MA 105: Calculus Lecture 17 . Prof. B.V. Limaye IIT Bombay - - PowerPoint PPT Presentation

. .

MA 105: Calculus Lecture 17

• Prof. B.V. Limaye

IIT Bombay Wednesday, 8 March 2017

B.V. Limaye, IITB MA 105: Lec-17

. . Double Integral on a Rectangle

The concept of the Riemann integral of a function was motivated by our attempt to find ‘the area under a curve’. We now look for a concept which will let us find ‘the volume under a surface’. We shall assume that the volume of a cuboid [a, b]×[c, d]×[p, q] is equal to (b − a)(d − c)(q − p). Let R := [a, b] × [c, d] be a rectangle in R2 with a < b, c < d. Let f : R → R be a bounded function. Define m(f ) :=inf{f (x, y):(x, y) ∈ R}, M(f ) :=sup{f (x, y):(x, y) ∈ R}. Let n, k ∈ N, and consider a partition P of R given by P := {(xi, yj) : i = 0, 1, . . . , n and j = 0, 1, . . . , k}, where a := x0 < x1 < · · · < xn := b; c = y0 < y1 < · · · < yk = d. The points in P divide the rectangle R into nk nonoverlapping subrectangles [xi−1, xi] × [yj−1, yj], i = 1, . . . , n; j = 1, . . . , k.

B.V. Limaye, IITB MA 105: Lec-17

Define mi,j(f ) := inf{f (x, y) : (x, y) ∈ [xi−1, xi] × [yj−1, yj]}, Mi,j(f ) := sup{f (x, y) : (x, y) ∈ [xi−1, xi] × [yj−1, yj]}. Clearly, m(f ) ≤ mi,j(f ) ≤ Mi,j(f ) ≤ M(f ) for i = 1, . . . , n; j = 1, . . . , k. Define the lower double sum and the upper double sum of f with respect to P by L(P, f ) :=

n

∑

i=1 k

∑

j=1

mi,j(xi − xi−1)(yj − yj−1), U(P, f ) :=

n

∑

i=1 k

∑

j=1

Mi,j(xi − xi−1)(yj − yj−1). Since ∑n

i=1(xi − xi−1) = b − a and ∑k j=1(yj − yj−1) = d − c,

we obtain m(f )(b−a)(d −c) ≤ L(P, f ) ≤ U(P, f ) ≤ M(f )(b−a)(d −c).

B.V. Limaye, IITB MA 105: Lec-17

y x z

Figure : Summands of lower and upper double sums.

B.V. Limaye, IITB MA 105: Lec-17

Let P1 and P2 be partitions of the rectangle R. A lower double sum increases and a upper double sum decreases when the partition is refined. Considering the common refinement of P1 and P2, we obtain L(P1, f ) ≤ U(P2, f ). Define L(f ) := sup{L(P, f ) : P is a partition of R}, U(f ) := inf{U(P, f ) : P is a partition of R}. L(f ) is called the lower double integral of f and U(f ) is called the upper double integral of f . Then L(f ) ≤ U(f ) by using the definitions of sup and inf. . Definition . . A bounded function f : R → R is said to be (double) integrable on R if L(f ) = U(f ).

B.V. Limaye, IITB MA 105: Lec-17

In this case, the double integral of f on R is the common value U(f ) = L(f ), and it is denoted by ∫ ∫

R

f

• r

∫ ∫

R

f (x, y)d(x, y). Let f : R → R be integrable and nonnegative. The double integral of f on R gives the volume of the solid Ef under the surface z = f (x, y) and above the rectangle R. Examples: (i) Let f (x, y) := 1 for all (x, y) ∈ R. Then L(P, f ) = U(P, f ) = (b − a)(d − c) for every partition P of

• R. Hence f is integrable on R, and its double integral on R is

equal to (b − a)(d − c). (ii) Define the bivariate Dirichlet function f : R → R by f (x, y) := { 1 if x and y are rational numbers, if x or y is an irrational number.

B.V. Limaye, IITB MA 105: Lec-17

Then f is a bounded function on R. For each partition P of R, mi,j(f ) = 0 and Mi,j(f ) = 1, i = 1, . . . , n; j = 1, . . . , k, and so L(P, f ) = 0 and U(P, f ) = (b − a)(d − c). Thus L(f ) = 0 and U(f ) = (b − a)(d − c). Since L(f ) ̸= U(f ), f is not integrable. (iii) Let ϕ : [a, b] → R be bounded, and define f : R → R by f (x, y) := ϕ(x) for (x, y) ∈ R. Then f is a bounded function. If P := {(xi, yj) : i = 0, 1, . . . , n and j = 0, 1, . . . , k} is any partition of R, then P1 := {x0, x1, . . . , xn} is a partition of [a, b], and mi,j(f ) = mi(ϕ), i = 1, . . . , n; j = 1, . . . , k, and so

n

∑

i=1 n

∑

j=1

mi,j(f )(xi−xi−1)(yj−yj−1) = (d−c)

n

∑

i=1

mi(ϕ)(xi−xi−1). Thus L(P, f )=(d −c)L(P1, ϕ). Also, U(P, f )=(d −c)U(P1, ϕ).

B.V. Limaye, IITB MA 105: Lec-17

Further, if Q := {x0, x1, . . . , xn} is a partition of [a, b], then Q = P1, where P := {(xi, yj) : i = 0, 1, . . . , n, j = 0, 1}. So L(f ) = (d − c)L(ϕ) and U(f ) = (d − c)U(ϕ). Hence f is integrable on R ⇐ ⇒ ϕ is Riemann integrable on [a, b], and in this case, the double integral of f on R is equal to (d − c) times the Riemann integral of ϕ on [a, b]. The following result allows us to test the integrability of a bounded function on R. . Theorem (Riemann condition) . . Let f : [a, b] → R be a bounded function. Then f is integrable if and only if for every ϵ > 0, there is a partition Pϵ of [a, b] such that U(Pϵ, f ) − L(Pϵ, f ) < ϵ. The proof is very similar to the proof of the result about the Riemann condition in the one variable case.

B.V. Limaye, IITB MA 105: Lec-17

. . Double Integration

The Riemann condition can be used to prove many useful results regarding double integration. .(Domain Additivity) . . Let R := [a, b]×[c, d], and let f : R → R be a bounded

• function. Let s ∈ (a, b), t ∈ (c, d). Then f is integrable on R

if and only if f is integrable on the four subrectangles [a, s]×[c, t], [a, s]×[t, d], [s, b]×[c, t] and [s, b]×[t, d]. In this case, the integral of f on R is the sum of the integrals

• f f on the four subrectangles.

We make the following conventions: If a = b or c = d, then ∫∫

[a,b]×[c,d] f := 0.

∫∫

[b,a]×[c,d] f := −

∫∫

[a,b]×[c,d] f =:

∫∫

[a,b]×[d,c] f , and

∫∫

[b,a]×[d,c] f :=

∫∫

[a,b]×[c,d] f .

B.V. Limaye, IITB MA 105: Lec-17

. . Integrable functions

. . Let f : R → R. (i) If f is monotonic in each of the two variables, then f is integrable on R. (ii) If f is bounded on R, and has at most a finite number of discontinuities in R, then f is integrable on R. Examples: (i) Let f (x, y) := [x] + [y] for (x, y) ∈ R. Since f is increasing in each variable, f is integrable. (ii) Let a, c > 0 and r, s ≥ 0. Define f (x, y) = xry s for (x, y) ∈ R. Since f is continuous on R, it is integrable. (iii) Let f (0, 0) := 0 and f (x, y) := xy/(x2 + y 2) if (x, y) ∈ [−1, 1]×[−1, 1] and (x, y) ̸= (0, 0). Since f is bounded on R, and it is discontinuous only at (0, 0), f is integrable.

B.V. Limaye, IITB MA 105: Lec-17

. . Albebraic and Order Properties

. . Let f , g : R → R be integrable functions. Then (i) f + g is integrable, and ∫∫

R(f + g) =

∫∫

R f +

∫∫

R g.

(ii) α f is integrable, and ∫∫

R α f = α

∫∫

R f for all α ∈ R.

(iii) f · g is integrable. (iv) If there is δ > 0 such that |f (x, y)| ≥ δ for all (x, y) ∈ R (so that 1/f is bounded), then 1/f is integrable. (v) If f ≤ g, then ∫∫

R f ≤

∫∫

R g.

(vi) |f | is integrable, and | ∫∫

R f | ≤

∫∫

R |f |.

Proof: (v) For any partition P of R, U(P, f ) ≤ U(P, g), and so ∫∫

R f = U(f ) ≤ U(g) =

∫∫

R g.

(vi) −|f | ≤ f ≤ |f | = ⇒ − ∫∫

R |f | ≤

∫∫

R f ≤

∫∫

R |f |.

B.V. Limaye, IITB MA 105: Lec-17

. . Evaluation of a Double Integral

Suppose a function is integrable on a rectangle R. How can we find its double integral? To evaluate the Riemann integral of an integrable function f

• n an interval [a, b], we used a powerful result known as the

fundamental theorem of calculus, Part II: If we find a function g on [a, b] whose derivative is equal to f on [a, b], then ∫ b

a

f (x)dx = ∫ b

a

g ′(x)dx = g(b) − g(a). Later in this course, we shall see some versions of the fundamental theorem of calculus for functions of two/three variables, known as the Green/Stokes’ theorem. For the present, we consider an easy and widely used method for evaluating double integrals, namely, reduction of the problem to a repeated evaluation of Riemann integrals.

B.V. Limaye, IITB MA 105: Lec-17

. Theorem (Fubini Theorem on a Rectangle) . . Let R := [a, b]×[c, d], let f : R → R be integrable, and let I denote the double integral of f on R. (i) If for each fixed x ∈ [a, b], the Riemann integral ∫ d

c f (x, y)dy exists, then the iterated integral

∫ b

a

(∫ d

c f (x, y)dy

) dx exists and is equal to I. (ii) If for each fixed y ∈ [c, d], the Riemann integral ∫ b

a f (x, y)dx exists, then the iterated integral

∫ d

c

(∫ b

a f (x, y)dx

) dy exists and is equal to I. (iii) If the hypotheses in both (i) and (ii) above hold, and in particular, if f is continuous on R, then ∫ b

a

(∫ d

c

f (x, y)dy ) dx = I = ∫ d

c

(∫ b

a

f (x, y)dx ) dy.

B.V. Limaye, IITB MA 105: Lec-17

The Fubini theorem can be proved by using the Riemann condition for a double integral and for a Riemann integral. Special case: Let ϕ : [a, b] → R and ψ : [c, d] → R be Riemann integrable. Define f : R → R by f (x, y) := ϕ(x)ψ(y), (x, y) ∈ R. Then f is integrable on R, and its double integral is equal to ∫ b

a

(∫ d

c

ϕ(x)ψ(y)dy ) dx = (∫ b

a

ϕ(x)dx ) (∫ d

c

ψ(y)dy ) . In particular, if r, s ∈ R with r ≥ 0 and s ≥ 0, then ∫∫

[a,b]×[c,d]

xry sd(x, y) = (br+1 − ar+1 r + 1 ) (ds+1 − cs+1 s + 1 ) , provided 0 < a < b and 0 < c < d.

B.V. Limaye, IITB MA 105: Lec-17

Geometrically, the Fubini theorem says that if f is a nonnegative integrable function on R := [a, b]×[c, d], then the volume of the solid D under the surface z = f (x, y) and above the rectangle R can be found by the slice method: Find the areas A(x) := ∫ d

c f (x, y)dy, x ∈ [a, b], of the cross-sections of

D perpendicular to the x-axis. Alternatively, find the areas B(y) := ∫ b

a f (x, y)dx, y ∈ [c, d], of the cross-sections of D

perpendicular to the y-axis. Then Vol (D) = ∫ b

a

A(x)dx = ∫ d

c

B(y)dy. Examples: (i) Let R := [0, 1]×[0, 1], and f (x, y):=(x + y)2, (x, y) ∈ R. Then f is continuous on R. The double integral of f on R is ∫ 1 (∫ 1 (x + y)2dx ) dy = 1 3 ∫ 1 (x + y)3 1

0 dy

= 1 3 ∫ 1 ( (1 + y)3 − y 3) dy = 7 6.

B.V. Limaye, IITB MA 105: Lec-17

(ii) Let R := [0, 1]×[0, 1], f (0, 0) := 0, and for (x, y) ̸= (0, 0), let f (x, y) := xy(x2 − y 2)/(x2 + y 2)3. For x ∈ [0, 1], x ̸= 0, A(x) := ∫ 1 f (x, y)dy = ∫ 1 xy(x2 − y 2) (x2 + y 2)3 dy = x 2(1 + x2)2, and also, A(0) = ∫ 1

0 0dy = 0. Hence

∫ 1 (∫ 1 f (x, y)dy ) dx = ∫ 1 A(x)dx = ∫ 1 x 2(1 + x2)2dx = 1 8. By interchanging x and y, ∫ 1 (∫ 1 f (x, y)dx ) dy = −1 8. Thus the two iterated integrals exist, but they are not equal. Note that since f (2/n, 1/n) = 6n2/125 for all n ∈ N, the function f is not bounded on R, and so it is not be integrable

• n R. Thus Fubini’s theorem is not applicable.

B.V. Limaye, IITB MA 105: Lec-17

. . Double Riemann Sum

If we are not able to evaluate the double integral exactly, we attempt to find its approximations. Given a bounded function f : R → R, and a partition P := {(xi, yj) : i = 0, 1, . . . , n and j = 0, 1, . . . , k}, of R := [a, b]×[c, d], a double sum of the form S(P, f ) :=

n

∑

i=1 k

∑

j=1

f (si, tj)(xi − xi−1)(yj − yj−1), where si ∈ [xi−1, xi], tj ∈ [yj−1, yj], is called a double Riemann sum for f corresponding to P. Note: L(P, f ) ≤ S(P, f ) ≤ U(P, f ) for any s1, . . . , sn ∈ [a, b] and t1, . . . , tk ∈ [c, d].

B.V. Limaye, IITB MA 105: Lec-17

Define the mesh of a partition P := {(xi, yj) : i = 0, 1, . . . , n and j = 0, 1, . . . , k} by µ(P) := max{x1 − x0, . . . , xn − xn−1, y1 − y0, . . . , yk − yk−1}. . . Theorem: Let f be integrable on R, and let ϵ > 0. Then there is δ > 0 such that U(P, f ) − L(P, f ) < ϵ for every partition P satisfying µ(P) < δ. Corollary: Let (Pn) is a sequence of partitions of R such that µ(Pn) → 0. Then U(Pn, f ) − L(Pn, f ) → 0. Further, if S(Pn, f ) is a double Riemann sum corresponding to Pn and f , then S(Pn, f ) → ∫∫

R f .

Proof: Given ϵ > 0, find δ > 0 as in the theorem, and let n0 ∈ N be such that µ(Pn) < δ for all n ≥ n0, so that U(Pn, f ) − L(Pn, f ) < ϵ. Thus U(Pn, f ) − L(Pn, f ) → 0. Since L(Pn, f ) ≤ S(Pn, f ) ≤ U(Pn, f ), and L(Pn, f ) ≤ L(f ) = ∫∫

R f = U(f ) ≤ U(Pn, f ), we see that

• S(Pn, f ) −

∫∫

R f

• ≤ U(Pn, f ) − L(Pn, f ) → 0.

B.V. Limaye, IITB MA 105: Lec-17

The above result also allows us to find the sums of certain

• series. For example, let

sn := 1 n4

n

∑

i=1 n

∑

j=1

( i + j)2 for n ∈ N. Then, as n → ∞, sn := 1 n2

n

∑

i=1 n

∑

j=1

( i n + j n )2 → ∫ 1 ∫ 1 (x + y)2d(x, y) = 7 6. Caution: If we define µ(P) := max{(xi − xi−1)(yj − yj−1) : 1 ≤ i ≤ n; 1 ≤ j ≤ k}, then we may not have S(Pn, f ) → ∫∫

R f as µ(Pn) → 0 for

every integrable function f on R. An example of this kind is given by the so-called bivariate Thomae function.

B.V. Limaye, IITB MA 105: Lec-17

. . Tutorial 9

• 1. Let D ⊂ R2. Show that D is closed if and only if

∂D ⊂ D.

• 2. Let E ⊂ R2. Then E is called open (in R2) if every point

in E is an interior point of E. Show that E is open ⇐ ⇒ the complement D := R2\E is closed.

• 3. Are the following subsets of R2 closed? Are they open in

R2? Find their boundaries. (i) {(x, y) ∈ R2 : 0 ≤ x < 1 and 0 < y ≤ 1}, (ii) {(x, y) ∈ R2 : 1 ≤ x2 + y 2 ≤ 2}, (iii) {(x, y) ∈ R2 : x ∈ Q and y ∈ Q}.

• 4. Let a, b > 0, D := {(x, y) ∈ R2 : b2x2 + a2y 2 ≤ a2b2},

and define f : D → R by f (x, y) := x2 − y 2. Does f attain its bounds on D? If so, find them. (Use the Orthogonal Gradient Theorem.)

B.V. Limaye, IITB MA 105: Lec-17

. . Tutorial 9

• 5. Maximize f (x, y, z) := 400 x y z subject to the constraint

x2 + y 2 + z2 = 1.

• 6. Let f (x, y, z) := x2 + y 2 + z2 for (x, y, z) ∈ R2. Does f

have constrained extrema subject to x + 2y + 3z = 6 and x + 3y + 4z = 9? If so, find them.

• 7. Let (x, y) ∈ [1, 3] × [2, 5]. Define f (x, y) := 1 if x and y

are both rational and f (x, y) := −1 otherwise. Is f integrable? Is |f | integrable?

• 8. Let f (x, y) := x2y 2 and g(x, y) := x2 + y 2 for

(x, y) ∈ [a, b] × [c, d]. Are f and g integrable on [a, b] × [c, d]? If so, find their double integrals.

• 9. Find the limit of the sequence (sn) if for n ∈ N,

sn := 1 n6

n

∑

i=1 n

∑

j=1

i2j2 and sn := 1 n4

n

∑

i=1 n

∑

j=1

(i2 + j2).

B.V. Limaye, IITB MA 105: Lec-17