpter 8 Applictions of Definite Integrls 8. Ares in the Plne Wht ou will lern out... Are Between Curves Are Enclosed Intersecting Curves Boundries with Chnging Functions Integrting with Respect to Sving Time with Geometric Formuls nd wh... The techniques of this section llow us to compute res of comple regions of the plne. Are Between Curves We know how to find the re of region etween curve nd the -is ut mn times we wnt to know the re of region tht is ounded ove one curve, = f(), nd elow nother, = g() (Figure 8.3). We find the re s n integrl ppling the first two steps of the modeling strteg developed in Section 8... We prtition the region into verticl strips of equl width nd pproimte ech strip with rectngle with se prllel to [, ] (Figure 8.). Ech rectngle hs re [f(c k ) - g(c k )] for some c k in its respective suintervl (Figure 8.5). This epression will e nonnegtive even if the region lies elow the -is. We pproimte the re of the region with the Riemnn sum [f(c k) - g(c k )]. Upper curve f() f() (c k, f (c k )) f (c k ) g(c k ) c k ower curve g() g() (c k, g(c k )) Figure 8.3 The region etween = f() nd = g() nd the lines = nd =. Figure 8. We pproimte the region with rectngles perpendiculr to the -is.. The limit of these sums s : is Figure 8.5 The re of tpicl rectngle is [f(c k ) - g(c k )]. [f() - g()] d. This pproch to finding re cptures the properties of re, so it cn serve s definition. DEFINITION Are Between Curves If f nd g re continuous with f() Ú g() throughout [, ], then the re etween the curves f() nd g() from to is the integrl of [f - g] from to, A = [f() - g()] d.

2 Section 8. Ares in the Plne 395 sec EXAMPE Appling the Definition Find the re of the region etween = sec nd = sin from = to = p>. SOUTION We grph the curves (Figure 8.6) to find their reltive positions in the plne, nd see tht = sec lies ove = sin on [, p>]. The re is therefore sin p> A = [ sec - sin ] d p> = c tn + cos d = units squred. Now Tr Eercise. Figure 8.6 The region in Emple. EXPORATION A Fmil of Butterflies = k k sin k = k sin k k = [, ] [, ] k = Figure 8.7 Two memers of the fmil of utterfl-shped regions descried in Eplortion. = = For ech positive integer k, let A k denote the re of the utterfl-shped region enclosed etween the grphs of = k sin k nd = k - k sin k on the intervl [, p>k]. The regions for k = nd k = re shown in Figure 8.7. A?. Find the res of the two regions in Figure Mke conjecture out the res A k for k Ú Set up definite integrl tht gives the re A k. Cn ou mke simple u-sustitution tht will trnsform this integrl into the definite integrl tht gives the re A?. Wht is lim k:q 5. If P k k denotes the perimeter of the kth utterfl-shped region, wht is lim k:q P k? (You cn nswer this question without n eplicit formul for.) Are Enclosed Intersecting Curves When region is enclosed intersecting curves, the intersection points give the limits of integrtion. EXAMPE Are of n Enclosed Region Find the re of the region enclosed the prol = - nd the line = -. SOUTION We grph the curves to view the region (Figure 8.8). The limits of integrtion re found solving the eqution P k [ 6, 6] [, ] Figure 8.8 The region in Emple. - = - either lgericll or clcultor. The solutions re = - nd =. continued

3 396 Chpter 8 Applictions of Definite Integrls Since the prol lies ove the line on [-, ], the re integrnd is - - (-). A = [ - - (-)] d - = c d - = 9 units squred Now Tr Eercise 5. = cos = [ 3, 3] [, 3] Figure 8.9 The region in Emple 3. EXAMPE 3 Using Clcultor Find the re of the region enclosed the grphs of = cos nd = -. SOUTION The region is shown in Figure 8.9. Using clcultor, we solve the eqution cos = - to find the -coordintes of the points where the curves intersect. These re the limits of integrtion. The solutions re = We store the negtive vlue s A nd the positive vlue s B. The re is NINT ( cos - ( - ),, A, B) This is the finl clcultion, so we re now free to round. The re is out.99. Now Tr Eercise 7. Finding Intersections Clcultor The coordintes of the points of intersection of two curves re sometimes needed for other clcultions. To tke dvntge of the ccurc provided clcultors, use them to solve for the vlues nd store the ones ou wnt. Boundries with Chnging Functions If oundr of region is defined more thn one function, we cn prtition the region into suregions tht correspond to the function chnges nd proceed s usul. EXAMPE Finding Are Using Suregions Find the re of the region R in the first qudrnt tht is ounded ove = nd elow the -is nd the line = -. SOUTION The region is shown in Figure 8.. Are d Are d (, ) B A Figure 8. Region R split into suregions A nd B. (Emple ) continued

4 Section 8. Ares in the Plne 397 While it ppers tht no single integrl cn give the re of R (the ottom oundr is defined two different curves), we cn split the region t = into two regions A nd B. The re of R cn e found s the sum of the res of A nd B. Are of R = d + [ - ( - )] d re of A = c 3 3> d re of B + c 3 3> - + d = 3 units squred Now Tr Eercise 9. Integrting with Respect to Sometimes the oundries of region re more esil descried functions of thn functions of. We cn use pproimting rectngles tht re horizontl rther thn verticl nd the resulting sic formul hs in plce of. For regions like these d f() d f() d g() f() g() c g() c c use this formul d A = [f () g()]d. c (g(), ) f() g() (, ) ( f (), ) Figure 8. It tkes two integrtions to find the re of this region if we integrte with respect to. It tkes onl one if we integrte with respect to. (Emple 5) EXAMPE 5 Integrting with Respect to Find the re of the region in Emple integrting with respect to. SOUTION We remrked in solving Emple tht it ppers tht no single integrl cn give the re of R, ut notice how ppernces chnge when we think of our rectngles eing summed over. The intervl of integrtion is [, ], nd the rectngles run etween the sme two curves on the entire intervl. There is no need to split the region (Figure 8.). We need to solve for in terms of in oth equtions: = - = ecomes ecomes = +, =, Ú. continued

5 398 Chpter 8 Applictions of Definite Integrls = 3, = +, 3 = + (, ) (c, d) We must still e creful to sutrct the lower numer from the higher numer when forming the integrnd. In this cse, the higher numers re the higher -vlues, which re on the line = + ecuse the line lies to the right of the prol. So, Are of R = ( + - ) d = c d = 3 units squred. Now Tr Eercise. [ 3, 3] [ 3, 3] Figure 8. The region in Emple 6. (c, d) (, ) EXAMPE 6 Mking the Choice Find the re of the region enclosed the grphs of = 3 nd = -. SOUTION We cn produce grph of the region on clcultor grphing the three curves = 3, = +, nd = - + (Figure 8.). This convenientl gives us ll of our ounding curves s functions of. If we integrte in terms of, however, we need to split the region t = (Figure 8.3). On the other hnd, we cn integrte from = to = d nd hndle the entire region t once. We solve the cuic for in terms of : = 3 ecomes = >3 [ 3, 3] [ 3, 3] Figure 8.3 If we integrte with respect to in Emple 6, we must split the region t =. To find the limits of integrtion, we solve >3 = -. It is es to see tht the lower limit is = -, ut clcultor is needed to find tht the upper limit d = We store this vlue s D. The cuic lies to the right of the prol, so Are = NINT ( >3 - ( - ),, -, D) = The re is out.. Now Tr Eercise 7. Sving Time with Geometr Formuls Here is et nother w to hndle Emple. (, ) Are Figure 8. The re of the lue region is the re under the prol = minus the re of the tringle. (Emple 7) EXAMPE 7 Using Geometr Find the re of the region in Emple sutrcting the re of the tringulr region from the re under the squre root curve. SOUTION Figure 8. illustrtes the strteg, which enles us to integrte with respect to without splitting the region. Are = d - ()() = 3 3> d - = 3 units squred Now Tr Eercise 35. The morl ehind Emples, 5, nd 7 is tht ou often hve options for finding the re of region, some of which m e esier thn others. You cn integrte with respect to or with respect to, ou cn prtition the region into suregions, nd sometimes ou cn even use trditionl geometr formuls. Sketch the region first nd tke moment to determine the est w to proceed.

6 Section 8. Ares in the Plne 399 Quick Review 8. (For help, go to Sections. nd 6..) Eercise numers with gr ckground indicte prolems tht the uthors hve designed to e solved without clcultor. In Eercises 5, find the re etween the -is nd the grph of the given function over the given intervl.. = sin over [, p]. = e over [, ] 3. = sec over [-p>, p>]. = - 3 over [, ] 5. = 9 - over [-3, 3] = + 6 In Eercises 6, find the - nd -coordintes of ll points where the grphs of the given functions intersect. If the curves never intersect, write NI. 6. = - nd 7. = e nd = + 8. = - p nd = sin 9. = + nd = 3. = sin nd = 3 Section 8. Eercises In Eercises 6, find the re of the shded region nlticll... 3 cos 5.. sec t (, 8) 8 (, 8) 3 3 t sin t 6. NOT TO SCAE 3. 3 (, )

7 Chpter 8 Applictions of Definite Integrls In Eercises 7 nd 8, use clcultor to find the re of the region enclosed the grphs of the two functions. 7. = sin, = - 8. = cos (), = - In Eercises 9 nd, find the re of the shded region nlticll. 9.. In Eercises nd, find the re enclosed the grphs of the two curves integrting with respect to.. = +, = 3 -. = + 3, = In Eercises 3 nd, find the totl shded re (, ). (, ) (, ) (3, 5) In Eercises 5 3, find the re of the regions enclosed the lines nd curves. 5. = - nd = 6. = - nd = -3 ƒ ƒ ƒ ƒ = ( >) + 7. = 7 - nd = + 8. = - + nd = 9. = -, 7, nd =. = nd 5 = + 6 (How mn intersection points re there?). = - nd. = nd = = nd - = 6. - = nd + = = nd + 3 = 6. + = nd - = 7. + = 3 nd + = 8. = sin nd = sin, p 9. = 8 cos nd = sec, -p>3 p>3 3. = cos (p>) nd = - 3. = sin (p>) nd = 3. = sec, = tn, = -p>, = p> 33. = tn nd = - tn, -p> p> 3. = 3 sin cos nd =, p> In Eercises 35 nd 36, find the re of the region sutrcting the re of tringulr region from the re of lrger region. 35. The region on or ove the -is ounded the curves = + 3 nd = 36. The region on or ove the -is ounded the curves = - nd = Find the re of the propeller-shped region enclosed the curve - 3 = nd the line - =. 38. Find the re of the region in the first qudrnt ounded the line =, the line =, the curve = >, nd the -is. 39. Find the re of the tringulr region in the first qudrnt ounded on the left the -is nd on the right the curves = sin nd = cos.. Find the re of the region etween the curve = 3 - nd the line = - integrting with respect to (), ().. The region ounded elow the prol = nd ove the line = is to e prtitioned into two susections of equl re cutting cross it with the horizontl line = c. () Sketch the region nd drw line = c cross it tht looks out right. In terms of c, wht re the coordintes of the points where the line nd prol intersect? Add them to our figure. () Find c integrting with respect to. (This puts c in the limits of integrtion.) (c) Find c integrting with respect to. (This puts c into the integrnd s well.). Find the re of the region in the first qudrnt ounded on the left the -is, elow the line = >, ove left the curve = +, nd ove right the curve = >.

8 Section 8. Ares in the Plne 3. The figure here shows tringle AOC inscried in the region cut from the prol = the line =. Find the limit of the rtio of the re of the tringle to the re of the prolic region s pproches zero. A (, ) C (, ) Stndrdized Test Questions 5. True or Flse The re of the region enclosed the grph of = + nd the line = is 36. Justif our nswer. 5. True or Flse The re of the region in the first qudrnt enclosed the grphs of = cos, =, nd the -is is given the definite integrl.739 ( - cos ) d. Justif our nswer. 5. Multiple Choice et R e the region in the first qudrnt ounded the -is, the grph of = +, nd the line =. Which of the following integrls gives the re of R? O (A) [ - ( + )] d (B) [( + ) - ] d. Suppose the re of the region etween the grph of positive continuous function f nd the -is from = to = is squre units. Find the re etween the curves = f() nd = f() from = to =. 5. Writing to ern Which of the following integrls, if either, clcultes the re of the shded region shown here? Give resons for our nswer. i. ( - (-)) d = d - - ii. (- - ()) d = - d - - (C) [ - ( + )] d - (E) [ - ( + )] d 53. Multiple Choice Which of the following gives the re of the region etween the grphs of = nd = - from = to = 3? (A) (B) 9 > (C) 3> (D) 3 (E) 7> 5. Multiple Choice et R e the shded region enclosed the grphs of = e -, = - sin (3), nd the -is s shown in the figure elow. Which of the following gives the pproimte re of the region R? You m use clcultor on this prolem. (A).39 (B).5 (C).869 (D). (E).3 (D) - [( + ) - ] d 6. Writing to ern Is the following sttement true, sometimes true, or never true? The re of the region etween the grphs of the continuous functions = f() nd = g() nd the verticl lines = nd = ( 6 ) is [f() - g()] d. Give resons for our nswer. 7. Find the re of the propeller-shped region enclosed etween the grphs of = nd = Find the re of the propeller-shped region enclosed etween the grphs of = sin nd = Find the positive vlue of k such tht the re of the region enclosed etween the grph of = k cos nd the grph of = k is. 55. Multiple Choice et f nd g e the functions given f() = e nd g() = >. Which of the following gives the re of the region enclosed the grphs of f nd g etween = nd =? (A) e - e - ln (B) ln - e + e (C) e - (D) e - e - (E) e - ln

9 Chpter 8 Applictions of Definite Integrls Eplortion 56. Group Activit Are of Ellipse An ellipse with mjor is of length nd minor is of length cn e coordintized with its center t the origin nd its mjor is horizontl, in which cse it is defined the eqution + =. () Find the equtions tht define the upper nd lower semiellipses s functions of. () Write n integrl epression tht gives the re of the ellipse. (c) With our group, use NINT to find the res of ellipses for vrious lengths of nd. (d) There is simple formul for the re of n ellipse with mjor is of length nd minor is of length. Cn ou tell wht it is from the res ou nd our group hve found? (e) Work with our group to write proof of this re formul showing tht it is the ect vlue of the integrl epression in prt (). Etending the Ides 57. Cvlieri s Theorem Bonventur Cvlieri (598 67) discovered tht if two plne regions cn e rrnged to lie over the sme intervl of the -is in such w tht the hve identicl verticl cross sections t ever point (see figure), then the regions hve the sme re. Show tht this theorem is true. Cross sections hve the sme length t ever point in [, ]. 58. Find the re of the region enclosed the curves = + nd = m, 6 m 6.