Chapter 4. Multiple Integrals. 1. Integrals on Rectangles

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1 Chapter 4. Multiple Integrals Let be a rectangle in I 2 given by 1. Integrals on ectangles = [a 1, b 1 [a 2, b 2 := {(x, y) I 2 : a 1 x b 1, a 2 y b 2 }. Let P 1 be a partition of [a 1, b 1 : P 1 : a 1 = x < x 1 < < x m = b 1, and let P 2 be a partition of [a 2, b 2 : P 2 : a 2 = y < y 1 < < y n = b 2. For i {1,..., m} and j {1,..., n}, let C ij be the rectangle [x i 1, x i [y j 1, y j. We will refer to C ij as the (i, j) call. The area of C ij is A ij = x i y j, where x i := x i x i 1 and y j := y j y j 1. The diameter of C ij is d(c ij ) = (x i x i 1 ) 2 + (y j y j 1 ) 2. Let P = P 1 P 2 be the collection { C ij : i = 1,..., m, j = 1,..., n }. Then P is a partition of. Now let f be a bounded real-valued function on the rectangle. Given a partition P = {C ij : i = 1,..., m, j = 1,..., n } of, let m ij := inf{f(x, y) : (x, y) C ij } and M ij := sup{f(x, y) : (x, y) C ij }. The upper sum U(f, P ) and the lower sum L(f, P ) for the function f and the partition P are defined by U(f, P ) := m n M ij A ij and L(f, P ) := i=1 j=1 m n m ij A ij. i=1 j=1 The upper integral U(f) of f over is defined by U(f) := inf{u(f, P ) : P is a partition of } and the lower integral L(f) of f over is defined by L(f) := sup{l(f, P ) : P is a partition of }. 1

2 If P = P 1 P 2 and P = P 1 P 2 are partitions of, then L(f, P ) U(f, P ). In order to prove this assertion, we let P 1 be a common refinement of P 1 and P 1, and let P 2 be a common refinement of P 2 and P 2. Then for P := P 1 P 2 we have L(f, P ) L(f, P ) U(f, P ) U(f, P ). Consequently, L(f) U(f). A bounded function f on is said to be iemann integrable if L(f) = U(f). In this case, we write f(x, y) da := L(f) = U(f). Theorem 1.1. A bounded function f on a rectangle is integrable if and only if for each ε > there exists a partition P of such that U(f, P ) L(f, P ) < ε. Proof. Suppose that f is integrable on. For ε >, there exist partitions P and P such that L(f, P ) > L(f) ε and U(f, P ) < U(f) + ε 2 2. Let P a common refinement of P and P. Then we have L(f) ε 2 < L(f, P ) L(f, P ) U(f, P ) U(f, P ) < U(f) + ε 2. Since L(f) = U(f), it follows that U(f, P ) L(f, P ) < ε. Conversely, suppose that for each ε > there exists a partition P of such that U(f, P ) L(f, P ) < ε. Then we have U(f) U(f, P ) = U(f, P ) L(f, P ) + L(f, P ) < ε + L(f, P ) ε + L(f). Since ε > is arbitrary, we conclude that U(f) L(f). Hence f is integrable. Theorem 1.2. If f is a continuous function on a rectangle, then it is integrable on. Proof. Consider ε >. Since f is uniformly continuous on, there exists some δ > such that (x, y), (x, y ) and (x x ) 2 + (y y ) 2 < δ imply f(x, y) f(x, y ) < ε A, where A is the area of the rectangle. Let P = {C ij : i = 1,..., m, j = 1,..., n} be a partition of such that d(c ij ) < δ for all i = 1,..., m and j = 1,..., n. Let 2

3 m ij := inf{f(x, y) : (x, y) C ij } and M ij := sup{f(x, y) : (x, y) C ij }. Since f attains its maximum and minimum on each closed cell C ij, we have M ij m ij < ε/a for i = 1,..., m and j = 1,..., n. Consequently, U(f, P ) L(f, P ) = m n (M ij m ij ) A ij < ε A i=1 j=1 m n A ij = ε. i=1 j=1 By Theorem 1.1 we conclude that f is integrable. The following elementary properties of the integral can be easily established. Theorem 1.3. Let f and g be integrable functions on a rectangle and let c be a real number. Then (1) cf is integrable on and (cf)(x, y) da = c f(x, y) da; (2) f + g is integrable on and (f + g)(x, y) da = f(x, y) da + g(x, y) da; (3) if f(x, y) g(x, y) for all (x, y), then f(x, y) da g(x, y) da. 2. epeated Integrals The following theorem demonstrates that the evaluation of the double integral of an integrable function on a rectangle can be reduced to repeated integrals. Theorem 2.1. Let f be a bounded, real-valued function that is integrable on a rectangle = [a 1, b 1 [a 2, b 2. Suppose that, for each y in [a 2, b 2, the function f(x, y) is an integrable function of x on [a 1, b 1. Then the function F (y) := b 1 a 1 function of y on [a 2, b 2 and f(x, y) da = b2 [ b1 a 2 a 1 f(x, y) dx dy. f(x, y) dx is an integrable Proof. Let ε > be given. Since the function f is integrable on, there exists a partition P of the rectangle such that U(f, P ) L(f, P ) < ε, by Theorem 1.1. Suppose that P = {C ij : i = 1,..., m, j = 1,..., n}, where each C ij = [x i 1, x i [y j 1, y j with a 1 = x < x 1 < < x m = b 1 and a 2 = y < y 1 < < y n = b 2. For i = 1,..., m and j = 1,..., n, let m ij := inf{f(x, y) : (x, y) C ij } and M ij := sup{f(x, y) : (x, y) C ij }. Moreover, let k j := inf{f (y) : y j 1 y y j } and K j := sup{f (y) : y j 1 y y j } for j = 1,..., n. If y j 1 y y j, then m ij f(x, y) M ij for x [x i 1, x i, i = 1,..., m. Hence, m m ij x i F (y) = i=1 b1 a 1 3 f(x, y) dx m M ij x i. i=1

4 It follows that L(f, P ) = n j= i= m m ij x i y j U(F, [a 2, b 2 ) n k j y j L(F, [a 2, b 2 ) j= n K j y j j= n j= i= m M ij x i y j = U(f, P ). Since U(f, P ) L(f, P ) < ε, it follows that U(F, [a 2, b 2 ) L(F, [a 2, b 2 ) < ε whenever ε >. Therefore, U(F, [a 2, b 2 ) = L(F, [a 2, b 2 ). In other words, F is integrable on [a 2, b 2. Consequently, L(f, P ) b2 F (y) dy U(f, P ) and L(f, P ) a 2 Since U(f, P ) L(f, P ) < ε, we conclude that ε < f(x, y) da b2 for any ε >. This shows that f(x, y) da = b 2 a 2 a 2 F (y) dy < ε F (y) dy. f(x, y) da U(f, P ). Of course, the symmetric version of the theorem holds. If f is integrable on the rectangle = [a 1, b 1 [a 2, b 2, and if for each x in [a 1, b 1, the function f(x, y) is an integrable function of y on [a 2, b 2. Then the function G(x) := b 2 a 2 function of x on [a 1, b 1 and f(x, y) da = b1 [ b2 a 1 a 2 f(x, y) dy dx. f(x, y) dy is an integrable xample. Let f(x, y) := xe xy for (x, y) = [, 2 [, 1. Since f is continuous on, we may evaluate the double integral by repeated integrals in either order. We have xe xy da = = 2 2 [ 1 xe xy dy dx = 2 [ e xy 1 dx (e x 1) dx = [e x x 2 = (e 2 2) (1 ) = e

5 3. iemann Domains Given a subset of a metric space (X, ρ), we use to denote the set of all interior points of. Then Bd() := \ is the set of all boundary points of. In what follows, by a closed rectangle we mean a set of the form [a 1, b 1 [a 2, b 2, where < a 1 < b 1 < and < a 2 < b 2 <. The area of a closed rectangle = [a 1, b 1 [a 2, b 2 is defined to be A() := (b 1 a 1 )(b 2 a 2 ). A subset of I 2 is called a null set if for any ε > there exists a finite collection { 1, 2,..., m } of closed rectangles such that m i=1 i and m i=1 A( i) < ε. In the above definition, the rectangles 1,..., m could be so chosen that m i=1 i. Indeed, if is a null set, then for any ε >, there exists a finite collection { 1, 2,..., m} of closed rectangles such that m i=1 i and m i=1 A( i ) < ε/2. For each rectangle i we can find a closed rectangle i such that i i and A( i) < 2A( i ). Consequently, m i=1 i and m i=1 A( i) < 2 m i=1 A( i ) < ε. The following properties of null sets can be deduced from the above definition at once. (1) The empty set is a null set. (2) If is a null set, then so is. (3) If is a null set and K, then K is a null set. (4) If and F are null sets, then F is a null set. A function g on a closed interval [a, b is said to be a Lipschitz function if there exists a real number M > such that g(s) g(t) M s t whenever a s, t b. Theorem 3.1. Let u 1 and u 2 be continuous functions on a closed interval [a, b. If one of u 1 and u 2 is a Lipschitz function, then K := {(u 1 (t), u 2 (t)) : a t b} is a null set. Proof. Without loss of any generality we may assume that u 1 is a Lipschitz function on [a, b. Thus, there exists a real number M > such that u 1 (s) u 1 (t) M s t whenever a s, t b. Let ε > be given. Since u 2 is uniformly continuous on [a, b, there exists δ > such that s, t [a, b and s t < δ imply u 2 (s) u 2 (t) < η := ε 4M(b a). Choose k in IN such that h := (b a)/k < δ. Partition the interval [a, b at the equally spaced points t j = a + jh, j =, 1,..., k. Let j := [u 1 (t j ) Mh, u 1 (t j ) + Mh [u 2 (t j ) η, u 2 (t j ) + η, j = 1,..., k. 5

6 If t j 1 t t j for some j, then t t j h < δ. Consequently, u 1 (t) u 1 (t j ) M t t j Mh and u 2 (t) u 2 (t j ) η. Hence, (u 1 (t), u 2 (t)) j. This shows that K k j=1 j. Moreover, we have k j=1 Therefore, K is a null set. A( j ) = k(2mh)(2η) < 4kM b a k ε 4M(b a) = ε. A bounded set in I 2 is called a iemann domain if its boundary Bd() is a null set. Clearly, a null set is a iemann domain. Theorem 3.2. If and F are iemann domains in I 2, then F, F, and \ F are all iemann domains. Proof. It suffices to show that the following three relations hold for two subsets and F of a metric space: Bd( F ) Bd() Bd(F ), Bd( F ) Bd() Bd(F ), Bd(\F ) Bd() Bd(F ). First, since F = F and ( F ) F, we have Bd( F ) = F \ ( F ) ( F ) \ ( F ) = ( \ ( F )) (F \ ( F )). But \ ( F ) \ = Bd() and F \ ( F ) F \ F = Bd(F ). This shows that Bd( F ) Bd() Bd(F ). Second, since F F and F = ( F ), we have Bd( F ) = F \ ( F ) ( F ) \ ( F ) = (( F ) \ ) (( F ) \ F ). But ( F ) \ \ = Bd() and ( F ) \ F F \ F = Bd(F ). This shows that Bd( F ) Bd() Bd(F ). Third, we observe that Bd( \ F ) = \ F \ ( \ F ) ( \ ) ( \ F ) (F \ F ) F. Since \ F is an open set and \ F \ F, we have \ F ( \ F ). Moreover, since F is an open set and F ( \ F ) F ( \ F ) =, we have F ( \ F ) c. This shows that ( \ F ) Bd( \ F ) = and F Bd( \ F ) =. Therefore, Bd( \ F ) ( \ ) (F \ F ) = Bd() Bd(F ). The proof of the theorem is complete. 6

7 4. Integrals on General Domains In this section we study integrals on general domains. First, we establish the following theorem which is an extension of Theorem 1.2. Theorem 4.1. Let f be a bounded, real-valued function defined on a closed rectangle. Let denote the set of points in where f is discontinuous. If is a null set, then f is integrable on. Proof. Since f is bounded, there exists a positive number M such that M f(x, y) M for all (x, y). Let ε > be given. Since is a null set, there exist closed rectangles 1,..., s such that s r=1 r and s r=1 A( r) < ε. Without loss of any generality we may assume that r for r = 1,..., s. Let K := \ s r=1 r. Then K is a bounded closed set. Since f is continuous on K, there exists some δ > such that (x, y), (x, y ) K and (x x ) 2 + (y y ) 2 < δ imply f(x, y) f(x, y ) < ε. Let P = {C ij : i = 1,..., m, j = 1,..., n} be a partition of such that d(c ij ) < δ for all i = 1,..., m and j = 1,..., n and that each r (r = 1,..., s) is the union of certain cells C ij. Let m ij := inf{f(x, y) : (x, y) C ij } and M ij := sup{f(x, y) : (x, y) C ij }. Then we have U(f, P ) L(f, P ) = m i=1 j=1 n (M ij m ij ) A ij = (i,j) Γ (M ij m ij ) A ij, where Γ is the index set {(i, j) : i = 1,..., m, j = 1,..., n}. Let Γ 1 be the set of all those indices (i, j) for which C ij is a subset of some r. Then Γ = Γ 1 Γ 2, where Γ 2 := Γ \ Γ 1. If (i, j) Γ 2, then C ij ( s r=1 r) =. In other words, C ij K whenever (i, j) Γ 2. Thus, M ij m ij < ε for (i, j) Γ 2. Consequently, (i,j) Γ 2 (M ij m ij ) A ij < ε (i,j) Γ 2 A ij εa, where A = A() is the area of the rectangle. Furthermore, since M f(x, y) M for all (x, y), we have But (i,j) Γ1 C ij s r=1 r. Hence, (i,j) Γ 1 (M ij m ij ) A ij 2M (i,j) Γ 1 A(C ij ) 7 (i,j) Γ 1 A(C ij ). s A( r ) < ε. r=1

8 Combining the above estimates, we obtain (i,j) Γ (M ij m ij ) A ij = (i,j) Γ 1 (M ij m ij ) A ij + Therefore, by Theorem 1.1 we conclude that f is integrable. (i,j) Γ 2 (M ij m ij ) A ij (A + 2M)ε. Let f be a bounded real-valued function defined on a bounded subset of I 2. Choose a closed rectangle such that. Let f be the function on given by f(x, y) := { f(x, y) for (x, y), for (x, y) \. If f is integrable on, then we say that f is integrable on and define f(x, y) da := f(x, y) da. vidently, the above definition is independent of the choice of the rectangle. If is a null set, then f(x, y) da =. To prove this assertion, we choose a closed rectangle such that. Let f be the function on defined by f(x, y) = f(x, y) for (x, y) and f(x, y) = for (x, y) \. Since \ is an open set, f is continuous on \. Moreover, is a null set because is a null set. Thus, the set of points in where f is discontinuous is a null set. In light of the proof of Theorem 4.1, f is integrable on and f(x, y) da =. The following theorem is an extension of Theorem 1.3 to integrals on general domains. Theorem 4.2. Let f and g be integrable functions on a bounded set in I 2 and let c be a real number. Then (1) cf is integrable on and (cf)(x, y) da = c f(x, y) da; (2) f + g is integrable on and (f + g)(x, y) da = f(x, y) da + g(x, y) da; (3) if f(x, y) g(x, y) for all (x, y), then f(x, y) da g(x, y) da. The following theorem gives a useful property of integrals. Theorem 4.3. Let f be a bounded function on = 1 2, where 1 and 2 are bounded sets in I 2 such that 1 2 is a null set. If f is integrable on both 1 and 2, then f is integrable on and f(x, y) da = f(x, y) da + 1 f(x, y) da. 2 8

9 Proof. Choose a closed rectangle such that. Let g, g 1, g 2, g 3 be the functions on defined as follows: g(x, y) := { f(x, y) for (x, y), for (x, y) \, g 1 (x, y) := g 2 (x, y) := g 3 (x, y) := { f(x, y) for (x, y) 1, for (x, y) \ 1, { f(x, y) for (x, y) 2, for (x, y) \ 2, { f(x, y) for (x, y) 1 2, for (x, y) \ ( 1 2 ). Then g = g 1 + g 2 + g 3. By our assumption, g 1 and g 2 are integrable on, g 1 (x, y) da = f(x, y) da 1 and g 2 (x, y) da = f(x, y) da. 2 Moreover, since 1 2 is a null set, g 3 is integrable on and g 3(x, y) da =. Therefore, g = g 1 + g 2 + g 3 is integrable on and f(x, y) da = This established the desired result. g(x, y) da = g 1 (x, y) da + g 2 (x, y) da. Theorem 4.4. Let be a bounded set and G an open set in I 2. Suppose that G and \ G is a null set. If f is a bounded function on and if f is continuous on G, then f is integrable on. Proof. Choose a closed rectangle such that. Let f be the function on given by f(x, y) := f(x, y) for (x, y) and f(x, y) := for (x, y) \. Let K be the set of those points in where f is discontinuous. By our assumption f is continuous on the open set G. Moreover, f(x, y) = for all (x, y) \. Hence, f is continuous on the open set \. Furthermore, f is continuous on Bd(). Therefore, K \ G. By our assumption, \ G is a null set. Consequently, K is a null set. By Theorem 4.1, f is integrable on. In other words, f is integrable on. 9

10 5. valuation of Double Integrals In this section we discuss how to reduce double integrals on iemann domains to repeated integrals. A set in I 2 is said to be y-simple if it can be represented as = { (x, y) I 2 : a x b, φ 1 (x) y φ 2 (x) }, where φ 1 and φ 2 are continuous functions on a bounded closed interval [a, b. It is easily seen that is a iemann domain. Let f be a bounded function on. If f is continuous on, then f is integrable on and f(x, y) da = b [ φ2 (x) a φ 1 (x) f(x, y) dy dx. In order to prove this formula, we choose = [a, b [c, d, where c := inf{φ 1 (x) : a x b} and d := sup{φ 2 (x) : a x b}. Let f be the function on defined by f(x, y) = f(x, y) for (x, y) and f(x, y) = for x \. By Theorem 2.1 we have f(x, y) da = b a [ d c f(x, y) dy dx. For each fixed x [a, b, f(x, y) is an integrable function of y on [c, d. But f(x, y) = f(x, y) for φ 1 (x) y φ 2 (x) and f(x, y) = for c y < φ 1 (x) or φ 2 (x) < y d. Hence, d c f(x, y) dy = φ2 (x) φ 1 (x) f(x, y) dy. Therefore we obtain f(x, y) da = f(x, y) da = b [ φ2 (x) a φ 1 (x) f(x, y) dy dx. A set in I 2 is said to be x-simple if it can be represented as = { (x, y) I 2 : c y d, ψ 1 (y) x ψ 2 (y) }, where ψ 1 and ψ 2 are continuous functions on a bounded closed interval [c, d. Let f be a bounded function on. If f is continuous on, then f is integrable on and d [ ψ2 (y) f(x, y) da = f(x, y) dx dy. c 1 ψ 1 (y)

11 xample 1. valuate the double integral (2xy + y 2 ) da, where is the triangle in I 2 with vertices (, ), (1, ), and (1, 2). Solution. The domain can be described as Thus, the double integral is evaluated as (2xy + y 2 ) da = = {(x, y) I 2 : x 1, y 2x}. = 1 [ 2x 1 (2xy + y 2 ) dy dx = 2 3 x3 dx = xample 2. valuate the repeated integral 1 1 y 2 ye x [ 5 3 x4 1 2 = 5 3. dx dy. 1 [xy 2 + y3 y=2x 3 dx y= Solution. Note that the integral e x2 dx cannot be expressed in a closed form. We may write this repeated integral as a double integral: 1 1 y 2 ye x 2 dx dy = ye x2 da, where = {(x, y) I 2 : y 1, y 2 x 1}. The domain is also y-simple and can be described as = {(x, y) I 2 : x 1, y x}. We therefore have x = = 1 y 2 ye x 2 dx dy = ye x2 dy dx = x 2 2 ex dx = ye x2 da 1 [ y 2 2 y= x 2 ex y= [ 1 4 ex2 1 = 1 (e 1). 4 dx xample 3. Let r 1 and r 2 be two real numbers such that < r 1 < r 2. valuate the double integral xy da, where = {(x, y) I2 : r1 2 x 2 + y 2 r2, 2 y }. 11

12 Solution. We observe that = 1 2 3, where 1 := { (x, y) I 2 : r 2 x r 1, y r2 2 x2}, 2 := { (x, y) I 2 : r 1 x r 1, r1 2 x2 y r2 2 x2}, 3 := { (x, y) I 2 : r 1 x r 2, y r2 2 x2}. By Theorem 4.3, xy da = xy da + xy da + xy da ach of the domains 1, 2, and 3 is y-simple. We have xy da = 1 xy da = 2 xy da = 3 r1 r 2 r1 r 1 r2 r 1 [ r 2 2 x2 [ r 2 2 x2 r 2 1 x 2 [ r 2 2 x2 r1 xy dy dx = xy dy dx = xy dy dx = r1 r x(r2 2 x 2 ) dx = (r2 2 r1) 2 2, 8 r x(r2 2 r 2 1) dx =, r2 r x(r2 2 x 2 ) dx = (r2 2 r1) The value of xy ds is the sum of these three numbers. Therefore xy da =. 6. Area The area of a iemann domain in I 2 is defined to be A() := 1 da. This integral is well defined, since the constant function 1 on a iemann domain is integrable. Clearly, if is a null set, then A() =. Moreover, if 1 and 2 are iemann domains, then A( 1 2 ) = A( 1 ) + A( 2 ) A( 1 2 ). Two sets 1 and 2 in I 2 are said to be nonoverlapping if ( 1 2 ) =. If 1 and 2 are two nonoverlapping iemann domains, then Bd( 1 2 ) is a null set and ( 1 2 ) = ; hence, A( 1 2 ) =. Consequently, A( 1 2 ) = A( 1 )+A( 2 ). More generally, if 1,..., m are mutually nonoverlapping iemann domains and = m i=1 i, then A() = m A( i ). i=1 12

13 Theorem 6.1. Let be a iemann domain in I 2. For any given ε > there exist two nonoverlapping closed sets G and H in I 2 such that (1) ach of the sets G and H is the union of finitely many nonoverlapping squares in I 2, (2) G G H, and (3) A(H) < ε. Proof. Let ε > be given. Since is a iemann domain, its boundary Bd() is a null set. Hence, there exists a finite collection { 1, 2,..., m } of closed rectangles such that Bd() m k=1 k and m k=1 A( k) < ε/4. Suppose that k = [a k, b k [c k, d k, k = 1,..., m. Choose h > such that h < min{(b k a k )/2, (d k c k )/2} for all k = 1,..., m. Consider squares Q ij := [ih, (i + 1)h [jh, (j + 1)h, (i, j) Z 2. These squares are mutually nonoverlapping and their union is I 2. For each k {1,..., m}, let I k be the set of those indices (i, j) for which Q ij k. Let H k := (i,j) Ik Q ij. If (x, y) Q ij and Q ij k, then a k h x b k + h and c k h y d k + h. This shows that H k [a k h, b k + h [c k h, d k + h. It follows that A(H k ) (b k a k + 2h)(d k c k + 2h) 2(b k a k )2(d k c k ) = 4A( k ). Let H := m k=1 H k. Then m m A(H) A(H k ) 4 A( k ) < ε. k=1 k=1 Now consider those indices (i, j) / I := m k=1 I k. If (i, j) / I, then Q ij Bd() Q ij ( m k=1 k ) =. Thus, Q ij does not contain any boundary point of. If Q ij, then Q ij does not contain any exterior point, for otherwise the line segment joining an interior point of and an exterior point of must intersect the boundary of. In other words, Q ij. Let G be the union of those squares Q ij for which (i, j) / I and Q ij. Then G and H are nonoverlapping, G and G H, as desired. For a vector v in I 2, we use T v to denote the mapping from I 2 to I 2 given by T v x = x + v, x I 2. We call T v the translation by v. The following theorem asserts that the area is invariant under translation. 13

14 Theorem 6.2. Let be a iemann domain in I 2, and let v I 2. Then T v () is a iemann domain and A ( T v () ) = A(). Proof. We observe that T v is a one-to-one continuous mapping from I 2 onto I 2 and its inverse mapping is T v. Thus, T v ( ) is the interior of T v () and T v (Bd()) is the boundary of T v (). We write + v for T v (). Let ε > be given. By Theorem 6.1, there exist two nonoverlapping closed sets G and H in I 2 such that each of the sets G and H is the union of finitely many nonoverlapping squares in I 2, G G H, and that A(H) < ε. If Q is a square, then Q + v is a square and A(Q + v) = A(Q). Since each of the sets G and H is the union of finitely many nonoverlapping squares in I 2, we have A(G + v) = A(G) and A(H + v) = A(H) < ε. But Bd( + v) = Bd() + v H + v. Consequently, Bd( + v) is a null set and hence + v is a iemann domain. Moreover, we have A(G) A() A(G H) < A(G) + ε and A(G + v) A( + v) A ( (G + v) (H + v) ) < A(G + v) + ε. Since A(G) = A(G + v), we deduce that for any ε >, ε < A( + v) A() < ε. Therefore, A( + v) = A(). A linear mapping L on I 2 has the form [ [ [ x1 a11 a L = 12 x1, x 2 a 21 a 22 x 2 [ x1 x 2 I 2. We have det L = a 11 a 12 a 21 a 22 = a 11a 22 a 12 a 21. If det L, then we say that L is nonsingular. Clearly, L is nonsingular if and only if L is a bijective mapping on I 2. If there is a real number λ such that [ x1 L = x 2 [ [ λ x1 1 x 2 or 14 [ x1 L = x 2 [ [ 1 x1, λ x 2

15 then L is called an elementary linear mapping of the first type. Let = [a, b [c, d be a closed rectangle, If L(x 1, x 2 ) = (λx 1, x 2 ), then L() = [λa, λb [c, d for λ > or L() = [λb, λa [c, d for λ <. In both cases we obtain This is also true if L(x 1, x 2 ) = (x 1, λx 2 ). A ( L() ) = λ (b a)(d c) = det L A(). A mapping L on I 2 is called an elementary linear mapping of the second type, if there is a real number µ such that [ [ [ x1 1 µ x1 L = 1 x 2 If L(x 1, x 2 ) = (x 1 + µx 2, x 2 ), we have x 2 or [ x1 L = x 2 [ [ 1 x1. µ 1 x 2 L() = {(x 1, x 2 ) : c x 2 d, a + µx 2 x 1 b + µx 2 }. This is a x-simple domain and its area is A ( L() ) = (b a)(d c) = det L A(). The above relation is also valid if L(x 1, x 2 ) = (x 1, x 2 + µx 1 ). A mapping L on I 2 is called an elementary linear mapping of the third type if L(x 1, x 2 ) = (x 2, x 1 ). In this case, L() = [c, d [a, b. It follows that A ( L() ) = (b a)(d c) = det L A(). To summarize, we have proved that for every elementary linear mapping L and every rectangle in I 2, A ( L() ) = det L A(). Note that a nonsingular linear mapping can be represented as a composition of finitely many elementary linear mappings. Theorem 6.3. Let be a iemann domain in I 2. If L is a linear mapping on I 2, then L() is a iemann domain and A ( L() ) = det L A(). Proof. If det L =, then L() is included in a line segment. Hence, L() is a null set and A(L()) = = det L A(). In what follows, we assume that det L, i.e., L is a nonsingular linear mapping. 15

16 Suppose that L is an elementary linear mapping on I 2. Since is a iemann domain, by Theorem 6.1 there exist two nonoverlapping closed sets G and H in I 2 such that each of the sets G and H is the union of finitely many nonoverlapping squares in I 2, G G H, and that A(H) < ε. If Q is a square, then A(L(Q)) = det L A(Q). Since each of the sets G and H is the union of finitely many nonoverlapping squares in I 2, we have A(L(G)) = det L A(G) and A(L(H)) = det L A(H) < det L ε. But Bd(L()) = L(Bd()) L(H). Consequently, Bd(L()) is a null set and hence L() is a iemann domain. Moreover, we have det L A(G) det L A() < det L A(G) + det L ε and det L A(G) = A(L(G)) A(L()) A ( L(G) L(H) ) < det L A(G) + det L ε. We deduce that for any ε >, det L ε < A(L()) det L A() < det L ε. Therefore, A(L()) = det L A(). Finally, suppose that L is a nonsingular linear mapping on I 2. represented as L = L k L 2 L 1, Then L can be where L 1, L 2,..., L k are elementary linear mappings on I 2. Let j := L j L 1 () for j = 1, 2,..., k. An inductive argument shows that 1, 2,..., k are iemann domains. Moreover, by what has been proved A( j ) = det L j A( j 1 ), j = 1,..., k, where :=. Consequently, A(L()) = A( k ) = det L k det L 2 det L 1 A() = det L A(), where we have used the fact det L = det L k det L 2 det L 1 to derive the last equality. A mapping S from I 2 to I 2 is called an affine mapping if there exist a linear mapping L on I 2 and a vector v in I 2 such that Sx = Lx + v, x I 2. If is a iemann domain in I 2, then Theorems 6.2 and 6.3 tell us that S() is a iemann domain and A ( S() ) = det L A(). 16

17 7. Smooth Mappings In this section we investigate the action of a continuously differentiable mapping on iemann domains. Let φ = (φ 1, φ 2 ) be a mapping from an open set U in I 2 to I 2. Suppose that the partial derivatives D 1 φ 1, D 2 φ 1, D 1 φ 2, and D 2 φ 2 exist and are continuous on U. Jacobian matrix of φ at x U is [ D1 φ Dφ(x) := 1 (x) D 2 φ 1 (x). D 1 φ 2 (x) D 2 φ 2 (x) If x, y U and Dφ(z) M for all z in the line segment [x, y, then the mean value theorem tells us that φ(x) φ(y) M x y. Let ρ be the metric of the uclidean plane I 2. For a subset of I 2 and x I 2, define ρ(x, ) := inf{ρ(x, y) : y }. For r >, let B r () := {x I 2 : ρ(x, ) < r}. Then B r () is an open set and B r () = {x I 2 : ρ(x, ) r}. It can be easily verified that B r () B r (Bd()). Moreover, if is a line segment of length b, then The A(B r ()) < (b + 2r)2r. Theorem 7.1. Let φ be a continuously differentiable mapping from an open set U in I 2 to I 2. If is a iemann domain in I 2 such that U, then φ() is a iemann domain and A ( φ() ) Jφ (x 1, x 2 ) da. Proof. Choose r > such that F := B r () U. Let M := sup{ Dφ(x) : x F }. Then M < because Dφ is continuous on the compact set F. For x, y F we have φ(x) φ(y) M x y. Let ε > be given. There exists δ > such that x, y F and x y < δ imply Dφ(x) Dφ(y) < ε. Let Q be a square of side length h such that < h < δ/ 2 and Q F. Choose an arbitrary point a Q. Let S a be the affine mapping on I 2 given by S a x := φ(a) + Dφ(a)(x a), x I 2. 17

18 Then S a (Q) is a parallelogram and its area A(S a (Q)) = J φ (a) A(Q), by Theorem 6.3. Let ψ := φ S a. Then ψ(a) = and Dψ(x) = Dφ(x) Dφ(a). For x Q we have x a 2h < δ; hence, Dψ(x) = Dφ(x) Dφ(a) < ε. It follows that φ(x) S a (x) = ψ(x) ψ(a) ε x a < ε 2h x Q. In other words, φ(x) B ε 2h (S a (Q)) for all x Q. Consequently, φ(q) B ε 2h (S a (Q)) S a (Q) B ε 2h (Bd(S a (Q))). Note that Bd(Q) is the union of four line segments of length h. Hence, S a (Bd(Q)) is the union of four line segments of length Mh. But Bd(S a (Q)) = S a (Bd(Q)). Therefore, the area of the iemann domain B ε 2h (Bd(S a (Q))) is less than 4(Mh + 2ε 2h)2ε 2h. We thereby obtain A(φ(Q)) J φ (a) h 2 + M εh 2 = [ J φ (a) + M ε A(Q), ( ) where M := 8 2(M + 2 2ε). Let N := sup{ J φ (x) : x F }. Then N <. Let ε, δ, M, M and N be given as above. Choose h such that < h < min{r, δ}/ 2. Consider squares of the form C ij := [ih, (i + 1)h [jh, (j + 1)h for (i, j) Z 2. Let Γ := {(i, j) Z 2 : C ij }. In light of our choice of h, C ij F whenever (i, j) Γ. Moreover, (i,j) Γ C ij. It follows that φ() (i,j) Γ φ(c ij ). For each (i, j) Γ, choose a ij C ij such that J φ (a ij ) = inf{ J φ (x) : x C ij }. By the estimate ( ) we have A(φ(C ij )) [ J φ (a ij ) + M ε A(C ij ). Let Γ 1 := {(i, j) Z 2 : C ij Bd() } and H := (i,j) Γ1 C ij. 18

19 Since is a iemann domain, \ = Bd() is a null set. Therefore, h > can be chosen so small that A(H) < ε. Consequently, [ A(φ(C ij )) Jφ (a ij ) + M ε A(C ij ) (N + M ε)a(h) < ε(n + M ε). (i,j) Γ 1 (i,j) Γ 1 It follows that A(φ(Bd())) < ε(n + M ε) whenever ε >. Hence, φ(bd()) is a null set. Let Γ 2 := Γ \ Γ 1. For (i, j) Γ 2 we have C ij. Furthermore, A(φ()) A(φ(C ij )) = A(φ(C ij )) + A(φ(C ij )). (i,j) Γ (i,j) Γ 1 (i,j) Γ 2 The first sum was estimated above. For the second sum we have the following estimate: A(φ(C ij )) J φ (a ij ) A(C ij ) + M ε A(C ij ). (i,j) Γ 2 (i,j) Γ 2 (i,j) Γ 2 Note that (i,j) Γ 2 A(C ij ) A(). Since J φ (a ij ) J φ (x) for all x C ij, we have J φ (a ij ) A(C ij ) J φ (x 1, x 2 ) da J φ (x 1, x 2 ) da. (i,j) Γ C ij 2 (i,j) Γ 2 Combining the above estimates, we obtain A(φ()) J φ (x 1, x 2 ) da + M εa() + ε(n + M ε). Since this estimate is valid whenever ε >, we conclude that A(φ()) J φ (x 1, x 2 ) da. It remains to show that φ() is a iemann domain. Let V := {x U : J φ (x) } and K := {x : J φ (x) = }. Then V is an open set, while K is a closed set. Clearly, \ K is an open set contained in V. Since J φ (x) for all x V, φ V is an open mapping, by the inverse mapping theorem. Thus, φ( \ K) is an open set contained in φ(). It follows that φ( \ K) (φ()). Moreover, since φ is a continuous mapping on the compact set, we have φ() = φ(). Consequently, Bd(φ()) = φ() \ (φ()) φ() \ φ( \ K) φ( \ ) φ(k). We have shown that φ( \ ) is a null set. To prove that φ(k) is a null set, we set Γ 3 := {(i, j) Z 2 : C ij K }. If (i, j) Γ 3, then J φ (a ij ) =. By the estimate ( ) we have A(φ(C ij )) M εa(c ij ). Hence, A(φ(K)) A(φ(C ij )) M ε A(C ij ) M εa(f ). (i,j) Γ 3 (i,j) Γ 3 This shows that φ(k) is a null set. Consequently, Bd(φ()) is a null set. 19

20 8. Change of Variables in Double Integrals In this section we establish a general formula for change of variables in double integrals. Theorem 8.1. Let U be an open set in I 2, and let be a closed iemann domain such that U. Suppose that φ is a continuously differentiable mapping from U to I 2. If f is a nonnegative continuous function on the domain φ(), then f(u 1, u 2 ) du 1 du 2 φ() f ( φ(x 1, x 2 ) ) J φ (x 1, x 2 ) dx 1 dx 2. Proof. Let ε > be given. Since f φ is continuous on the compact set, there exists δ > such that x, y and x y < δ imply f(φ(x)) f(φ(y)) < ε. Partition the domain into mutually disjoint iemann domains 1, 2,..., n such that = n j=1 j and the diameter of each domain j is less than δ. For j = 1,..., n, let M j := sup{f(φ(x)) : x j } and m j := inf{f(φ(x)) : x j }. Since the diameter of each domain j is less than δ, we have M j m j ε for j = 1,..., n. By our assumption, f is nonnegative. Moreover, φ() n j=1 φ( j). Hence, φ() f(u 1, u 2 ) du 1 du 2 n By Theorem 7.1 we assert that n M j A(φ( j )) j=1 φ( j ) j=1 j=1 n f(u 1, u 2 ) du 1 du 2 n M j A(φ( j )). j=1 j M j J φ (x 1, x 2 ) dx 1 dx 2. Since M j m j + ε f(φ(x 1, x 2 )) + ε for all (x 1, x 2 ) j, we obtain φ() n f(u 1, u 2 ) du 1 du 2 M j A(φ( j )) j=1 [ f ( φ(x1, x 2 ) ) + ε J φ (x 1, x 2 ) dx 1 dx 2. The desired result follows after letting ε +. Theorem 8.2. Suppose that φ is a one-to-one mapping from an open set U in I 2 onto an open set V in I 2. Let be a closed iemann domain such that U. If both φ and 2

21 φ 1 are continuously differentiable, and if f is a continuous function on the domain φ(), then φ() f(u 1, u 2 ) du 1 du 2 = f ( φ(x 1, x 2 ) ) J φ (x 1, x 2 ) dx 1 dx 2. Proof. First, consider the case that f is nonnegative. By Theorem 8.1, φ() f(u 1, u 2 ) du 1 du 2 f ( φ(x 1, x 2 ) ) Jφ (x 1, x 2 ) dx1 dx 2. On the other hand, applying Theorem 8.1 to the mapping φ 1 and the function (f φ) J φ on = φ 1 (φ()), we obtain φ 1 (φ()) φ() f ( φ(x 1, x 2 ) ) Jφ (x 1, x 2 ) dx1 dx 2 f ( φ φ 1 (u 1, u 2 ) ) Jφ (φ 1 (u 1, u 2 )) Jφ 1(u 1, u 2 ) du1 du 2. But φ φ 1 is the identity mapping on V. By the chain rule we have Jφ (φ 1 (u 1, u 2 )) Jφ 1(u 1, u 2 ) = 1. Consequently, f ( φ(x 1, x 2 ) ) J φ (x 1, x 2 ) dx 1 dx 2 φ() f(u 1, u 2 ) du 1 du 2. Thus, the change of variable formula as stated in the theorem is valid for the case that f is nonnegative. For the general case, we may write f = f + f, where f + := ( f + f)/2 and f := ( f f)/2. Then both f + and f are nonnegative continuous functions on φ(). The change of variable formula is valid for both f + and f. Therefore, we conclude that it is also valid for f. For the special case that f = 1 on φ(), Theorem 8.2 yields the following result: A(φ()) = J φ (x 1, x 2 ) dx 1 dx 2. The following stronger version of Theorem 8.2 is often used in applications of change of variables for double integrals. 21

22 Theorem 8.3. Suppose that φ is a one-to-one mapping from an open iemann domain U in I 2 onto an open iemann domain V in I 2. Let f be a bounded continuous function on V. If both φ and φ 1 are continuously differentiable, and if J φ is bounded on U, then f(u 1, u 2 ) du 1 du 2 = V U f ( φ(x 1, x 2 ) ) Jφ (x 1, x 2 ) dx1 dx 2. Proof. Let I 1 and I 2 denote the integral on the left and the right of the above equation respectively. By our assumption, there exists a positive real number M such that f(u 1, u 2 ) M for all (u 1, u 2 ) V and ( f φ(x1, x 2 ) ) J φ (x 1, x 2 ) M for all (x1, x 2 ) U. Let ε > be given. Since U and V are open iemann domains, there exists a compact iemann domain K of U such that A(U \ K) < ε and A(V \ φ(k)) < ε. By Theorem 8.2 we obtain f(u 1, u 2 ) du 1 du 2 = φ(k) On the other hand we have I 1 f(u 1, u 2 ) du 1 du 2 = φ(k) A similar argument shows that I 2 U K V \φ(k) f ( φ(x 1, x 2 ) ) Jφ (x 1, x 2 ) dx1 dx 2. f(u 1, u 2 ) du 1 du 2 MA(V \ φ(k)) < Mε. f ( φ(x 1, x 2 ) ) Jφ (x 1, x 2 ) dx1 dx 2 MA(U \ K) < Mε. Consequently, I 1 I 2 < 2Mε for any ε >. Therefore, I 1 = I 2. We are in a position to investigate double integrals in polar coordinates. Consider the mapping φ : (r, θ) (x, y) from I 2 to I 2 given by x = r cos θ, y = r sin θ, (r, θ) I 2. The Jcobian determinant of φ is J φ (r, θ) = x r y r x θ y θ = cos θ sin θ r sin θ r cos θ = r. Thus J φ (r, θ) for r. But φ is not one-to-one on I 2. Let r 1, r 2, θ 1 and θ 2 be real numbers such that r 1 < r 2 and θ 1 < θ 2 θ 1 + 2π. It is easily seen that φ is one-to-one 22

23 on the open domain U := {(r, θ) : r 1 < r < r 2, θ 1 < θ < θ 2 }. Hence, if f is a bounded continuous function on V := φ(u), then by Theorem 8.3 we obtain θ2 r2 f(x, y) dx dy = f(r cos θ, r sin θ) r dr dθ. V θ 1 r 1 In particular, if r 1 =, θ 1 =, and θ 2 = 2π, then V = V 1 \ {(x, y) : x 1, y = }, where V 1 is the open disk {(x, y) I 2 : x 2 + y 2 r 2 2}. Hence, for a bounded continuous function f on the disk V 1 we have f(x, y) dx dy = f(x, y) dx dy = V 1 V 2π xample 1. valuate the double integral e x2 y 2 dx dy, V r2 where V is the open disk {(x, y) : x 2 + y 2 < b 2 } with b >. f(r cos θ, r sin θ) r dr dθ. Solution. By using polar coordinates we obtain 2π b f(x, y) dx dy = e r2 r dr dθ = 2π [ e r2 /2 b = π(1 ). e b2 V xample 2. valuate the double integral xy(x 2 + y 2 ) dx dy, where is the domain in the first quadrant bounded by the curves x 2 y 2 = 1, x 2 y 2 = 4, xy = 1, and xy = 2. Solution. Let ψ : (x, y) (u, v) be the mapping from I 2 to I 2 given by u = x 2 y 2, v = 2xy, (x, y) I 2. The mapping ψ is not one-to-one on I 2, but it is one-to-one on the first quadrant. Indeed, we have x 2 + y 2 = u 2 + v 2. Hence, if x > and y >, then x = ( u 2 + v 2 + u)/2 and y = ( u 2 + v 2 u)/2. Let φ be the mapping (u, v) (x, y) as given above. Then φ : ψ() is the inverse of the mapping ψ : ψ(). Note that ψ() = {(u, v) : 1 u 4, 2 v 4}. We have u u J ψ (x, y) = x y = 2x 2y 2y 2x = 4(x2 + y 2 ). v x v y 23

24 It follows that Therefore, J φ (u, v) = xy(x 2 + y 2 ) dx dy = ψ() 1 J ψ (x, y) = 1 4(x 2 + y 2 ) = 1 4 u 2 + v 2. v u2 + v u 2 + v du dv = xample 3. For a >, find the area of the domain Q := {(x, y) I 2 : x 2/3 + y 2/3 a 2/3 }. 4 1 [ 4 2 v dv du = 9 4. Solution. Consider the mapping φ : (r, t) (x, y) given by x = r cos 3 t, y = r sin 3 t, (r, t) I 2. Let := {(r, t) I 2 : r a, t 2π}. It is easily seen that φ() = Q. The Jacobian determinant of φ is J φ (r, t) = x r y r x t y t = cos 3 t sin 3 t Therefore, the area of the domain Q is It follows that A(Q) = A(φ()) = J φ (r, t) dr dt = 3r cos 2 t sin t 3r sin 2 t cos t = 3r sin2 t cos 2 t. 2π [ a 3r dr sin 2 t cos 2 t dt. A(Q) = 3a2 2 2π sin 2 (2t) 4 dt = 3a2 8 2π 1 cos 4t 2 dt = 3a2 [ 2π t sin 4t/4 = πa2. 24