The Double Integral. The Riemann sum of a function f (x; y) over this partition of [a; b] [c; d] is. f (r j ; t k ) x j y k

Size: px

Start display at page:

Download "The Double Integral. The Riemann sum of a function f (x; y) over this partition of [a; b] [c; d] is. f (r j ; t k ) x j y k"

Amy Stephens
5 years ago
Views:

1 The Double Integrl De nition of the Integrl Iterted integrls re used primrily s tool for omputing double integrls, where double integrl is n integrl of f (; y) over region : In this setion, we de ne double integrls nd begin emining how they re used in pplitions. To begin with, set of numbers f ; j ; r j g ; j = ; : : : ; m; is sid to be tgged prtition of [; b] if = < < < : : : < m = b nd if j r j j for ll j = ; :::; m: Moreover, if we let j = j j, then the prtition is sid to be h- ne if j h for ll j = ; : : : ; n: If f ; j ; r j g ; j = ; : : : ; m; is n h- ne tgged prtition of [; b] ; nd if fy ; y k ; t k g ; k = ; : : : ; n is l- ne tgged prtition of [; d] ; then the retngles [ j ; j ] [y k ; y k ] prtition the retngle [; b] [; d] nd the points (r j ; t k ) re inside the retngles [ j ; j ] [y k ; y k ] : The iemnn sum of funtion f (; y) over this prtition of [; b] [; d] is mx j= k= nx f (r j ; t k ) j y k We then de ne the double integrl of f (; y) over [; b] [; d] to be the limit s h; l pproh of iemnn sums over h; l ne prtitions: [;b][;d] f (; y) da = lim lim h! l! mx j= k= nx f (s j ; t k ) j y k To de ne the double integrl over bounded region other thn retngle,

2 we hoose retngle [; b] [; d] tht ontins ; nd we de ne g so tht g (; y) = f (; y) if (; y) is in nd g (; y) = otherwise. The double integrl of f (; y) over n rbitrry region is then de ned to be f (; y) da = g (; y) da [;b][;d] It then follows from the de nition tht the double integrl stis es the following properties: [f (; y) + g (; y)] da = f (; y) da + g (; y) da () [f (; y) g (; y)] da = f (; y) da g (; y) da () kf (; y) da = k f (; y) da (3) where k is onstnt. EXAMPLE Evlute the integrl of f + g over if f (; y) = 3 nd g (; y) = (4) Solution: We use property () to write [f (; y) + g (; y)] da = f (; y) da+ g (; y) da = 3+ = 5 Chek your eding: Wht is the integrl of f g over given (4)?

3 Volume If f (; y) on [; b] [; d] ; then the f (r j ; t k ) j y k is the volume of "bo" over retngle determined by the prtitions of [; b] nd [; d] ; respetively. Consequently, the iemnn sum is n pproimtion of the volume of the solid 3

under z = f (; y) nd over the retngle [; b] [; d] : Thus, if f (; y) over ; then the volume of the solid below z = f (; y) nd bove is V = f (; y) da It follows from the previous setion tht if is type

4 under z = f (; y) nd over the retngle [; b] [; d] : Thus, if f (; y) over ; then the volume of the solid below z = f (; y) nd bove is V = f (; y) da It follows from the previous setion tht if is type I region bounded by = ; = b; y = h () ; y = g () ; then f (; y) da = Z b Z g() h() f (; y) dyd nd if is type II region bounded by y = ; y = d; = q (y) ; = p (y), then f (; y) da = Z d Z p(y) q(y) f (; y) ddy EXAMPLE over the region Find the volume of the region below z = y nd : = y = = y = 4

Solution: Sine the region is type I region, we obtin V = y da = y dyd = = = 5 y d 4 d In generl, if f (; y) g (; y) over region ; then the volume of the solid between z = f (; y) nd z = g (; y) over

5 Solution: Sine the region is type I region, we obtin V = y da = y dyd = = = 5 y d 4 d In generl, if f (; y) g (; y) over region ; then the volume of the solid between z = f (; y) nd z = g (; y) over is V = [f (; y) g (; y)] da (5) If is type I or type II, then (5) n be evluted by reduing to either type I or type II integrl, respetively. EXAMPLE 3 Find the volume of the solid between z = + y nd z = y over the region : y = = y y = = y Solution: Aording to (5), the volume of the solid is V = (( + y) ( y)) da = y da 5

whih trnsforms into the type II iterted integrl V = Z y y y ddy Evluting the inside integrl results in V = It then follows tht V = yj y y dy = y y y y dy y y 3 dy = 6 Chek your eding: Wht type of

6 whih trnsforms into the type II iterted integrl V = Z y y y ddy Evluting the inside integrl results in V = It then follows tht V = yj y y dy = y y y y dy y y 3 dy = 6 Chek your eding: Wht type of region is the region given in emple 4? Converting Iterted Integrls into Di erent Type Mny regions n be desribed s either type I or type II. As result, type I integrl over suh region n be onverted into double integrl, whih n in turn be onverted into type II integrl. This llows us to evlute mny iterted integrls tht nnot be evluted diretly. EXAMPLE 4 Evlute the iterted integrl sin y dyd (6) Solution: Sine the ntiderivtive of sin y nnot be epressed in losed form, the iterted integrl (6) nnot be evluted s type I integrl. Insted, we onvert (6) to double integrl sin y dyd = sin y da nd notie tht the region between = ; = ; y = ; nd y = n lso be desribed s type II region. As Type I = y = = y = As Type II y = = y = = y 6

As result, we n rest the originl integrl s type II integrl, thus leding to sin y dyd = sin y Z y da = sin y ddy Not only did the desription of the region hnge, but lso the order of the di erentils

7 As result, we n rest the originl integrl s type II integrl, thus leding to sin y dyd = sin y Z y da = sin y ddy Not only did the desription of the region hnge, but lso the order of the di erentils hnged. Sine sin y is onstnt with respet to ; we now hve Z y sin y ddy = = sin y y dy y sin y dy The substitution u = y ; du = ydy; u () = ; u () = then results in sin y dyd = sin (u) du = EXAMPLE 5 Evlute the iterted integrl jyj sinh y 3 os ddy (7) Solution: The iterted integrl (7) nnot be evluted in losed form, so we insted onvert (7) to double integrl: sinh y 3 os ddy = sinh y 3 os da jyj The region of integrtion is both type I nd type II: As Type II y = = jyj y = = As Type I = y = = y = Consequently, when trnsformed into type I region we hve sinh y 3 os Z da = sinh y 3 os dyd = 7 os Z sinh y 3 dy d

The resulting integrl lso nnot be evluted in losed form, but beuse sinh y 3 is odd, we hve Z sinh y 3 dy = Thus, the entire integrl must be zero, whih mens tht jyj sinh y 3 os ddy = Chek your eding:

8 The resulting integrl lso nnot be evluted in losed form, but beuse sinh y 3 is odd, we hve Z sinh y 3 dy = Thus, the entire integrl must be zero, whih mens tht jyj sinh y 3 os ddy = Chek your eding: Why n (7) not be evluted in losed form? Fubini s Theorem nd Additionl Properties The de nition of the double integrl implies mny other properties. For emple, if f (; y) g (; y) on ; then f (; y) da g (; y) da nd likewise, if f (; y) on nd S ; then f (; y) da f (; y) da S Moreover, suppose tht nd S re non-overlpping regions i.e., tht nd S do not interset eept possibly on the boundry: Then s will be shown in the eerises, we must hve f (; y) da = f (; y) da + f (; y) da (8) [S S where [ S denotes the union of the regions nd S: 8

9 Finlly, properties of the double integrl lso follow from their reltionship to iterted integrls.. For emple, sine the retngle [; b] [; d] is both type I nd type II region, we must hve Z b Z d f (; y) da = f (; y) dyd nd f (; y) da = [;b][;d] [;b][;d] As result, the two iterted integrls re the sme. This result is known s Fubini s theorem, whih sys tht if ; b; nd d re onstnt nd if the double integrl of f (; y) eists, then Z b Z d f (; y) dyd = Z d Z b f (; y) ddy Tht is, the order of integrtion my be swithed if the limits of integrtion re onstnt. Z d Z b f (; y) ddy EXAMPLE 6 Use Fubini s theorem to evlute Z os () sin y dyd Solution: Fubini s theorem implies tht Z os () sin y dyd = Z As result, we integrte os () to obtin Z os () sin y dyd = = = sin y os () ddy sin y sin ()j sin y ( ) dy dy Eerises: Find the volume of the solid between the grphs of the given funtions over the given region:. f (; y) = y; g (; y) =. f (; y) = + y; g (; y) = = ; = ; y = ; y = = ; = ; y = ; y = 6 3. f (; y) = + y ; g (; y) = 4. f (; y) = 3 + y ; g (; y) = y = ; y = ; = y; = y = ; y = ; = y; = y 5. f (; y) = + y; g (; y) = + y 6. f (; y) = y; g (; y) = 4 = ; = ; y = ; y = y = ; y = ; = y; = 7. f (; y) = sin () ; g (; y) = ; 8. f (; y) = os ; g (; y) = ; = ; = ; y = ; y = = ; = ; y = ; y = 9

10 Evlute the iterted integrl by hnging it from type I to type II or vie vers: os y dyd. y y sin 3 ddy sin(y) y dyd. jyj sin y 3 ddy = sin () s (y) dyd 4. 4 e =py dyd py +y ddy 6. 4 e y 4 y dyd Evlute using Fubini s theorem sin (y) dyd 8. 3 ey dyd. sin y ddy. 3 3 sin y ddy sinh (y) ddy =4 tn (y) tn () ddy =4 Use the properties of the double integrls nd the double integrls f (; y) da = 5 f (; y) da = 7 g (; y) da = S to evlute the double integrls below: 3. 7f (; y) da [f (; y) + g (; y)] da [S f (; y) da [S g (; y) da S g (; y) da 3. [f (; y) g (; y)] da [f (; y) 3f (; y)] da 7f (; y) da [f (; y) + g (; y)] da [S [S S g (; y) da 3. Find the volume of the solid bound between the surfes z = + y nd z = 9: 3. Find the volume of the solid bound between the surfes z = +y nd z = : (hint: integrte over the region whose boundry urve is the intersetion of the two surfes). 33. Show tht for ll (; y) in [; ] [; ] tht nd then use this result to estimte sin () + os sin () (y) sin () + os (y) dyd 34. Let D denote the unit irle. Eplin why D e +y da nd then evlute this lst integrl. e +y dyd

11 35. Suppose tht f () over [; b] nd rell tht the surfe of revolution obtined by revolving the grph of f bout the -is is given by r (u; v) = hv; f (v) os (u) ; f (v) sin (u)i for u in [; ] nd v in [; b] : Show tht the volume of the resulting solid of revolution is Z b Z f() q [f ()] y dyd f() nd then ompute the innermost integrl using the trigonometri substitution y = f () sin () 36. Suppose tht f () > for ll in (; b) nd suppose tht f () = f (b) =. Wht is the volume of the solid enlosed by the surfe y + z = [f ()] 37. Use the iemnn de nition of the double integrl to prove (3). 38. Use the iemnn de nition of the double integrl to prove (). 39. Write to Lern: Suppose tht f (; y) is integrble over two bounded, non-overlpping regions nd S: Let g (; y) = f (; y) if (; y) is in nd g (; y) = if (; y) is not in : Similrly, let g (; y) = f (; y) if (; y) is in S nd let g (; y) = otherwise. Write short essy in whih you show tht f (; y) da = [g (; y) + g (; y)] da [S [;b][;d] where [; b] [; d] ontins [ S: Then in tht essy use this result to prove (8). 4. Write to Lern: Write short essy in whih you show tht Z b Z " d Z # " d Z # b f () g (y) ddy = f () d g (y) dy

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e Green s Theorem. Let be the boundry of the unit squre, y, oriented ounterlokwise, nd let F be the vetor field F, y e y +, 2 y. Find F d r. Solution. Let s write P, y e y + nd Q, y 2 y, so tht F P, Q. Let