Mathematical Analysis II - Part B
2 Integral calculus in several variables
In this section we approach the problem of integrating real functions of several real varia- bles. The knowledge of Riemann integral for functions of one real variable is required.
2.1 Lebesgue versus Riemann integral: motivation and a brief presentation
A theory of Riemann integration for functions of several real variables (i.e. f : A R, ! with A Rn) could be constructed by adapting in a natural way the already known theory ⇢ for one variable: the role of intervals of R will be taken by the “n-dimensional intervals” I = I1 In,wheretheIj are intervals of R with extremities included or not; there ⇥···⇥ will still be the simple functions (or “step functions”, “staircase functions”) from which to start the definition of integral, the upper and lower sums, and so on.
Figure 2.1: The integral of a function of two real variables.
The problem of defining the integral is substantially equivalent to the one of providing a measure to subsets of Rn: one would say that E Rn is elementarily measurabile ⇢ (or measurable `ala Peano-Jordan) when its indicator function E is Riemann-integrable, calling “measure of E” (or also “area”, “volume” in the cases n =2, 3) the value of such integral. Obvious good qualities of this theory are its simplicity and naturality; however, the drawbacks already observed in the case of one variable are still present. Let us mention three of them. Riemann-integrable functions (and, in parallel, elementarily measurable subsets) are • relatively few when one has to enlarge its scope beyond continuous functions, or rat- her beyond the functions having a “reasonable number of singular points”.(13) Na-
(13)This is what happens since more than one century in the progress of exact sciences, as new important
Corrado Marastoni 29 Mathematical Analysis II - Part B
mely, on a compact n-dimensional interval a Riemann-integrable function must be bounded, and continuous “almost everywhere”:(14) for example, this fact says that the Dirichlet function (which is discontinuous in all points [0, 1]) cannot be Q [0,1] Riemann-integrable. \ The Riemann integral has no satisfactory interaction with the limit.Givenasequence • of integrable functions fn which converges to a function f in a su ciently weak way (e.g. pointwise, or even less), we would like also f to be integrable and that it could be possible to “pass to limit under the sign of integral”, i.e. that f =lim fn. Now, in such a generality, this request is rather optimistic, because a weak convergence cannot R R control “mass dispersion” phenomena in the domain, as the following example shows.
n n (n 1) Example. Let us define fn : R R as 2 on the interval [2 , 2 ], and 0 elsewhere. Any of ! the fn is Riemann-integrable with fn = 1, and the sequence (fn) converges pointwise to the zero
function f 0, which is obviously Riemann-integrable as well; but f =0=lim fn =1. ⌘ R 6 R R To prevent such phenomena, a reasonable request is that this should be true at least for sequences “dominated” by a fixed integrable function, i.e. such that there exists an integrable function g such that fn g for any n N. However, for the Riemann | | 2 integral also this is not enough: in fact the convergence should be uniform, because pointwise convergence could not work.
Example. Let q : N Q be any bijection (we know that Q is countable), and define fn : R R ! ! as 1 onlu on the rationals qm with m n, and 0 elsewhere: any of the fn is Riemann-integrable (it has finitely many discontinuity points) with fn =0,butthesequence(fn) converges pointwise (not
uniformly!) to Dirichlet function f = [0,1] , which is not Riemann-integrable. Q\ R
The Riemann integral is linked to the properties of real numbers. Riemann’s costruction • depends a lot on the properties of real numbers (in particular on the total order), and the extendibility to the case of real functions defined on more general sets than Rn appears to be di cult.
These reasons, among some others, suggested at the end of XIXth century to look for a more e↵ective integration theory, being prepared to pay a price in terms of simplicity but without loosing too much of its naturality: among various attempts, the one by Henri tools (for example, Fourier series) could generate highly irregular functions. (14)The precise meaning of this “almost everywhere” is —as we shall see later— “everywhere excepted a null set”, where “null set” means a set of points whose Lebesgue measure is zero: the exact statement is n that a bounded function with compact support f : R R is Riemann-integrable if and only if the set of its ! discontinuity points is a null set. The Lebesgue measure, as we shall see, is much larger than the elementary measure of Peano-Jordan. For example, all countable sets are Lebesgue-measurable (with measure zero), 1 while they could be not elementarily measurable: for example E = n : n N is elementarily measurable 1 1 { 12 } (if sn : R R is the simple function with values 1 on [0, n ] n 1 ,..., 2 , 1 and zero elsewhere, it holds ! 1 [{ } E sn and hence sn = is a upper sum for E ), while the Dirichlet set Q [0, 1] is not. A classical R n \ example of Riemann-integrable function whose discontinuity set is Lebesgue-null but not elementarily R p measurable is f :[0, 1] R defined as 0 on [0, 1] Q and as 1 on Q [0, 1] with p, q 0 coprime, ! \ q q 2 \ whose discontinuity set is the Dirichlet set Q [0, 1]. \
Corrado Marastoni 30 Mathematical Analysis II - Part B
Lebesgue’s (1875-1941) eventually turned out to be the most successful. Let us discuss why by making reference to the just mentioned Riemann’s drawbacks. When compared to Riemann theory, Lebesgue’s considerably enlarges the family of • measurable sets (and, in parallel, the family of measurable functions, possibly integra- ble), so that it is very complicated to describe non Lebesgue-measurable subsets of Rn; in particular, the Dirichlet function is Lebesgue-integrable with integral zero. Q [0,1] The role played by continuity in Riemann\ integral is played here by the property of “measurability” which is extremely weak and in practice satified all the time. Lebes- gue’s construction, which nowadays keeps having a large agreement(15), is in a certain sense the result of a process of completion of Riemann’s, similar to the construction of real numbers starting from rationals. The Lebesgue integral works much better under limit with respect to Riemann’s: if • we have a sequence of Lebesgue-integrable functions fn which converges pointwise “almost everywhere”(16) to a function f, in two very frequent cases (a monotone or a dominated sequence) also f is Lebesgue-integrable and f =lim fn. In particular
the example with f = [0,1] is fixed (the sequence fn converges monotonically to f, Q\ R R which is Lebesgue-integrable and 0 = =lim f ), while the example with Q [0,1] n the “mass dispersion” keeps resisting (actually\ the sequence f converges pointwise R R n to f = 0 but neither in a monotone nor in a dominated way). Lebesgue theory is largely independent from particular properties (topological or not) • of the domain, and hence it naturally suits to real functions defined on any domain: in fact, to present the theory in full generality does not create more di culties with respect to Rn (on the contrary, it rather helps to better clarify the situation). Anyway, we should point out that a large part of the results which are valid for • Lebesgue theory are valid also for Riemann’s: but usually the hypotheses in the latter are some heavier (and, after all, quite unessential).
For what has been said, it should appear convenient to make reference to Lebesgue integral from now on. Let us briefly recall the main aspects of this theory, without proofs.
Let X be any set, and let us plan to introduce a measure µ of the subsets of X: in other words, to a A X we want to associate a measure µ(A), that we expect to be a number ⇢ 0, possibly + . How can we do? The first requirement would be, at one side, to be able 1 to measure as many subsets of X as possible; but, on the other side, the measure should satisfy some natural properties, that we now list.
(15)We should however say that, while Lebesgue theory fixes many problems, some unexpected others sin x appear: for example x is not Lebesgue-integrable on R although it has a generalized Riemann-integral (called “Dirichlet integral”); in fact it is not absolutely Riemann-integrable. (16)i.e., as said above, “excepted a null set”: it is clearly a rather weak notion of convergence.
Corrado Marastoni 31 Mathematical Analysis II - Part B
– Enough stability for the set-theoretical operations . We aim that the family of µ- measurable subsets, beyond than being rich of elements, would be stable under the main set-theoretical operations, in particular under the union and the intersection (even better, under countable union and intersection) and under complementation in X. In other words: if A, B X are µ-measurable, we aim that also A B, A B and ⇢ [ \ {X A = X A be so; and that, if An is a countable family of µ-measurable subsets, \ (17) also An and An be so. – Isotony .IfA B, it would be natural that µ(A) µ(B). T ⇢S – Additivity . The measure should be additive on disjoint unions, better on countable ones. In other words, if A, B X are µ-measurable and disjoint, it should hold that ⇢ µ(A B)=µ(A)+µ(B); moreover, even better, if A is a countable family of µ- [ n measurable pairwise disjoint subsets, then µ( An)= µ(An). S P In fact, Lebesgue theory is developed in full generality (and without problems, rather with a better clarity and language) in the framework of a measure space (X, ,µ), where X Measure space M is any set, a family of parts of X which contains the same X and is stable under com- M plementation and countable union,(18) and µ : [0, + ] is a function non identically M ! 1 equal to + and countably additive, i.e. such that given a countable family An of 1 (19) 2M pairwise disjoint subsets of X it holds µ( An)= µ(An). We shall say that µ is a (positive) measure on X, and the elements of (called a -algebra of parts of X)willbe Measure S M P called the µ-measurable subsets of X. Measurable sets
Examples. (0) (Measure zero) The first (trivial) example is the measure identically 0. (1) (Measure which counts the elements) Choose = P (X), and for A X define µ(A)asthenumberofelementsofA (set M ⇢ µ(A)=+ if A is not finite). (2) (Measure concentrated in a point) Choose = P (X); fixed an element 1 M x0 X, for A X define µ(A)=0ifx0 / A and µ(A)=1ifx0 A. 2 ⇢ 2 2
From now on let us concentrate on X = Rn, with the aim of constructing the Lebesgue n measure n on it, without forgetting that on R we already have Riemann’s elementary measure to be preserved as far as possible. Hence, in this particular framework, to the previous desired properties we should also add the following familiar ones. – Extension of elementary measure . We aim the elementarily measurable subsets of Rn be also n-measurable, and that the two measures coincide on them. In particular, n for a bounded interval I = j=1 ]aj,bj[ (with or without extremities) it should be n n(I)= (bj aj). j=1 Q – Invariance under isometries . The measure of A Rn should not change by applying Q ⇢ (17)The measurability of countable unions would be a first evident progress with respect to Riemann: recall that the single points are elementarily measurable (with measure zero), but their countable union Q [0, 1] (the Dirichlet set) is not. \ (18)As a consequence, there is stability also under countable intersection (by de Morgan duality) and obviously also under finite intersection; but also under di↵erence, because A B = A (X B). (19) \ \ \ It is then easy to show that one also has µ(?)=0,andthatalsothefiniteadditivity(i.e.ifA1,...,An n 2 are pairwise disjoint then µ(A1 An)= µ(Aj )) and isotony (i.e. if A, B with A B M [···[ j=1 2M ⇢ then µ(A) µ(B)). P
Corrado Marastoni 32 Mathematical Analysis II - Part B
to A any isometry of Rn, i.e. a function of Rn into itself which preserves the distances (e.g. the translation by a fixed vector).
Let us start the construction of n (in particular, of the -algebra on which it is defined). n The first step is to define n on the intervals: if I = j=1 ]aj,bj[ (possibly half open or n also closed; and possibly degenerated), we set n(I):= (bj aj) , as expected. The Q j=1 second is, then, to define a quantity (called external measure) which makes sense for External measure n⇤ Q any subset: given A Rn,weset ⇢ + + 1 n 1 n⇤ (A):=inf n(Ik): (Ik)k a sequence of intervals of R such that A Ik . R 2N ⇢ ⇢ k=0 k=0 P S In other words,e we consider the various at most countable covers of A made by n-dimen- sional intervals, compute the sum of the measures of the intervals of the cover, and then take the infimum of such sums among the various covers.(20) n Proposition 2.1.1. The external measure ⇤ : P (R ) [0, + ] : n ! 1 (a) respects the measure of the intervals (i.e. if I is an interval then n⇤ (I)= n(I));
(b) it holds n⇤ (?)=0; (c) is isotone (i.e. if A B then (A) (B)); ⇢ n⇤ n⇤ (d) is countably subadditive (i.e.: given (Ak)k it holds n⇤ ( Ak) n⇤ (Ak)). 2N Unfortunately the external measure n⇤ is not countably additive,S becauseP there exist (but we shall not prove it) some very technical examples of sequences (Ak) of pairwise disjoint subsets for which n⇤ ( Ak) < n⇤ (Ak). However, since it is unreasonable to give up this n property, the idea is to restrict ⇤ to a selected family of subsets of R where everything S P n works well. It turns out that this selected family is very large and satisfactory, as the following result shows. n Theorem - Definition 2.1.2. (Carath´eodory) Let n be the family of subsets of R M which “decompose additively the external measure”:
n n n = A R : ⇤ (E)= ⇤ (E A)+ ⇤ (E A) for any E R . M { ⇢ n n \ n \ ⇢ } Then: (a) is a -algebra containing all intervals, all subsets having zero external measure Mn (among them, the finite or countable ones), and in general all open and all closed euclidean subsets of Rn;
(b) the restriction of n⇤ to n is a measure, in particular is countably additive (i.e.: M n given in n a countable family (Ak)k of pairwise disjoint subsets of R it holds M 2N n⇤ ( Ak)= n⇤ (Ak)). Such measure extends Peano-Jordan’s, and is invariant under isometries (in particular under translations).(21) S P Lebesgue- measurable (20)Note that it is not necessary to require that the covers should be formed by pairwise disjoint intervals (we could also require it: in any case, after taking the infimum, the result would not change). (21)For the notion of isometry see also at p. 42.
Corrado Marastoni 33 Mathematical Analysis II - Part B
n The elements of n are called the Lebesgue-measurable subsets of R , and the restriction M n of ⇤ to n (coherently denoted by n)iscalledLebesgue measure on R . Lebesgue measure n M
n The family n of Lebesgue-measurable subsets of R is very large, so large that it is really M complicated to construct non measurable subsets: in practice, measurability is always verified.
Of particular importance is the role played by the subsets of Rn which are measurable with measure zero, called (Lebesgue-)null sets (or n-null sets). A property P(x) of the Null sets n points x of a subset E R is said to be true almost everywhere (often shortened into Almost ⇢ everywhere (a.e.) a.e.)inE when the set x E :P(x) is false is a null set. { 2 }
Example. A function f : E R is “continuous a.e. in E ”ifthesetofdiscontinuitypointsinE is a null ! set: so are e.g. the functions “entire part” and “fractional part” in R (the discontinuity points are those of Z,whichisanullsetinR). Proposition 2.1.3. The null sets have the following properties. (a) A countable union of null sets is a null set. (b) A subset of a null set is a null set. (c) The proper (i.e. of dimension n 1) submanifolds of Rn are null sets in Rn, as well as their finite or countable unions.
Examples. (1) AsinglepointisnullinR, and the same holds for the countable sets (as Z,orQ,orthe 2 Dirichlet set Q [0, 1]). (2) In R the single points and the regular curves are null, as well as their finite \ 2 or countable unions, are null. (3) In R the single points, the regular curves and the regular surfaces are null, as well as their finite or countable unions.
n A function f : R R is said to be measurable if the “over-level sets” Measurable function ! 1 n e f (]↵, + ]) = x R : f(x) >↵ 1 { 2 } are measurable for any ↵ R. 2 Since the family of measurable subset is very large, also the family of measurable functions is very large: in practice all functions are measurable. The following proposition shows that this property is stable enough.
Proposition 2.1.4. Measurable function enjoy the following properties. (a) The definition of measurability could be modified, without changing the class of functi- ons enjoying it, by substituting > with <, or , or also by requiring that the inverse image of any open subset (or of any closed subset) of R be measurable.
Corrado Marastoni 34 Mathematical Analysis II - Part B
(b) If f is measurable, also f , f + and f are so.(22) | | (c) If f and g are measurable, also f + g and fg are measurable.
(d) If fn is a sequence of measurable functions which converges pointwise to a function f, then also f is measurable.
The Lebesgue integral of a measurable function f : Rn R on a measurable set E Rn ! ⇢ is defined as follows. e If f : Rn R is measurable and 0 on E,thenthetrapezoid (see Figure 2.2(1)) • ! Trap (f)= (x, t) E R :0 t f(x) e E { 2 ⇥ } is a measurable subset of Rn+1, and the integral is defined in the following natural way (the value could be + ): 1
f(x) d n := n+1(TrapE (f)) . ZE
This is a simplified (but correct) version of the definition of Lebesgue integral. The usual definition uses the approximation of a positive measurable function by means of an increasing sequence of measurable simple functions, defined as those functions having only finitely many values; and the integral of a positive simple function is defined by summing on value strata. In fact Lebesgue theory, unlike Riemann’s, focuses the attention more on the codomain than on the domain (this was already visible in the previous definition of measurable function): loosely speaking, while Riemann divides the domain in intervals and then approximates the volume between the intervals and the graph by means of rectangles, Lebesgue divides the codomain in strata and then, for any level, measures the set of the elements of the domain whose image goes beyond that level. A simple pictorial representation of this idea can be seen in Figure 2.2(2). However, as one can imagine, in general such “over-level sets” could have a complicate structure, and to assign them a measure is much more challenging than assigning it to intervals: hence it is natural that the first interest of Lebesgue was to enlarge n as much as possible the class of measurable subsets of R .
n + If f : R R is any measurable function and at least one out of f (x) d n and • ! E f (x) d is finite, we set E n R e R + f(x) d := f (x) d f (x) d n n n ZE ZE ZE (this value could be + , or ). 1 1 + In particular, f is said to be (Lebesgue-)integrable on E when both E f (x) d n Integrable function 1 and f (x) d exist finite, and we shall write f L (E). Space L1(E) E n 2 R (22) + Recall thatR the positive part f and the negative part f of a function f are the functions defined as f +(x)=sup(f(x), 0) and f +(x)=sup( f(x), 0) (in other words, f + coincides with f where f is positive, and is zero elsewhere; and f coincides with f where f is negative, and is zero elsewhere). Note that both + + + f and f are positive functions (in spite of the name of f ), and that f = f f and f = f + f . | |
Corrado Marastoni 35 Mathematical Analysis II - Part B
1 Figure 2.2: (1) The trapezoid of f(x)=3+sinx over E =[ 2 , 3]. (2) The ideas of Riemann and Lebesgue integral.
Let us list some facts about Lebesgue integral. Beyond the expected properties (as the three —linearity, isotony, fundamental inequality— already known from Riemann’s), the most interesting features are the frequent presence of the condition “a.e.” (or “almost- everywhere”, which allows one to check that a hypothesis is satisfied up to null sets) and the theorems about the passage to the limit, which are more natural and e↵ective than the Riemann’s analogous ones.
Proposition 2.1.5. The following properties and results about Lebesgue integral hold true.
(a) (Countable additivity in the domain) Let Ek : k N be a countable family of { 2 } measurable pairwise disjoint subsets of Rn, f : Rn R a function, and denote ! E = E .Iff is measurable and positive, or if f L1(E), then k k 2 S fd n = fd n . E Ek Z Xk Z
(b) (Linearity) Given a measurable subset E of Rn, the set of functions L1(E) is a vector space on R, and the integral is a linear form on it. In other words: if f,g L1(E) and 2 ↵, R then ↵f + g L1(E),and 2 2
(↵f + g) d n = ↵ fd n + gd n . ZE ZE ZE (c) (Isotony) If f,g L1(E) and f g a.e. in E, then fd gd . 2 E n E n (d) (Fundamental inequality) Let f be measurable. Then f L1(E) if and only if R 2 R f L1(E) , and in such case it holds fd f d . | |2 E n E | | n 1 1 (e) (Comparison) Let f,g be measurable with R f g a.e.R . If g L (E) then f L (E). | | 2 2
Corrado Marastoni 36 Mathematical Analysis II - Part B
(f) (Integral and null sets) One has f =0 if and only if f is zero a.e. in E. E | | In general, two measurable functions which coincide a.e. have the same integral.(23) R (g) (Integral and boundedness) A function measurable and bounded a.e. (e.g.o, continuous on a compact subset) is integrable on any set of finite measure.
(h) (Monotone convergence for positive functions) Let E be measurable, and let fk be an increasing sequence of positive measurable functions. Then, called f the pointwise limit (24) of the fk, it holds fd n =limk + fk d n (could be + ). E ! 1 E 1 (i) (Monotone convergenceR for integrable functions)R Let E be measurable, and let fk be a monotone sequence of integrable functions. Then the sequence fk converges pointwise a.e. to a function f integrable on E if and only if the sequence E fd n is bounded; and in that case fd n =limk + fk d n R . E ! 1 E 2 R (l) (Dominated convergence)R Let E be measurable,R and be fk be a sequence of measurable functions which converges a.e. to a measurable function f. If there exists g L1(E) 1 2 such that fk g a.e., then f L (E) and fd n =limk + fk d n . | | 2 E ! 1 E (m) (Integration of series of positive functions) LetR E be measurable, andR let fk be a se- quence of positive measurable functions. Setting f(x)= k fk(x) (possibly having values in [0, + ]),itholds fd n = f d n . 1 E k E k P R P R We now describe more carefully the relation between Riemann and Lebesgue integrals.
Proposition 2.1.6. Riemann and Lebesgue integrals are related by the following facts. (a) (Integral on compact subsets) Let K be a elementarily measurable compact subset of Rn (e.g. an interval), and let f : K R.Iff is Riemann-integrable then f is also ! Lebesgue-integrable —i.e. f L1(K)— and the two integrals coincide. 2 Conversely, if f is measurable and bounded (hence in particular Lebesgue-integrable), then f is Riemann-integrable if and only if it is continuous a.e. in K. (b) (Generalized Riemann integral and Lebesgue integral) Let I be an interval of R and f : I R a locally Riemann-integrable function. Then f is Lebesgue-integrable on ! I if and only if it is absolutely Riemann-integrable in generalized sense on I,andin such case the Lebesgue integral of f on I is equal to the generalized Riemann integral of f on I. In particular, if f is Riemann-integrable in generalized sense on I but not absolutely,(25) then f is not Lebesgue-integrable on I. (c) (Fundamental Theorem of Calculus) If f :[a, b] R is Lebesgue-integrable, setting x ! F (x)= a fd 1 it holds F 0(x)=f(x) a.e. in [a, b].
(23)(Of course,R we mean, whenever the integral makes sense: hence when we are dealing with positive functions, or more generally when at least one of the integrals of positive and negative parts is finite.) The essential point of the statement is that, altering a measurable function only on a null subset of the domain (e.g. by setting there the function to zero), its integral does not change. In fact, for the Lebesgue integration two functions which coincide a.e. are indistinguishable: an evident example is the Dirichlet function , which is a.e. zero and hence has zero integral. Q [0,1] (24) \ The limit f(x):=limk fk(x)existsin[0, + ] for any x E,because(fk(x))k is positive and increasing. (25) sin x 1 2 for example, the function Dirichlet x on I = R.
Corrado Marastoni 37 Mathematical Analysis II - Part B
Conversely, let F :[a, b] R be derivable in any point of [a, b], and set f = F 0.Iff ! x is Lebesgue-integrable on [a, b], then it holds F (x) F (a)= fd . a 1 R
2.2 Integral calculus on a ne space
We now present the most relevant results for the applications in calculus, starting with a concrete example.
Let f(x, y)=x +2y and D = (x, y) R2 : x2 + y2 < 4 ,y>0 (see Figure 2.3). { 2 } Since f is continuous and D is measurable and bounded, the integral D f(x, y) d 2(x, y) (for which from now on we shall better use the more convenient notation f(x, y) dx dy) R D exists finite: let us compute it. R
Figure 2.3: (1) The set D,andthey-slice Dy at y =0,7. (2) The graph of f(x, y).
Now, Fubini’s theorem (see Proposition 2.2.1(a) below) says that this integral can be computed “by integrating on one variable at time”, or “by iterated integration”, by cutting the integration set D in slices where some variables are constant. For example, if we let y 2 2 vary from 0 to 2, then the “y-slice” of D is Dy = x R : 4 y x 4 y ,so that { 2 } p p 2 2 x=p4 y2 (x +2y) dx dy = dy (x +2y) dx = dy (x +2y) dx 2 D 0 Dy 0 x= p4 y Z Z Z Z Z 2 2 x=p4 y2 = ( 1 x2 +2xy] dy = ( 1 (4 y2)+2y 4 y2) ( 1 (4 y2) 2y 4 y2) dy 2 x= p4 y2 2 2 0 0 Z 2 Z 3 p p 2 4 2 2 2 32 32 = 4y 4 y dy =( 3 (4 y ) ]0 = (0) ( 3 )= 3 . 0 Z p
Corrado Marastoni 38 Mathematical Analysis II - Part B
Changing the order of integration would not have changed the result: if we let x vary from 2 2 to 2, then the “x-slice” of D is Dx = y R :0 y p4 x , so that { 2 } 2 2 y=p4 x2 (x +2y) dx dy = dx (x +2y) dy = dx (x +2y) dy D 2 D 2 y=0 Z Z Z x Z Z 2 2 2 y=p4 x2 2 2 = (xy + y ]y=0 dx = (x 4 x +4 x ) (0) dx 2 2 Z Z 1 2 3 1 3 2 p8 8 32 =( (4 x ) 2 +4x x ] =(8 ) ( 8+ )= . 3 3 0 3 3 3
But in this case the particular shape of D and of f suggests to use polar coordinates instead than the cartesian ones, i.e. (x, y)= (r, ✓)=(r cos ✓, r sin ✓). The theorem of change of variables (see Proposition 2.2.1(c) below) explains how to pass from a cartesian 1 integration to a polar one: it is enough to describe D in polar coordinates (i.e. (D)), compose f by , and also multiply by the absolute value of the determinant of the jacobian 0(r, ✓), which is r. Hence (using Fubini) we obtain once again
⇡ 2 (x +2y) dx dy = f( (r, ✓)) rdrd✓ = d✓ r2(cos ✓ +2sin✓) dr 1 ZD Z (D) Z0 Z0 ⇡ ⇡ 1 3 r=2 8 32 = ( 3 r (cos ✓ +2sin✓)]r=0 d✓ = 3 (cos ✓ +2sin✓) d✓ = 3 . Z0 Z0
Now we turn to precise statements, in the general case.
In the following we mean that n = n + n with n ,n N, and M N M N 2 n n x =(x ,x )withx =(x1,...,xn ) R M ,x=(xn +1,...,xn) R N ; M N M M 2 N M 2 n n n we shall denote the natural projections of R on R M and R N respectively with
n n n n ⇡ : R R M ,⇡(x)=x ,⇡: R R N ,⇡(x)=x ; M ! M M N ! N N n n moreover, given a subset E R and a function f : E R, for x R M we define the ⇢ ! M 2 x -section of E and of f setting respectively M
n Ex = x R N :(x ,x ) E ,fx : Ex R with fx (x )=f(x ,x ) M { N 2 M N 2 } M M ! M N M N
(analogous definitions are given for Ex and fx : Ex R). N N N ! Proposition 2.2.1. The following facts hold true. (a) (Fubini’s Reduction Theorem) Let E Rn be measurable and f an integrable function ⇢ on E. Then fx is integrable on Ex for a.e. x ⇡ (E); the function (defined a.e. M M M 2 M
Corrado Marastoni 39 Mathematical Analysis II - Part B
on ⇡ (E)) given by x f(x ,x ) d n (x ) is integrable on ⇡ (E);and M M Ex M N N N M 7! M R
fd n = f(x ,x ) d n (x ) d n (x ) . M N N N M M E ⇡ (E) Ex ! Z Z M Z M In particular we mean that all iterated integrals at the right-hand side, which could be obtained by varying the possible decompositions n = n + n and of the possible M N subfamilies of variables x and x give the same result. M N (b) (Tonelli’s Integrability Theorem) Let E Rn be measurable and f a measurable func- ⇢ tion on E. If any iterated integral of the modulus
f(x ,x ) d n (x ) d n (x ) M N N N M M ⇡ (E) Ex ! Z M Z M exists finite, then f is integrable on E (and for the computation one can apply Fubini). (c) (Change of variables) Let : B A be a di↵eomorphism between open subsets of ! Rn, and let E be a measurable subset of A. Given a function f : A R, we have that ! f L1(E) if and only if (f ) det L1( 1(E)) (where is the jacobian of 2 | 0|2 0 ),and
f(x) d n(x)= f( (⇠)) det 0(⇠) d n(⇠) . 1 | | ZE Z (E) Proof. Omitted.
Before providing some examples, let us make a couple of remarks.
Fubini’s Theorem is usually stated in the following alternative way (which we shall • n preferably use from now on), where the measure n on R is denoted by dx and where it is suggested that this measure could be interpreted as the “product measure” of n n the measures n = dx on M and n = dx on N : M M R N N R
f(x ,x ) dx dx = f(x ,x ) dx dx . M N M N M N N M E ⇡ (E) Ex ! Z Z M Z M Moreover, for the sake of simplicity it is customary to write the integral at the right-hand side also as
dx f(x ,x ) dx , M M N N ⇡ (E) Ex Z M Z M where we mean that f(x ,x ) is integrated first with respect to x on Ex (for a M N N M generic x ⇡ (E)) obtaining a function of x which will be then integrated on M 2 M M ⇡M(E).
Corrado Marastoni 40 Mathematical Analysis II - Part B
As we already said, on a set of finite measure any function measurable and bounded • a.e. (e.g. a continuous function) is Lebesgue-integrable. Beyond that case (hence when one deals with unbounded domains, or unbounded functions), the Theorem of Tonelli provides the most useful Lebesgue-integrability criterion: if an iterated integral of the modulus of the function exists finite then the function is integrable, and to compute its integral one can use Fubini’s Theorem. However one should be careful about the correct application of Tonelli’s result (hence by using the modulus), otherwise the conclusion could be false: some examples will be shown here below.
The most important changes of variables are the following ones. • 2 2 – Polar coordinates in R .If A = R (x, 0) : x 0 with coordinates (x, y), Polar \{ } coordinates and A =]0, + [ ] ⇡,⇡[ with coordinates (r, ✓), we set 0 1 ⇥ (x, y)= (r, ✓)=(r cos ✓, r sin ✓) .
cos ✓ r sin ✓ Therefore it holds 0(r, ✓)= sin ✓ rcos ✓ , and det 0 = r . ✓ ◆
Figure 2.4: (a) Polar coordinates in the plane. (b) Cylindrical and (c) spherical coordinates in the space.
3 3 – Cylindrical coordinates in R .If A = R (x, 0,z):x 0 with coordinates Cylindrical \{ } coordinates (x, y, z), and A0 =]0, + [ ] ⇡,⇡[ R with coordinates (r, ✓, z), we set 1 ⇥ ⇥ (x, y, z)= (r, ✓, z)=(r cos ✓, r sin ✓, z) .
cos ✓ r sin ✓ 0 Then it holds 0(r, ✓, z)= sin ✓rcos ✓ 0 , and det 0 = r . 0 0011 3@ 3 A – Spherical coordinates in R .If A = R (x, 0,z):x 0 with coordinates Spherical \{ } coordinates (x, y, z), and A =]0, + [ ] ⇡,⇡[ ]0,⇡[ with coordinates (r, ✓,'), we set 0 1 ⇥ ⇥ (x, y, z)= (r, ✓,')=(r cos ✓ sin ', r sin ✓ sin ', r cos ') .
cos ✓ sin ' r sin ✓ sin 'rcos ✓ cos ' 2 Then 0(r, ✓,')= sin ✓ sin 'rcos ✓ sin 'rsin ✓ cos ' , and det 0 = r sin ' . 0 cos ' 0 r sin ' 1 | | @ A
Corrado Marastoni 41 Mathematical Analysis II - Part B
(26) – A ne transformations, isometries, homotheties. In an euclidean space V , A ne an “a ne transformation” T : V V is a function of type T (v)=↵(v)+v , transformations ! 0 where ↵ : V V is a linear function (also called “linear part” of T )whilev is a ! 0 fixed vector which represents a translation. On the other hand an “isometry” is Isometries a self-di↵eomorphism : V V which preserves euclidean distances, i.e. such ! that (v) (w) = v w for any v, w V : it is known that isometries are || || || || 2 exactly the a ne transformations whose linear part is orthogonal (hence such t 1 (27) that ↵ = ↵ ) , which describes a rotation possibly followed by a reflection with respecto to an hyperplane of V . Another important example of a ne
transformation is given by the “homothety” of ratio µ>0, i.e. Tµ = µ idV :the e↵ect is clearly to dilate/contract all distances by a factor µ. In coordinates (thinking at V = Rn with the usual euclidean scalar product) an a ne transformation T : Rn Rn is therefore of type T (x)=Ax + b where b ! is a fixed vector of Rn and A is ae n n matrix: if E is a measurable bounded n ⇥ subset of R (hence its measure n(E) is finite), by the change of variables one has (T (E)) = det A (E) . The transformation is an isometry if and only n | | n if det A = 1, i.e. det A = 1: in this case the volume E is preserved. As for | | ⌥ n n the homothety of ratio µ one has det A = µ ,hence n(Tµ(E)) = µ n(E) (as it is reasonable to expect, if the distances are dilated by a factor µ then n-dimensional volumes are dilated by a factor µn).
Examples. (1) The function f(x, y)=3x2y 1 is continuous, hence integrable on any compact subset of 2 R : let us compute the integrals on the rectangle D1 =[ 1, 2] [0, 1] and on the triangle D2 of vertexes ⇥ (0, 0), (2, 0) and (1, 1). The projection of D1 on x is [ 1, 2], and for a generic x [ 1, 2] the x-section • 2 2 1 2 of D1 on y is always [0, 1]: therefore by Fubini one has f(x, y) dx dy = dx (3x y 1) dy = D1 1 0 2 3 2 2 y=1 2 3 2 1 3 x=2 1 3 1( 2 x y y]y=0 dx = 1( 2 x 1) dx =(2 x x]x=R 1 =(2) ( 2 )=R2 . InvertingR the order of integrations the result must be the same: namely the projection of D1 on y is [0, 1], and for a generic R R 1 2 2 y [0, 1] the y-section of D1 on x is always [ 1, 2], and we find f(x, y) dx dy = dy (3x y 2 D1 0 1 1 3 x=2 1 9 2 y=1 3 3 1) dx = (x y x]x= 1 dy = (9y 3) dy =( y 3y]y=0 =( ) (0) = . The projection of 0 0 2 R 2 2 • R R D2 on y is [0, 1], and for a generic y [0, 1] the y-section of D2 on x is [y, 2 y]: therefore by Fubini R 1 2 y R2 2 1 3 x=2 y 1 4 3 2 f(x, y) dx dy = dy (3x y 1) dx = (x y x]x=y dy = ( 2y +6y 12y +10y 2) dy = D2 0 y 0 0 2 5 3 4 3 2 1 1 ( y + y 4y +5y 2y] = . Also in this case let us invert the order. The projection of D2 on x is R 5 2 R R 0 10 R R [0, 2]; for a generic x [0, 1] the x-section of D2 on y is [0,x], while for a generic x [1, 2] is [0, 2 x]: we 2 1 x 2 2 2 x 2 21 3 2 2 y =x then obtain f(x, y) dx dy = dx (3x y 1) dy + dx (3x y 1) dy = ( x y y] dx + D2 0 0 1 0 0 2 y=0 2 3 2 2 y=2 x 1 3 4 2 3 2 2 ( x y y] dx = ( x x) dx + ( x (2 x) (2 x)) dx, which eventually gives once more 1 2 R y=0 0 2 R R 1 2 R R R 1 + 3 = 1 . (2) (Area of a plane region described in polar coordinates) Let us consider a region of R 5 10 10 R R the cartesian plane in polar coordinates given by D = (⇢,✓):✓ [↵, ],⇢1(✓) ⇢ ⇢2(✓) for certain { 2 } 0 ↵< 2⇡ and certain functions 0 ⇢1(✓) ⇢2(✓). Using the change of polar variables we obtain ⇢2(✓) 1 2 ⇢2(✓) 1 2 2 Area(D)= dx dy = d✓ ⇢d⇢ = ( ⇢ ] d✓ = ⇢2(✓) ⇢1(✓) d✓. 2 ⇢1(✓) 2 ZD Z↵ Z⇢1(✓) Z↵ Z↵ For example, the cardioid (Figure 2.5(a)) is the polar curve in the plane given by r(✓)=a(1 + cos ✓), where a>0and✓ [0, 2⇡]: the area of the portion of plane enclosed by the curve is then 1 2⇡ a2(1 + 2 2 0 (26)Recall that “euclidean” means that V is finite-dimensional and endowed with a scalar product.R (27) Given a linear function ↵ : V1 V2 where V1 and V2 are euclidean spaces, the transposed function t ! is the linear function ↵ : V2 V1 associating to a vector v2 V2 the unique vector v1 V1 such that ! t 2 2 v1 v = ↵(v) v2 for any v V1: the function ↵ is then represented in coordinates by the matrix transposed · · 2 to the one of ↵.
Corrado Marastoni 42 Mathematical Analysis II - Part B
cos ✓)2 02 d✓ = 1 a2(✓ +2sin✓ + 1 (✓ +sin✓ cos ✓)]2⇡ = 1 a2((3⇡) (0)) = 3 a2⇡.Anotherexample:the 2 2 0 2 2 region E = (x, y):2ax x2 + y2 2bx, 0 y x , where 0 a
Corrado Marastoni 43 Mathematical Analysis II - Part B
+ 1 xy x=2k⇡ + 1 + e 2k⇡y 1 1 1 [0,2k⇡] [0,+ [ f(x, y) dx dy = 0 ( 1+y2 e (cos x+y sin x)]x=0 dy = 0 1+y2 dy 0 1+y2 dy = ⇥ 12k⇡y 2k⇡y ⇡ + e 2k⇡ sin x ⇡ + e R 1 2 dy. By comparingR the two equalities we get dxR = 1 R 2 dy for any 2 0 1+y 0 x 2 0 1+y + e 2k⇡y + 2k⇡y 1 1 1 k NR: passing to the limit for k + and noting that 0 R 1+y2 dy 0 e R dy = 2k⇡ tends to 2 ! 1+ sin x ⇡ zero, we finally find the desired value 1 dx = . 0 x 2 R R R
2 2 Figure 2.5: (a) Cardiod ⇢(✓) a(1 + cos ✓). (b) E = (x, y):2ax x + y 2bx, 0 y x . (c) E1 = (x, y): 2 2 { 2 2 } { 0 x 1, x y x +1 . (d) E2 = (x, y, z):x + y min z, a , 0 z b . } { { } }
Other results which are useful from a practical point of view are the following ones. Proposition 2.2.2. The following facts hold. n n (a) (Integration of products with separate variables) Let E0 R M and E R N be ⇢ N ⇢ measurable sets, and g : E0 R and h : E R be measurable functions. Then ! N ! the function f : E0 E R defined by f(x ,x )=g(x ) h(x ) is integrable on ⇥ N ! M N M N E E if and only if g and h are integrable respectively on E and E ,anditholds 0 ⇥ N 0 N g(x ) h(x ) d = g(x ) d h(x ) d . M N n M n N n E E E M E N Z 0⇥ N ✓Z 0 ◆✓Z N ◆ (b) (Symmetries) Let T be an isometry of Rn, E Rn be of the form (up to null sets) ⇢ E = E+ E where E+ and E are measurable sets such that T (E )=E . Given t ± ⌥ f L1(E),iff T = f then fd =2 fd ;andiff T = f then 2 E n E+ n fd =0. E n R R Proof.R (a) Let us bound to su ciency. The function f is measurable because it is a product of measu- rable functions; moreover one can prove that E0 E is measurable. Then, since g and h are integra- ⇥ N ble on E0 and EN,also g and h are: since the iterated integral E E g(x ) h(x ) d n d n = | | | | 0 N | M N | M N ⇣ ⌘ E E g(x ) h(x ) d n d n = E E g(x ) d n h(xR ) dR n = E g(x ) d n E h(x ) d n 0 N | M || N | M N 0 N | M | M | N | N N | M | M 0 | N | N is finite, by Tonelli we get that g(x ) h(x ) is integrable, and then the equality follows from Fubini by R ⇣R ⌘ M R N ⇣R ⌘ ⇣R ⌘ R repeating these computations without modulus. (b) Considering the change of variables given by T we obtain fd n = + fd n depending on f T = f:since fd n = + fd n + fd n,the E ± E ± E E E statement is proven. R R R R R
+ x2 x2 Examples. (1) (Gaussian integral) Let us compute the generalized integral 1 e dx:sincee 1 x2 is (absolutely) Riemann-integrable, it will be also Lebesgue-integrable, with integral e dx equal R R R
Corrado Marastoni 44 Mathematical Analysis II - Part B
x2 y2 to the generalized integral. For the computation let us start from the double integral e d 2: R2 x2 y2 x2 y2 noting that e = e e , by Proposition 2.2.2(a) such double integral is finiteR and has value x2 y2 x2 2 ( e dx)( e dy)=( e dx) . On the other hand, using polar coordinates we easily compute R R R x2 y2 2⇡ + r2 2⇡ 1 r2 + x2 2 e d 2 = d✓ 1 re dr =(✓]0 ( e ] 1 = ⇡,henceweget e dx = p⇡. (2) R R 0 0R 2 0 R (Simple symmetries) Let us consider the square D =[1, 3] [ 2, 2], and compute f(x, y) d 2 and R R R ⇥ R D g(x, y) d with f(x, y)=x(1 + y2)andg(x, y)=y(1 + x2). The set D is clarly symmetric with respect D 2 R Rto the x-axis: in other words, considered the isometry T (x, y)=(x, y) (the reflexion with respect to the x- + axis) and the decompositione D = D D with D± = (x, y) D : y 0 (where we neglect the segment t { 2 ? } drawn by D on the x-axis, a null set) we have T (D±)=D⌥.Sincef T = f (namely f(x, y)=f(x, y)) 3 2 2 1 2 3 1 3 2 112 we can say that D f(x, y) d 2 =2 D+ f(x, y) d 2 =2 1 dx 0 x(1 + y ) dy =2(2 x ]1 (y + 3 y ]0 = 3 . On the other hand g T = g (namely g(x, y)= g(x, y)), hence we can say that g(x, y) d 2 =0. R R R R D R If a solid body E occupies a measurable and bounded region (which we also denote 3 by E)inR with mass density µ(x) (the total mass is then m = E µ(x) d 3(x)), Barycenter 1 the barycenter of E is defined as the point of coordinates x µ(x) d (x) for Moment of inertia m E j R 3 j =1, 2, 3, and the moment of inertia of E with respect to a certain a ne line r is 2 R given by I = E(distr(x)) µ(x) d 3(x), wheredistr(x) is the function of distance from r. In the homogeneos case (µ constant, hence m = µ 3(E)where 3(E)= E d 3(x)is R 1 the volume of E) one obtains the geometrical barycenter of coordinates xj d 3(x) 3(E) RE 2 for j =1, 2, 3, and the moment of inertia becomes I = µ E(distr(x)) d 3(xR). R 2 2 2 Exercise. Draw the sets E1 = (x, y):0 x 1, x y x +1 and E2 = (x, y, z):x + y { } { min z,a2 , 0 z b (ove a, b > 0 and b>a2); compute the area (resp. the volume), the barycenter and { } } the moment of inertia with respect to a generic vertical axis.
2 1 x +1 1 x2+1 Solution. E1 is the “ax” of Figure 2.5(c). The area is E dx dy = 0 dx x dy = 0 [y] x dx = 1 1(x2 +x+1)dx = 11 . The barycenter has coordinates x = 1 xdxdy = 6 1(x3 +x2 +x) dx = 13 0 6 GR 11/6 E1 R R 11 0 R 22 1 6 1 1 2 x2+1 3 1 4 2 23 Rand yG = 11/6 E ydxdy = 11 0 [ 2 y ] x dx = 11 0 (x + x R+1)dx = 55 .IfR µ if the mass density 1 11 (by unit of area),R the total massR is m = 6 µ and theR moment of inertia with respect to the y-axis is 2 1 4 3 2 47 (28) Iy = µ x dx dy = µ (x + x + x ) dx = µ; then, using the Theorem of Huygens-Steiner, the E1 0 60 2 moment of inertia Iy,x with respect to a generic vertical axis x = x0 is Iy,x = Iy,x +m(x0 x ) = Iy R 0 R 0 G G 2 2 mxG +m(x0 xG ) = Iy+mx0(x0 2xG ). E2 is the “bullet” of Figure 2.5(d). Using cylindrical coordinates, 2 2⇡ • a pz 2⇡ b a 2 1 2 the volume will be dx dy dz = d✓ dz rdr + d✓ 2 dz rdr = ⇡a (b a ). By E2 0 0 0 0 a 0 2 1 symmetry the barycenter will surely belong to the z-axis, at height zG = 2 1 2 E zdxdydz = R R R R R R R ⇡a (b 2 a ) 2 2 2 2⇡ a pz 2⇡ b a 3b2 a4 R ⇡a2(2b a2) ( 0 d✓ 0 zdz 0 rdr+ 0 d✓ a2 zdz 0 rdr)= 3(2b a2) .Ifµ is the mass density (by volume 2 1 2 unit), the total mass is m = ⇡a (b 2 a )µ, and the moment of inertia with respect to the barycentric axis z R R R R R2 R 2 2 2⇡ a pz 2 2⇡ b a 2 1 4 2 is Iz = µ (x +y ) dx dy dz = µ( d✓ dz r rdr+ d✓ 2 dz r rdr)= ⇡µa (3b 2a ); E2 0 0 0 0 a 0 6 (28) once moreR by Huygens-Steiner, Rthe momentR ofR inertia withR respectR to aR generic vertical axis (x, y)= 2 2 (x0,y0)willbeIz + m(x0 + y0 ). To compute the volumes obtained by rotating a bounded and measurable plane set around
(28) 3 The Theorem of Huygens-Steiner states that the moment of inertia of a body in R with respect to a 2 certain axis r is equal to r = + md , where is the moment of inertia of the budy with respect I Ir(G) Ir(G) to the axis r(G) parallel to r and passing through the barycenter, m is the total mass of the body and d is the distance between the axes r and r(G).
Corrado Marastoni 45 Mathematical Analysis II - Part B a certain axis, the following classical result is very useful.
Proposition 2.2.3. (Guldinus’ Theorem for the volumes of the solids of revolution) Given a bounded and measurable plane set D contained in the half plane (x, z) with x>0, let it rotate of an angle ↵ around the z-axis. Then the volume described in R3 by this revolution is ↵ D xd 2 , also expressible as ↵ 2(D) xG where G and 2(D) are respectively the geometric barycenter and the area of D. R Proof. In cylindrical coordinates (r, ✓, z), the solid of rotation described above is given by ((r, ✓, z): ↵ { (r, z) D,✓ [0,↵] ,hencethevolumeis d✓ rdrdz = ↵ rdrdz, as stateda. For the rest of the 2 2 } 0 D D statement, just recall the definition of barycenter. R R R
3 Examples. (1) (Volume of the sphere) The volume of the sphere of radius R in R has already been 4 3 computed previously as volume integral, with value 3 ⇡R . Let us compute it again by Guldinus’ Theorem, by viewing the sphere ad described by the rotation around the y-axis of the half disk D = (x, y):x2 +y2 < { R, x > 0 . Instead than computing only the integral ↵ xd 2, let us find the geometric barycenter of } D D, which by symmetry will belong to the x-axis: using polar coordinates we get x = 1 xd = R G R2⇡/2 D 2 ⇡ ⇡ ⇡ 1 2 R 2 2 R 2 2 2 1 3 R 4 R2⇡/2 ⇡ d✓ 0 r cos ✓rdr = R2⇡ ⇡ cos ✓d✓ 0 r dr = R2⇡ (sin ✓] ⇡ ( 3 r ]0 = 3⇡ R,hencethevolumeR 2 2 2 4 R2⇡ 2 3 4 3 of a portionR ofR sphere of angle ↵ is R↵ 3⇡ R 2 =R 3 ↵R , and for ↵ =2⇡ we recover the volume 3 ⇡R . (2) (Volume of the torus) (Figure 2.7, p. 50) Let R>r>0. The rotation of a disk of radius r centered in (R, 0) all around the y-axis generates a torus; the volume of a portion of angle ↵ is ↵R (⇡r2)=⇡Rr2↵, hence the volume of the whole torus is 2⇡2Rr2.
2.3 Integral calculus on manifolds
Let M be a manifold (at least of class 1) of dimension k in Rn, and let f : M R be a C ! function. How could one define in a suitable way a notion of k-dimensional integral of f on M, by using what has been done for the integration in Rk? In the case k =1(whereM is a curve) the answer is already known, as it is the arc integral : we just have to generalize it to a manifold of higher dimension.
The idea of the Riemann integral (then refined by Lebesgue) in Rk is as follows. Given a function with compact support f : Rk R,wedivideRk into k-dimensional cubes along (k) (k) ! N the k coordinate directions e1 ,...,ek , of side length 2 (whose k-dimensional measure N k kN is then (2 ) =2 ); taken a reference point xC into each cube C,wedefine
kN f(x) d k := lim f(xC ) k(C)= lim 2 f(xC ) . k N + C N + C ZR ! 1 ! 1 X X This sum, where C runs over all k-dimensional cubes of Rk, is in fact finite (since f has compact support) and rapresents the integral of the step function costructed with the given choice of the reference points xC :iff is Riemann-integrable (for example, if it is
Corrado Marastoni 46 Mathematical Analysis II - Part B continuous) the limit exists and does not depend on this choice, and this is what we call “(Riemann) integral of f ”. So let us develop this idea in the more general framework of manifolds.
Figure 2.6: Volume integral on manifolds.
Assume that f : M R has compact support, and for the sake of simplicity that M ! is described —up to k-null sets— by a single parametrization : V M with V open k ! in R : it is then natural to construct f(x) d M (x) as lim f( (vC )) k( (C)), M N + C N ! 1 where the sum is done on the cubes C ofR side length 2 containedP in V , vC is a reference point in C and k( (C)) represents the k-dimensional measure of the image of C by . N Now, since C is a small cube of side length 2 around vC whose jth side is parallel (k) to ej , and since at the neighborhood of vC the map is well approximated bt the di↵erential d , the image (C) can be seen as the small parallelogram whose jth side vC is d (e(k))= @ (v )multipliedby2 N , hence its measure ( (C)) will be 2 kN vC 1 @vj C k times the measure of such parallelogram. Hence we need to recall how to compute the n k-dimensional measure of the parallelogram generated by k vectors u1,...,uk in R ,i.e. x = u + + u :0 1 . { 1 1 ··· k k j } Proposition 2.3.1. Given a square matrix A Mk(R),thek-dimensional measure of the 2 parallelogram generated in Rk by the k column vectors of A is det A . | | More generally: given a rectangular matrix A Mn,k(R) with k n,thek-dimensional 2 measure of the parallelogram generated in Rn by the k column vectors of A is det(AtA), where At denotes the transposed matrix of A. p Proof. If the k column vectors of the square matrix A Mk(R) are linearly dependent then the k- 2 dimensional measure of the parallelogram generated by them is zero, and at the same time the matrix A is singular and hence det A = 0. On the other hand, if the vectors are linearly indipendent, then A represents k the linear change of coordinates of R which sends the vectors of the canonical basis into the column vectors of A, and then the result follows immediately from the formula of change of variables for multiple integrals (Proposition 2.2.1(c)). Now let us deal with the general case. Given a rectangular matrix A Mn,k(R) • n 2 with k n,letu1,...,uk be its column vectors in R . Taken any orthonormal family of vectors w1,...,wk n in R such that u1,...,uk w1,...,wk , for what has been said before the k-dimensional measure of the h i✓hn i parallelogram generated in R by the vectors u1,...,uk is det B , where B Mk(R) is the square matrix | | 2 whose jth column is made by the components of uj with respect to the orthonormal basis w1,...,wk (in
Corrado Marastoni 47 Mathematical Analysis II - Part B
t t other words, Bi,j = wi uj ). Since det B =detB we can also write det B = det(B B): but now it is ·t | | enough to note that (B B)i,j = ui uj (the scalar product does not depend on the choice of orthonormal · p basis), and hence BtB = AtA.
n If u1,...,uk are the k column vectors in R of the matrix A Mn,k(R), the square matrix t 2 A A Mk(R) (called Gramian matrix of u1,...,uk) has elements ui uj; its determinant Gramian matrix, 2 · gramian is usually called gramian of u1,...,uk. It is useful to recall the following method for computing the gramian:
Proposition 2.3.2. (Cauchy-Binet identity) The gramian det(AtA) is also equal to the sum of the squares of the determinants of all minors of order k of A.
Thanks to Proposition 2.3.1 we have understood what should be the measure k( (C)) of the previous discussion, and we can give a precise definition of what we mean by volume integral on M. Given a function f : M R, we say that f is integrable on M if Volume integral ! on a manifold f det(( )t ) is integrable on V (where is the jacobian of ), and we set | | 0 0 0 p t (2.1) f(x) d M (x):= f( (v)) det( 0(v) 0(v)) d k(v) . M V Z Z p t The quantity d M (x):= det( 0(v) 0(v)) d k(v) (sometimes other notations are used, Volume element as d which recalls the idea of surface) is called volume element of the manifold M in its p point x = (u): when integrated on M (hence by choosing f 1), this integral should give ⌘ the usual k-dimensional volume of M, whenever this makes sense. We should also check that:
Proposition 2.3.3. The above definition of M f(x) d M (x) does not depend on the pa- rametrization chosen on M. R 1 Proof. Let : V M be another parametrization, and set := : V ⇠ V (hence = ). We ! !k then have (using the formula of change of variables for multiple integrals in R (Proposition 2.2.1(c)) in the last equality):e e e e e
t t f( (v)) det( 0(v) 0(v)) d k(v)= f( ( (v))) det ( 0( (v)) 0(v)) 0( (v)) 0(v) d k(v) ZV ZV p q t = f( e( (v))) det e( (v)) ( (v))e det 0(v) d k(v) 0 0 | | ZV q t = f( e(v)) det 0(ve) 0(v) de k(v) . ZV q e e e e e e e e
Let us emphasize the following particular cases.
(1) Case k =1. As has been said, in the case of a curve the previous definition recovers the arc integral (and the curve length).
3 (2) Case k =2and n =3. M is a surface in R : denoting by v =(v1,v2) the parameters 2 @ @ of V R , the surface element is d = (v) (v) dv1 dv2, the norm of the ⇢ k @v1 ⇥ @v2 k vector product of the two derivatives (which is orthogonal to M in x = (v)).
Corrado Marastoni 48 Mathematical Analysis II - Part B
n (3) Case of cartesian graphs. Written x =(x1,...,xn 1), let M = x R : xn = M { 2 f(x ) for a suitable f: in this case the la jacobian matrix of the obvious cartesian M } 1n 1 parametrization (x )=(x ,f(x )) is , hence the surface element is d = M M M f r 2 1+ f d n 1(x ). (Of course the choice of xn as dependent variable is just for kr k M notational convenience: the same conclusion holds for graphs of other components.) p
Examples. (1) (Cosine law for the areas) Consider a bounded region D↵ on the plane z =(tg↵) x which projects over the bounded region D of the horizontal plane (x, y): then it holds Area(D)=
Area(D↵)cos↵ .NamelyD↵ is the graphs above D of f(x, y)=(tg↵) x, hence the surface element is 2 2 1 1 d = 1+ f dx dy = 1+tg ↵dxdy= dx dy, and this yields Area(D↵)= dx dy = kr k cos ↵ D cos ↵ 1 dx dy = 1 Area(D). (2) Let us compute the area of the spherical half surface of radius R using cos ↵ Dp cos ↵ p R threeR di↵erent parametrizations. (a) Using polar coordinates 1(✓,')=(R cos ✓ sin ',R sin ✓ sin ',R cos '), R sin ✓ sin 'Rcos ✓ cos ' ⇡ R cos ✓ sin 'Rsin ✓ cos ' where (✓,') ]0, 2⇡[ ]0, 2 [. We have 10 = , hence the surface element is 2 ⇥ 0 0 R sin ' 1 ⇡ 2 @ 2 A 2 2⇡ 2 2 d = R sin 'd✓d' and the desired area is ]0,2⇡[ ]0, ⇡ [ R sin 'd✓d' = R 0 d✓ 0 sin 'd' =2⇡R ⇥ 2 2 2 2 (a well-known result). (b) Using cartesian coordinatesR 2(x, y)=(x, y, R Rx yR), where (x, y) runs in the disk D centered in (0, 0) with radius R. Here the half spherep can be interpreted as the graph z = f(x, y)= R2 x2 y2,henced = 1+ f 2 dx dy = R dx dy,andusingthepo- pR2 x2 y2 ||r || lar coordinatesp in the plane we get once morep R dx dy = R ⇢d⇢d✓ = D pR2 x2 y2 ]0,2⇡[ ]0,R[ pR2 ⇢2 ⇥ 2⇡ R ⇢ 2 2 R 2 R d✓ d⇢ =2⇡R( R ⇢ ]0 =2R ⇡R . (c) Using the parametrizationR by cylindrical 0 0 pR2 ⇢2 2 2 2 2 coordinatesR R 3(✓, z)=(pR z cosp ✓, pR z sin ✓, z), where (✓, z) ]0, 2⇡[ ]0,R[. We have 30 = R2 z2 sin ✓ z cos ✓ 2 ⇥ R2 z2 p 0 R2 z2 cos ✓ q z sin ✓ 1,hence d = Rd✓dz and the area is again Rd✓dz = 2 2 ]0,2⇡[ ]0,R[ B R z C ⇥ B p C B 01q C R @ 2⇡ R 2 A R 0 d✓ 0 dz =2⇡R . (3) Let D be the disk centered in (0, 0) of radius r in the plane (x, y), and consider the straight cylinder and cone with base D and height h along the z-axis. In cylindrical coordinates (✓, z) R R 2 ]0, 2⇡[ ]0,h[ they are parametrized respectively by (R cos ✓,R sin ✓, z)and(R(1 z )cos✓,R(1 z )sin✓, z); ⇥ h h the surface elements are Rd✓dz and R a(1 z ) d✓dz (where a = pR2 + h2 is the apothem of the cone), h h hence the side areas of cylinder and cone are respectively 2⇡Rh and Ra⇡ , as it is well known from elementary geometry.
Also in the framework of material surfaces in R3 it is possible to define total mass, barycen- ter and moment of inertia by using the integrals just defined. For a surface M in R3 with mass density µ(x) (for unit of area; the total mass is then m = M µ(x) d ), the bary- Barycenter center of M is defined as the point with coordinates 1 x µ(x) d for j =1, 2, 3, m M j R and the moment of inertia of M with respect to a certain a ne line r is given by Moment of inertia 2 R I = M (distr(x)) µ(x) d ,wheredistr(x) is the function of distance from r.Inthe homogeneous case (µ constant) we obtain the geometrical barycenter with coordinates R 1 x d for j =1, 2, 3(where (M)= d is the area —i.e. the 2-dimensional 2(M) M j 2 M 2 measure—R of M), and the moment of inertia becomesR I = µ M (distr(x)) d . R Example. For ↵ [0,⇡]letS↵ be the portion of spherical surface of radius R with latitude ' bet- 2 ween 0 and ↵, with uniform mass density µ: let us compute the geometrical barycenter and the mo- ment of inertia with respect to any line parallel to the z-axis. For evident reasons of symmetry the •
Corrado Marastoni 49 Mathematical Analysis II - Part B
barycenter of S↵ belongs to the z-axis, hence the only coordinate to compute is z. Parametrizing S↵ 2⇡ ↵ 2 2 as (R cos ✓ sin ',R sin ✓ sin ',R cos ')wehave 2(S↵)= 0 d✓ 0 R sin 'd' =2⇡R (1 cos ↵), da 1 1 2⇡ ↵ 2 1+cos ↵ cui zd 2(x)= 2 d✓ (R cos ')R sin 'd' = R. By Huygens-Steiner 2(S↵) S↵ 2⇡R (1 cos ↵) 0 0 R R 2 (28) • Theorem R the moment of inertia of SR↵ withR respect to a line parallel and at distance d from the z- 2 axis is Iz + md , where Iz denotes the moment of inertia with respect to the z-axis: and the latter is 2 2 2⇡ ↵ 2 2 2 4 ↵ 3 4 1 3 Iz = µ (x + y ) d 2(x)=µ d✓ (R sin ')R sin 'd'=2⇡µR sin 'd'=2⇡µR ( cos ' S↵ 0 0 0 3 ↵ 2⇡ 4 3 2 cos ↵ cos2 ↵ 2 cos ']0 R= 3 µR (2 3cos↵ +cosR ↵R)= 3 mR , where weR have introduced the total mass 2 m = µ 2(S↵)=2⇡R (1 cos ↵)µ. There is also an analogous version of Guldinus’ Theorem for volumes (Proposition 2.2.3) which describes the area of the surfaces of revolution obtained by rotating a rectifiable curve around a given axis.
Proposition 2.3.4. (Guldinus’ Theorem for surfaces of revolution) Given a rectifiable plane curve contained in the half plane (x, z) with x>0, let it rotate of an angle ↵ around the z-axis. Then the area of the surface described in R3 by this revolution is
↵ xd`, also expressible as ↵xG `( ) where G and `( ) are respectively the geometric barycenter and the length of . R 2 Proof. Let :[a, b] Rx,z be a parametrization of , with x(t)= 1(t) > 0: then the surface of revolution ! 3 described above is parametrized by :[a, b] [0,↵] R with (t, ✓)=( 1(t)cos✓, 1(t)sin✓, 2(t)). It 0 (t)cos✓ 1(t)sin✓ ⇥ ! 1 2 2 holds 0 = 0 (t)sin✓ 1(t)cos✓ , with surface element given by d = 1(t) 0 (t) + 0 (t) dt d✓ = 0 1 1 1 2 0 20 (t) | | @ A ↵ b x(t) 0(t) dt d✓, so that the area is x(t) 0(t) dt d✓ = d✓ x(t) p0(t) dt = ↵ xd`.For k k [a,b] [0,↵] k k 0 a k k the rest of the statement, just recall the definition⇥ of barycenter. R R R R
Figure 2.7: Atorus.
Examples. (1) (Torus surface) (Figure 2.7) Let R>r>0, and consider the circle of radius r centered in (R, 0). The rotation of the circle all around the y-axis generates a torus surface; the area of a portion of angle ↵ is ↵R (2⇡r)=2⇡Rr↵, hence the area of the whole surface is 4⇡2Rr. Alternatively we could also use the parametrization given by (x, y, z)= (✓, )=((R + r cos )cos✓, (R + r cos )sin✓, r sin ): t 2⇡ 2⇡ since d = det(( 0) 0) d✓d = r(R + r cos ) d✓d we obtain again 0 d✓ 0 r(R + r cos ) d = 2r⇡(R + r sin ]2⇡ =4⇡2Rr. (2) Consider the arc of sinusoid = (x, y):x =cosy, 0 y ⇡ ,and p 0 { R R 2 } let it rotate of an angle ↵ around the y-axis. Setting sin y =sinh⇠ (hence cos ydy =cosh⇠d⇠)weget ⇡ p p p p 2 2 log( 2+1) 2 ⇠+sinh ⇠ cosh ⇠ log( 2+1) 2+log( 2+1) xd` = 0 cos y 1+sin ydy = 0 cosh ⇠d⇠ =( 2 ]0 = 2 ,hence p2+log(p2+1) theR area ofR the surfacep of revolutionR is 2 ↵.
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