Leture 16 Double integrls Dn Nihols nihols@mth.umss.edu MATH 233, Spring 218 University of Msshusetts Mrh 27, 218 (2) iemnn sums for funtions of one vrible Let f(x) on [, b]. We n estimte the re under the urve y f(x) on this intervl by doing iemnn sum: rete series of retngles with equl widths whose totl re pproximtes the re under the urve. Eh retngle hs width x b n So the right endpoints re x i + i x Height of the i-th retngle is f(x i ). Sum of res is n n f(x i ) x i1 x x 1 x 2 x 3 b This is right-endpoint iemnn sum with 3 retngles. If we used more retngles, we would get better estimte of the re. y x
(3) The definite integrl for funtions of one vrible The ext re under the urve y f(x) on [, b] is the definite integrl where x b n. ˆ b f(x) dx lim n n f(x i ) x, i1 This is the limit of sequene of iemnn sums s n (number of retngles) goes towrds infinity. We sy f(x) is integrble on [, b] if this limit exists. Continuous funtions re integrble, nd some disontinuous funtions too. We don t tully ompute definite integrls using this limit formul. We use the FToC to ompute extly or we use numeril methods to get n pproximtion. (4) The double integrl over retngle Multivrite version of this problem: ompute the volume under surfe z f(x, y) over losed retngle. Think of it s finding the volume of room. Surfe z f(x, y) is the roof of the room. (Assume z for now) Nottion for retngle in 2 : [, b] [, d] { (x, y) 2 : x b, y d } this defines the wlls of the room xy-plne (z ) is the floor
(5) The double integrl over retngle We ll pproximte the volume of the room by filling it up with retngulr boxes. Volume of the room is the sum of the volumes of the boxes. (6) The double integrl over retngle First we divide into eqully-sized retngles. Eh one will be the bse of 3D box. Divide intervl [, b] into m intervls of width x b m Divide intervl [, d] into n intervls of width y d n Totl: mn retngles. Eh hs re A x y. x d y b
(7) The double integrl over retngle Cll the i-th retngle in the j-th row ij. Lbel endpoints of retngles like this: x d x i + i x y j + j y So ij [x i 1, x i ] [y j 1, y j ]. Set height of box over ij to be f(x i, y j ) (vlue of f t top-right orner of retngle). Volume of box i, j is V ij f(x i, y j ) x y 1m. 12 y 11 2m. 22 21 x 1 x 2. y 3 b y 2 y 1 (8) The double integrl over retngle So double iemnn sum of f(x, y) on [, b] [, d] using m, n subdivisions is: m,n m n f(x i, y j ) A i1 j1 where A x y (b )(d ) mn is the re of the retngles. Definition The double integrl of f(x, y) over the retngle [, b] [, d] is f(x, y) da lim m lim n m i1 j1 n f(x i, y j ) A, where A (b )( d) mn, if this limit exists. If so, we sy f(x, y) is integrble on. We will lmost never do this mnully (without omputer). It s horrible!
(9) The double integrl over retngle Let f(x, y) be funtion tht is integrble on retngle. If f(x, y), then f(x, y) da is the volume of the solid under the surfe z f(x, y) on. If f(x, y) is sometimes positive nd sometimes negtive, then f(x, y) da V + V where V + is the volume under the positive prt of the surfe nd V is the volume over the negtive prt of the surfe. (1) The double integrl over retngle: exmple Exmple 1: Use double iemnn sums to estimte the vlue of the double integrl ( x 3 + xy y 2) da where [, 1] [, 2]. Compute iemnn sums for different vlues of m nd n nd ompre. Suppose we wnt to use n n n double iemnn sum. x 1/n nd y 2/n endpoints of the retngles re x i i x i/n nd y j j y 2j/n So the double iemnn sum is n,n n i1 j1 n i1 j1 n [ x 3 i + x i y j y j 1 2] n [ 1 n 3 i3 + 2 n 2 ij 4 n 2 j2 ].
(11) The double integrl over retngle: exmple Exmple 1: (ont.) Using omputer: 3,3 1.48148148 5,5 1.36 1,1 1.265 5,5 1.1866 1,1 1.17664999 5,5 1.168666 2,2 1.16716662 The tul vlue is ( x 3 + xy y 2) da 7 6 1.16666... (12) ules for double integrls ules for working with double integrls: 1. [f(x, y) + g(x, y)] da f(x, y) da + g(x, y) da 2. f(x, y) da f(x, y) da 3. 1 da re of 4. If we split up entirely into olletion of smller retngles 1, 2,..., k, then f(x, y) da f(x, y) da +... + f(x, y) da 1 k k f(x, y) da i i1 3 4 1 2
(13) Iterted integrls We tully ompute double integrls by writing them s iterted integrls. An iterted integrl looks like this: ˆ b ˆ d f(x, y) dy dx Or like this: ˆ b [ ˆ d ] f(x, y) dy dx. The thing in the middle is prtil integrl. (14) Prtil integrls Definition The prtil integrl of f(x, y) with respet to x on the intervl [, b] is written ˆ b f(x, y) dx. To ompute this, pretend y is onstnt nd integrte s if f were single-vrible funtion of x only. esult is funtion of y. eltion to prtil derivtives: prtil FToC ˆ t ˆ t f(x, y) dx f(t, y) t ( ) f(x, y) dx f(t, y) f(, y). x Similrly, the prtil integrl with respet to y on [, d] is d f(x, y) dy.
(15) Prtil integrls: exmple Exmple 2: Compute the prtil integrl 2 (x + y) dy. emember, tret x s onstnt nd pretend y is the only vrible. ˆ 2 (x + y) dy (xy + 12 ) 2 y2 (2x + 2) ( + ) 2x + 2. Wht is the mening of this funtion? If we plug in x, we get 2. How should we interpret this? (16) Iterted integrls nd volume Let S be the solid under the surfe z f(x, y) on [, b] [, d]. In the plne x x, f beomes single-vrible funtion f(x, y) on [, d]. The prtil integrl of f(x, y) with respet to y t x x is the integrl of this funtion: (ˆ d f(x, y) dy) xx ˆ d f(x, y) dy. This the re under the urve z f(x, y) on [, d] Or: re of the ross-setion of S in plne x x Importnt: When you ompute d f(x, y) dy nd plug in x x, you get the re of the ross-setion of the solid S in the plne x x.
(17) Iterted integrls nd volume Let S be the solid under the surfe z f(x, y) on [, b] [, d]. Define the ross-setion re s funtion of x: A(x) ˆ d f(x, y) dy. For eh x in [, b], A(x ) is the re of the ross setion of S in plne x x. emember in lulus II we lulted volume by integrting ross-setion re funtion... Let S be the solid under the surfe z f(x, y) on [, b] [, d]. (18) Iterted integrls We n ompute the volume of S by integrting the ross-setion re funtion: V ˆ b ˆ b A(x) dx [ ˆ d f(x, y) dy ] dx This is lled n iterted integrl In this se y is the inner vrible nd x is the outer vrible Integrte f(x, y) with respet to the inner vrible (y) to get funtion of just the outer vrible (x) Then integrte tht funtion of the outer vrible (x) to get the volume More omptly, we n leve out the brkets nd just write it s ˆ b ˆ d f(x, y) dy dx.
(19) Iterted integrls: exmple Exmple 3: Compute the iterted integrl: ˆ 2 emember, work from the inside out. ˆ 2 (x + y) dy dx (x + y) dy dx. [ˆ 2 ] (x + y) dy [ ( xy + 1 2 y2 [2x + 2 ] dx ( x 2 + 2x ) 1 3. dx ) ] 2 dx (2) Iterted integrls: exmple Exmple 4: Compute the iterted integrls. Notie tht the funtions nd the retngles re the sme, but the order of the vribles is different. ˆ 3 ( x 2 y + xe y) dy dx [ ( x 2 y + xe y) ] dy dx [ (1 ) ] 1 2 x2 y 2 + xe y dx [ ] 1 2 x2 + x(e 1) dx ( 1 6 x3 + e 1 ) 3 2 x2 ˆ 3 ˆ 3 ˆ 3 9e 2. ˆ 3 [ˆ 3 [ (1 ( x 2 y + xe y) dx dy ( x 2 y + xe y) ] dx dy 3 x3 y + 1 ) ] 3 2 x2 e y dy [9y + 92 ey ] ( 9 2 y2 + 9 ) 1 2 ey 9e 2. dy
(21) Iterted integrls: Fubini s Theorem We ve shown tht the volume under the surfe z f(x, y) on retngle (.k.. the double integrl of f on ) is equl to the iterted integrl of f on. More formlly: Theorem (Fubini s Theorem) Suppose f(x, y) is integrble on the retngle [, b] [, d]. Then f(x, y) da ˆ b ˆ d f(x, y) dy dx ˆ d ˆ b f(x, y) dx dy. We n ompute double integrl by writing it s n iterted integrl...... nd the order doesn t mtter. You n integrte x then y, or y then x. But mke sure to mth up the vribles nd endpoints orretly! f(x, y) dx dy is n integrl over different retngle. b d (22) Iterted integrls emember, lwys work from the inside out. To integrte ˆ b ˆ d f(x, y) dy dx, 1. Compute the prtil integrl of f(x, y) with respet to y on [, d], treting x like onstnt. The result is funtion g(x) whih doesn t depend on y. 2. Now integrte g(x) with respet to x on [, b] For some funtions, dy dx might be esier thn dx dy or vie vers.
(23) Iterted integrls: exmple Exmple 5: Integrte eh funtion over the given region using iterted integrtion. f(x, y) sin(x + y) f(x, y) ye x+y [, π/2] [, π] ˆ π ˆ π/2 ˆ π ˆ π sin(x + y) dx dy [ ] π/2 os(x + y) dy [os(y) os(π/2 + y)] dy π (sin y sin(π/2 + y)) 2. ˆ 3 (e 3 1) [, 3] [, 1] ye x+y dx dy [ ] 3 ye x+y dy [ ye 3+y ye y] dy [ye y ] dy 1 (e 3 1)(y 1)e y (e 3 1). (24) Integrtion tehniques We lerned mny integrtion tehniques in Cl II. Most importntly: Integrls nd derivtives of ommon funtions like ln x, se x, et. u-substitution Integrtion by prts: b u dv uv b b v du Prtil frtions See Chpters 5,6,7 of the textbook if you need to refresh your memory.
(25) Iterted integrls In speil ses where f(x, y) n be written like this: f(x, y) g(x)h(y) i.e. produt of funtion of x nd funtion of y, we n rewrite the iterted integrl s ˆ b ˆ d (ˆ b ) (ˆ d ) g(x)h(y) dy dx g(x) dx h(y) dy. Exmple 6: Compute the iterted integrl: (ˆ 2 ˆ 2 ˆ π y sin x dx dy. ) (ˆ π ) y dy sin x dx 2 2 4. (26) Volume between surfes Suppose f(x, y) g(x, y) on retngle. To find the volume of the solid bounded between f(x, y) nd g(x, y) on, integrte f(x, y) g(x, y). Exmple 7: Find the volume of the solid bounded below by the plne 2x y z 3, bove by the prboloid z x 2 + xy + y 2, nd bounded by the xz-plne, the yz-plne, nd the plnes x 1 nd y 1. Integrte roof minus floor : V (prboloid plne) da [ (x 2 + xy + y 2 ) (2x y 3) ] da Wlls re the plnes x, x 1, y, y 1... so [, 1] [, 1].
(27) Volume between surfes: exmple Exmple 7: (ont.) [ (x 2 + xy + y 2 ) (2x y 3) ] da, where [, 1] [, 1]. Write s n iterted integrl (in either order) nd ompute: [ (x 2 + xy + y 2 ) (2x y 3) ] dx dy [ ( x 2 + (y 2)x + y 2 + y + 3 ) ] dx dy [ (1 3 x3 + (y 2) 1 ) ] 1 2 x2 + (y 2 + y + 3)x dy [( 1 3 + 1 ) ] 2 (y 2) + y2 + y + 3 dy [ 7 3 + 3 ] ( 7 2 y + y2 dy 3 y + 3 4 y2 + 1 ) 1 3 y3 41 12. (28) In the future On Thursdy we ll lern how to integrte over ertin lmost-retngulr regions. Next week we ll lern how to integrte over some other types of regions using polr oordintes. At the end of the semester we ll lern Green s theorem, powerful tool for integrting over the region bounded by simple losed urve.
(29) Summry The double integrl f(x, y) da is the volume under the surfe z f(x, y) on the retngle. It is defined s the limit of sequene of double iemnn sums. Suppose [, b] [, d]. Fubini s theorem sys you n express double integrl over s n iterted integrl: f(x, y) da ˆ b ˆ d ˆ d ˆ b f(x, y) dy dx f(x, y) dx dy. To evlute n iterted integrl, do the inside first, then the outside. (3) Homework Pper homework #16 due Thursdy WebAssign homework 14.7, 14.8 due Wednesdy, 11:59 PM Midterm 2: next Thursdy, April 5th, 7-9 PM Setion 2: Hsbrouk 126 Setion 4: Hsbrouk 134 Covers everything up through Thursdy: 14.1-14.8 nd 15.1-15.3 Prtie problems hve been posted on the setion website Mth lub tomorrow t 5 PM in LGT 1634: fun tlk by Ben Orlin, who runs website lled mth with bd drwings. Free pizz!