CHAPTER 4 MULTIPLE INTEGRALS

Transcription

1 CHAPTE 4 MULTIPLE INTEGAL The objects of this chpter re five-fold. They re: (1 Discuss when sclr-vlued functions f cn be integrted over closed rectngulr boxes in n ; simply put, f is integrble over iff there is unique rel number, to be denoted I (f or f when it exists, which is cught between the upper nd lower sums reltive to ny prtition P of. ( how tht ny continuous function f cn be integrted over. (3 Discuss Fubini s theorem, which when pplicble, llows one to do multiple integrls s iterted integrls, i.e, integrte one vrible t time. (4 how tht bounded functions with negligible sets of discontinuities cn be integrted over. (5 Discuss integrls of continuous functions over generl compct sets. 4.1 Bsic notions We will first discuss the question of integrbility of bounded functions on closed rectngulr boxes, nd then move on to integrtion over slightly more generl regions. ecll tht in one vrible clculus, the integrl of function over n intervl [, b] ws defined s the limit, when it exists, of certin sums over finite prtitions P of [, b] s P becomes finer nd finer. To try to trnsport this ide to higher dimensions, we need to generlize the notions of prtition nd refinement. In this chpter, will lwys denote closed rectngulr box in n, written s [, b] = [, b 1 ] [ n, b n ], where j, b j, for ll j with j < b j. Definition. A prtition of is finite collection P of subrectngulr (closed boxes 1,,..., r such tht (i = r j=1 j, nd (ii the interiors of i nd j hve no intersection for ll i j. Definition. A refinement of prtition P = { j } r j=1 of is nother prtition P = { k }m k=1 with ech k contined in some j. It is cler from the definition tht given ny two prtitions P, P of, we cn find third prtition P which is simultneously refinement of P nd of P. Now let f be bounded function on, nd let P = { j } r j=1 prtition of. Then f is certinly bounded on ech j, i.e., f( j is bounded subset of. It ws proved in Chpter tht every bounded subset of dmits sup (lowest upper bound nd n inf (gretest lower bound. 1 Typeset by AM-TEX

2 CHAPTE 4 MULTIPLE INTEGAL Definition. The upper (resp. lower sum of f over reltive to the prtition P = { j } r j=1 is given by U(f, P = ( resp. L(f, P = vol( j sup(f( j j=1 j=1 vol( j inf(f( j. Here vol( j denotes the volume of j. Of course, we hve L(f, P U(f, P for ll P. More importntly, it is cler from the definition tht if P = { k }m k=1 of P, then L(f, P L(f, P nd U(f, P U(f, P. is refinement Put nd L(f = {L(f, P P prtition of } U(f = {U(f, P P prtition of }. Lemm. L(f dmits sup, denoted I(f, nd U(f dmits n inf, denoted I(f. Proof. Thnks to the discussion in Chpter, ll we hve to do is show tht L(f (resp. U(f is bounded from bove (resp. below. o we will be done if we show tht given ny two prtitions P, P of, we hve L(f, P U(f, P s then L(f will hve U(f, P s n upper bound nd U(f will hve L(f, P s lower bound. Choose third prtition P which refines P nd P simultneously. Then we hve L(f, P L(f, P U(f, P U(f, P. Done. Definition. A bounded function f on is integrble iff I(f = I(f. When such n equlity holds, we will simply write I(f (or I (f if the dependence on needs to be stressed for I(f (= I(f, nd cll it the integrl of f over. ometimes we will write I(f = f or f(x 1,..., x n dx 1... dx n. Clerly, when n = 1, we get the integrl we re fmilir with, often written s b 1 f(x 1 dx 1. This definition is hrd to understnd, nd useful criterion is given by the following

3 CHAPTE 4 MULTIPLE INTEGAL 3 Lemm. f is integrble over iff for every ε > we cn find prtition P of such tht U(f, P L(f, P < ε. Proof. If f is integrble, I (f is rbitrrily close to the sets of upper nd lower sums, nd we cn certinly find, given ny ε >, some P such tht I (f L(f, P < ε/ nd U(f, P I (f < ε/. Done in this direction. In the converse direction, since L(f, P I(f I(f U(f, P for ny P, if U(f, P L(f, P < ε, we must hve I(f I(f < ε. ince ε is n rbitrry positive number, I(f must equl I(f, i.e., f is integrble over. 4. tep functions The obvious question now is to sk if there re integrble functions. One such exmple is given by the constnt function f(x = c, for ll x. Then for ny prtition P = { j }, we hve L(f, P = U(f, P = c vol( j = c vol(. o I(f = I(f nd f = c vol(. This cn be jzzed up s follows. Definition. A step function on is function f on which is constnt on ech of the subrectngulr boxes j of some prtition P. j=1 Lemm. Every step function f on is integrble. Proof. By definition, there exists prtition P = { j } r j=1 of nd sclrs {c j} such tht f(x = c j, if x j. Then, rguing s bove, it is cler tht for ny refinement P of P, we hve L(f, P = U(f, P = c j vol( j. Hence, I(f = I(f. j=1 4.3 Integrbility of continuous functions The most importnt bounded functions on re continuous functions. (ecll from Chpter 1 tht every continuous function on compct set is bounded, nd tht is compct. The first result of this chpter is given by the following

4 4 CHAPTE 4 MULTIPLE INTEGAL Theorem. Every continuous function f on closed rectngulr box is integrble. Proof. Let be ny closed rectngulr box contined in. Define the spn of f on to be spn f ( = sup(f( inf(f(. A bsic result bout the spn of continuous functions is given by the following: The mll pn Theorem. For every ε >, there exists prtition P = { j } r j=1 of such tht spn f ( j < ε, for ech j r. Let us first see how this implies the integrbility of f over. ecll tht, by Lemm of section 4.1, it suffices to show tht, given ny ε >, there is prtition P of such tht U(f, P L(f, P < ε. Now by the smll spn theorem, we cn find prtition P = { j } such tht spn f ( j < ε, for ll j, where ε = ε/ vol(. Then clerly, Done. U(f, P L(f, P < ε vol( = ε. It now remins to supply proof of the smll spn theorem. We will prove this by contrdiction. uppose the theorem is flse. Then there exists ε > such tht, for every prtition P = { j } of, spn f ( j ε for some j. For simplicity of exposition, we will only tret the cse of rectngle = [, b 1 ] [, b ] in. The generl cse is very similr, nd cn be esily crried out long the sme lines with bit of bookkeeping. Divide into four rectngles by subdividing long the bisectors of [, b 1 ] nd [, b ]. Then for one of these four rectngles, cll it 1, we must hve tht for every prtition { j } of 1 there is j so tht spn f ( j ε. Do this gin nd gin, nd we finlly end up with n infinite sequence of nested closed rectngles =, 1,,..., such tht, for every m, the spn of f is t lest ε for ny prtition of P m = { j,m } of m on some j,m. Let z m = (x m, y m denote the southwestern corner of m, for ech m. Then the sequence {z m } m is bounded, nd so we my find the lest upper bound (sup α (resp. β of x m (resp. y m. Put γ = [α, β]. Then γ s the northestern corner of is clerly n upper bound of the z m. ince f is continuous t γ, we cn find non-empty closed rectngulr subbox of contining γ such tht spn f ( < ε. But by construction m will hve to lie inside if m is lrge enough, sy for m m. This gives contrdiction to the spn of f being ε on some open set of every prtition of m. Thus the smll spn theorem holds for (f,. 4.4 Bounded functions with negligible discontinuities One is very often interested in being ble to integrte bounded functions over which re continuous except on very smll subset. To be precise, we sy tht subset Y of n hs content zero if, for every ε >, we cn find closed rectngulr boxes Q 1,..., Q m such tht (i Y m i=1 Q i, nd (ii m i=1 vol(q i < ε.

5 CHAPTE 4 MULTIPLE INTEGAL 5 Exmples. (1 A finite set of points in n hs content zero. (Proof is obvious! ( Any subset Y of which contins non-empty open intervl (, b does not hve content zero. Proof. It suffices to prove tht (, b hs non-zero content for < b in. uppose (, b is covered by finite union of closed intervls I i, 1 i m in. Then clerly, := m i=1 length(i i length(, b = b. o we cn never mke less thn b. (3 The line segment L = {x, < x < b 1 } in hs content zero. (Compring with (, we see tht the notion of content is very much dependent on wht the mbient spce is, nd not just on the set. Proof. For ny ε >, cover L by the single closed rectngle = { (x, y ε 1 x b 1, 4(b 1 y ε 4(b 1 }. Then vol( = (b 1 ε (b 1 = ε < ε, nd we re done. The third exmple leds one to sk if ny bounded curve in the plne hs content zero. The best result we cn prove here is the following Proposition. Let ϕ : [, b] be continuous function. Then the grph Γ of ϕ hs content zero. Proof. Note tht Γ = {(x, y x b, y = ϕ(x}. Let ε >. By the smll spn theorem, we cn find prtition = t < t 1 < < t r = b of [, b] such tht spn ϕ ([t i, t i ] <, for every i = 1,..., r. Thus the piece of Γ lying between ε (b x = t i nd x = t i cn be enclosed in closed rectngle i of re less thn ε(t i t i (b. Now consider the collection { i } 1 i r which covers Γ. Then we hve re( j < j=1 ε (b (t i t i = ε. i=1 Theorem. Let f be bounded function on which is continuous except on subset Y of content zero. Then f is integrble on. Proof. Let M > be such tht f(x M, for ll x. ince Y hs content zero, we cn find closed subrectngulr boxes 1,..., m of such tht (i Y m i=1 i, nd (ii m i=1 vol( i < ε 4M.

6 6 CHAPTE 4 MULTIPLE INTEGAL Extend { 1,..., m } to prtition P = { 1,..., r }, m < r, of. Applying the smll spn theorem, we my suppose tht m+1..., r re so chosen tht (for ech i m + 1 ε spn f ( i < vol(. (We cn pply this theorem becuse f is continuous outside the union of 1,..., m. o we hve U(f, P L(f, P M m vol( i + i=1 ( ε < (M + M i=m+1 spn f ( i vol( i ε vol( i=m+1 vol( i. But the right hnd side is ε + ε = ε, becuse r i=m+1 vol( i vol(. Exmple. Let = [, 1] [, 1] be the unit squre in, nd f : the function defined by f(x, y = x + y if x y nd x y if x y. how tht f is integrble on. Let D = {(x, x x 1} be the digonl in. Then D hs content zero s it is the grph of the continuous function ϕ(x = x, x 1. Moreover, f is discontinuous only on D. o f is continuous on D with D of content zero, nd consequently f is integrble on. emrk. We cn use this theorem to define the integrl of continuous function f on ny closed bounded set B in n if the boundry of B hs content zero. Indeed, in such cse, we my enclose B in closed rectngulr box nd define function f on by mking it equl f on B nd on B. Then f will be continuous on ll of except for the boundry of B, which hs content zero. o f is integrble on. ince f is outside B, it is resonble to set f = f. B It is often useful to consider finer notion thn content, clled mesure. Before giving this definition recll tht set X is countble iff there is bijection (or s some would sy, one-to-one correspondence, between X nd subset of the set N of nturl numbers. Check tht Z nd Q re countble, while is not. A subset Y of n is sid to hve mesure zero if, for every ε >, we cn find countble collection of closed rectngulr boxes Q 1, Q,..., Q m,... such tht (i Y i 1 Q i, nd (ii i 1 vol(q i < ε. One cn use open rectngulr boxes insted of closed ones, nd the resulting definition will be equivlent. Exmples. (1 A countble set of points in n hs mesure zero. ( Any subset Y of which contins non-empty open intervl (, b does not hve mesure zero.

7 CHAPTE 4 MULTIPLE INTEGAL 7 (3 A countble union of lines in [, 1] [, 1] hs mesure zero. We will stte the following result without proof: Theorem. Let be closed rectngulr box in n, nd f bounded function on which is continuous except on subset Y of mesure zero. Then f is integrble on. 4.5 Fubini s theorem o fr we hve been meticulous in figuring out when given bounded function f is integrble on. But if f is integrble, we hve developed no method whtsoever to ctully find wy to integrte it except in the relly esy cse of step function. We propose to melorite the sitution now by describing very resonble nd computtionlly helpful result. We will stte it in the plne, but there is nturl nlog in higher dimensions s well. In ny cse, mny of the intriccies of multiple integrtion re present lredy for n =, nd it is wise ide to understnd this cse completely t first. Theorem (Fubini. Let f be bounded, integrble function on = [, b 1 ] [, b ]. For x in [, b 1 ], put A(x = b f(x, y dy nd ssume the following (i A(x exists for ech x [, b 1 ], i.e., the function y f(x, y is integrble on [, b ] for ny fixed x in [, b 1 ]; (ii A(x is integrble on [, b 1 ]. Then f(x, y dxdy = b1 ( b f(x, y dy dx. In other words, once the hypotheses (i nd (ii re stisfied, we cn compute f by performing two 1-dimensionl integrls in order. One cnnot lwys reverse the order of integrtion, however, nd if one wnts to integrte over x first, one needs to ssume the obvious nlog of the conditions (i, (ii. Proof. Let P 1 = {B i 1 i l} (resp. P = {C j 1 j m} be prtition of [, b 1 ] (resp. [, b ], with B i, C j closed intervls in. Then P = P 1 P = {B i C j } is prtition of. By hypothesis (i, we hve L(f x, P A(x U(f x, P, where f x is the one-dimensionl function y f(x, y. Then pplying hypothesis (ii, we get L(L(f x, P, P 1 b1 A(x dx U(U(f x, P, P 1.

8 8 CHAPTE 4 MULTIPLE INTEGAL But we hve L(L(f x, P, P 1 = = l length(b i inf(l(f x, P (B i i=1 l m length(b i length(c j inf(f(b i C j = L(f, P. i=1 j=1 imilrly for the upper sum. Hence L(f, P b 1 A(x dx U(f, P. Given ny prtition Q of, we cn find prtition P of the form P 1 P which refines Q. Thus L(f, Q b 1 A(x dx U(f, Q, for every prtition Q of. Then by the uniqueness of f, which exists becuse f is integrble, b 1 A(x dx is forced to be f. emrk. The reson we denote b f(x, y dy by A(x is the following. The double integrl f(x, y dxdy is the volume subtended by the grph Γ = {(x, y, f(x, y 3 } over the rectngle. (Note tht Γ is surfce since f is function of two vribles. When we fix x t the sme point x in [, b 1 ], the intersection of the plne {x = x } with Γ in 3 is curve, which is none other thn the grph Γ x of f x in the (y, z-plne shifted to x = x. The re under Γ x over the intervl [, b ] is just b f x (y dy; whence the nme A(x. Note lso tht s x goes from to b 1, the whole volume is swept by the slice of re A(x. A nturl question to sk t this point is whether the hypotheses (i, (ii of Fubini s theorem re stisfied by mny functions. The nswer is yes, nd the prime exmples re continuous functions. Theorem. Let f be continuous function on = [, b 1 ] [, b ]. Then f cn be computed s n iterted integrl in either order. To be precise, we hve f(x, y dxdy = b1 [ b ] f(x, y dy dx = b [ b1 ] f(x, y dx dy. Proof. ince f is continuous on (the compct set, it is certinly bounded. Let M > be such tht f(x, y M. We hve lso seen tht it is integrble. For ech x, the function y f(x, y is integrble on [, b ] becuse of continuity on [, b ]. o we get hypothesis (i of Fubini. To get hypothesis (ii, it suffices to show tht A(x = b f(x, y dy is continuous in x. For h smll, we hve A(x + h A(x = b b (f(x + h, y f(x, y dy f(x + h, y f(x, y dy. By the smll spn theorem we cn find prtition { j } of with spn f ( j < ε/(b. If h is smll enough so tht (x + h, y nd (x, y lie in the sme box for ll y

9 CHAPTE 4 MULTIPLE INTEGAL 9 (which we cn chieve since x is fixed nd there re only finitely mny boxes we hve f(x+h, y f(x, y < spn f ( j < ε/(b where j is box contining both points. Note tht this rgument lso works if (x, y lies on the verticl boundry between two boxes: for positive h we lnd in one box nd for negtive h in the other. Hence b f(x + h, y f(x, y dy < ε for h sufficiently smll. This shows tht A(x is continuous nd hence integrble on [, b 1 ]. We hve now verified both hypotheses of Fubini, nd hence f(x, y dxdy = b1 [ b ] f(x, y dy dx. To prove tht f is lso computble using the itertion in reverse order, ll we hve to do is note tht by symmetricl rgument, the integrl b 1 f(x, y dx mkes sense nd is continuous in y, hence integrble on [, b ]. The Fubini rgument then goes through. emrk. We will note the following extension of the theorem bove without proof. Let f be continuous function on closed rectngulr box = [, b 1 ] [ n, b n ]. Then the integrl of f over is computble s n iterted integrl b1 [ [ bn [ bn n n f(x 1,..., x n dx n ] ] ] dx n... dx 1. Moreover, we cn compute this in ny order we wnt, e.g., integrte over x first, then over x 5, then over x 1, etc. Note tht there re n! possible wys here of permuting the order of integrtion. 4.6 Integrtion over specil regions Let Z be compct set in n. ince it is bounded, we my enclose it in closed rectngulr box. If f is bounded function on Z, we my define n extension f to by setting f(x to be f(x (resp. for x in Z (resp. in Z. Let us sy tht f is integrble over Z if f is integrble over, nd put f = f. Z It is cler tht this definition is independent of the choice of. In 4.4, where we introduced the notion of content, we remrked tht if the boundry of Z hd content zero nd if f is continuous, then f would be integrble on. The sme ide esily proves the following

10 1 CHAPTE 4 MULTIPLE INTEGAL Theorem. Let Z be compct subset of n such tht the boundry of Z hs content zero. Then ny function f on Z which is continuous on Z is integrble over Z. In fct, one cn replce content by mesure in this Theorem. Now we will nlyze the simplest cses of this phenomenon in. Definition. A region of type I in is set of the form { x b, ϕ 1 (x y ϕ (x}, where ϕ 1, ϕ re continuous functions on [, b]. A region of type II in is set of the form {c y d, ψ 1 (y x ψ (y}, where ψ 1, ψ re continuous functions on [c, d]. A region of type III in is subset which is simultneously of type I nd type II. emrk. Note tht circulr region is of type III. Theorem. Let f be continuous function on subset of. ( uppose is region of type I defined by x b, ϕ 1 (x y ϕ (x, with ϕ 1, ϕ continuous. Then f is integrble on nd f = b ( ϕ (x ϕ 1 (x f(x, y dy dx. (b uppose is region of type II defined by c y d, ψ 1 (y x ψ (y, with ψ 1, ψ continuous. Then f is integrble on nd f = d c ( ψ (y ψ 1 (x f(x, y dx dy. Proof. We will prove ( nd leve the symmetricl cse (b to the reder. ( Let = [, b] [c, d], where c, d re chosen so tht contins. Define f on s bove (by extension of f by zero outside. By the Proposition of 4.4, we know tht the grphs of ϕ 1 nd ϕ re of content zero, since ϕ 1, ϕ re continuous. Thus the min theorem of 4.4 implies tht f is integrble on s its set of discontinuities is contined in the boundry of. It remins to prove tht f (= f is given by the iterted integrl b ( ϕ (x ϕ 1 (x f(x, y dy dx. For ech x (, b, the integrl d f(x, y dy c exists s the set of discontinuities in [c, d] hs t most two points. Moreover, the function x d f(x, y, dy is integrble on [, b]. Hence (the proof of Fubini s theorem pplies c in this context nd gives f = b ( d c f(x, y dy dx. ince the inside integrl (over y is none other thn ϕ (x ϕ 1 (x theorem follows. f(x, y dy, the ssertion of the

11 CHAPTE 4 MULTIPLE INTEGAL Exmples (1 Compute f, where is the closed rectngle [, 1] [, 3] nd f the function (x, y x y x cos πy. ince f is continuous on, we my pply Fubini s theorem nd compute I s the iterted integrl 1 ( 3 I = (x y x cos πy dy dx. ecll tht in 3 (x y x cos πy dy, x is treted like constnt, hence equls 3 3 ( ( ] 3 x y dy x cos πy dy = x 3 sin πy x = 5 π x. I = 5 1 x dx = 5 ( ] x 3 1 = We could lso hve computed it in the opposite order to get 3 [ 1 ] I = (x y x cos πy dx dy 3 ( ] x 3 1 ] 1 = = 3 ( y ( y 3 3 dy = y 3 ( x cos πy ] 3 = 5 3. dy ( Find the volume of the tetrhedron T in 3 bounded by the plnes x =, y =, z = nd x y z =. Note first tht the bse of T is tringle defined by x, y x + 1. Given ny (x, y in, the height of T bove it is simply given by z = x y + 1. Hence we get by the Theorem of 4.6, ( x+1 vol(t = (x y + 1 dxdy = (x y + 1 dy dx = = (xy y + y (x + 1 dx = x ] x+1 1 dx u du = 1 6. (3 Fix, b >, nd consider the region inside the ellipse defined by x + y b. Compute I = x dxdy. = 1 in

12 1 CHAPTE 4 MULTIPLE INTEGAL Note tht is region of type I s we my write it s { x, b 1 x y b 1 x ince the function (x, y x is continuous, we cn pply the min theorem of 4.6. We obtin ( b 1 x x I = = b = 4 b 4 b 3 b 1 x ( x dx = b = 8 b 3. dy dx ( x x3 3 ] }. 4.8 Applictions Let be thin plte in with mtter distributed with density f(x, y (= mss/unit re. The mss of is given by m( = f(x, y dxdy. The verge density is m( re = f(x, y dxdy dxdy. The center of mss of is given by z = ( x, ȳ, where x = 1 x f(x, y dxdy m( nd ȳ = 1 y f(x, y dxdy. m( When the density is constnt, the center of mss is clled the centroid of. uppose L is fixed line. For ny point (x, y on, let δ = δ(x, y denote the (perpendiculr distnce from (x, y to L. The moment of inerti bout L is given by I L = δ (x, y f(x, y dxdy. When L is the x-xis (resp. y-xis, it is customry to write I x (resp. I y. Note tht the center of mss is liner invrint, while the moment of inerti is qudrtic. An interesting use of the centroid occurs in the computtion of volumes of revolutions. To be precise we hve the following

13 CHAPTE 4 MULTIPLE INTEGAL 13 Theorem (Pppus. Let be region of type I, i.e., given s { x b, ϕ 1 (x y ϕ (x}, with ϕ 1, ϕ continuous. uppose tht min x ϕ 1 (x >, so tht lies bove the x-xis. Denote by V the volume of the solid M obtined by revolving bout the x-xis, nd by z = ( x, ȳ the centroid of. Then where ( is the re of. V = πȳ (, Proof. Let V i denote the volume of the solid obtined by revolving {(x, ϕ 1 (x x b} bout the x-xis. Then b V i = π ϕ i (x dx. (This is result from one-vrible clculus. But clerly, V = V V 1. o we hve V = π b [ϕ (x ϕ 1 (x ] dx. On the other hnd, we hve by the definition of the centroid, ȳ = 1 y dxdy. ( ince y is continuous nd region of type I, we hve ȳ = 1 b ( ϕ (x y dy dx ( = 1 ( b The theorem now follows immeditely. ϕ 1 (x 1 [ϕ (x ϕ 1 (x ] dx. Exmples. (1 Let be the semi-circulr region { x 1, y 1 x }. Compute the centroid of. ince is of type I, we hve 1 1 x ( = dxdy = dx dy 1 1 = 1 x dx = 1 x dx. Put x = sin t, t π. Then dx = cos t dt nd 1 x = cos t. o we get π π ( ( = cos 1 + cos t t dt = dt [ ] π ] π sin t = + = π 4 4.

14 14 CHAPTE 4 MULTIPLE INTEGAL Of course, we could hve directly resoned by geometry tht the re of semi-circulr region of rdius 1 is π. Let z = ( x, ȳ be the centroid. x = 1 x dxdy = ( π = π 1 1 x 1 x dx = π ( 1 x π π dy x dx sin t cos t dt =, since the integrnd is n odd function. Agin, the fct tht x = cn be directly seen by geometry. The key thing is to compute ȳ. We hve ȳ = π = 1 π 1 (x x3 3 o the centroid of is (, 4 3π. ( 1 x dx ] 1 y dy = 1 1 (1 x dx π = π 3π = 4 3π. ( Find the volume V of the torus π obtined by revolving bout the x-xis circulr region of rdius r (lying bove the x-xis. The re ( is πr, nd the centroid ( x, ȳ is locted t the center of (esy check!. Let b be the distnce from the center of to the x-xis. Then by Pppus theorem, V = πȳ ( = πb(πr = π r b.