Multivariable Calculus

Sartaj Ul Hasan

Department of Mathematics
Indian Institute of Technology Jammu
Jammu, India

Email: sartaj.hasan@iitjammu.ac.in

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 Multivariable Calculus
(February 21, 2022)
Lecture 29

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 Multiple Integral

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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.

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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
n X
X k
L(P, f ) := mi,j (f )(xi − xi−1 )(yj − yj−1 ),
i=1 j=1

n X
X k
U(P, f ) := Mi,j (f )(xi − xi−1 )(yj − yj−1 ).
i=1 j=1
Pn Pk
Since i=1 (xi − xi−1 ) = b − a and j=1 (yi − yi−1 ) = d − c, we obtain

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

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 z

Figure:
Figure:Summands
Summands of
of lower andupper
lower and upperdouble
double sums.
sums.

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Let P1 and P2 be partitions of the rectangle R. A lower double sum
increases and a upper double sum decreases when a partition is refined.
Let P ? be a common refinement of P1 and P2 . Then
L(P1 , f ) ≤ L(P ? , f ) ≤ U(P ? , 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 ).

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In this case, the double integral of f on R is the common value
U(f ) = L(f ), and it is denoted by
Z Z Z Z
f or f (x, y )d(x, y ).
R R

Let f : R → R be integrable and nonnegative. The double integral of f on

R gives the volume of the solid
Ef := {(x, y , z) ∈ R3 : (x, y ) ∈ R and 0 ≤ z ≤ f (x, y )} 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 ) = (b − a)(d − c) = U(P, f ) 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
(
1 if x and y are rational numbers,
f (x, y ) :=
0 if x or y are rational numbers.
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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 ) 6= 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. Let
P := {(xi , yj ) : i = 0, 1, . . . , n and j = 0, 1, . . . , k} be 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 X
X n X
mi,j (f )(xi − xi−1 )(yj − yj−1 ) = (d − c) mi (φ)(xi − xi−1 ).
i=1 j=1 i=1

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

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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 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 gives a necessary and sufficient condition for the
integrability of a bounded function on R.

Theorem (Riemann condition)

Let f : [a, b] × [c, d] → R be a bounded function. Then f is integrable if
and only if for every  > 0, there is a partition P of [a, b] × [c, d] such
that U(P , f ) − L(P , f ) < .

The proof is very similar to the proof of the result about the Riemann
condition in the one variable case.

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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 of f on the four
subrectangles.

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Some conventions

Let a, b, c, d ∈ R with a ≤ b and c ≤ d. In view of domain additivity, we

make the following conventions:
Z Z
f := 0 if a = b or c = d
[a,b]×[c,d]
Z Z Z Z
f := − f
[b,a]×[c,d] [a,b]×[c,d]
Z Z Z Z
f := − f
[a,b]×[d,c] [a,b]×[c,d]
Z Z Z Z
f := f
[b,a]×[d,c] [a,b]×[c,d]

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Integrable functions

Let R := [a, b] × [c, d], and 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 ) = x r y 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 /(x 2 + y 2 ) if
(x, y ) ∈ [−1, 1] × [−1, 1] and (x, y ) 6= (0, 0). Since f is bounded on
R, and it is discontinuous only at (0, 0), f is integrable.

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Algebraic and Order Properties
Let R := [a, b] × [c, d]. If f , g : R → R are integrable, then
RR RR RR
(i) f + g is integrable, and R (f + g ) = R f + R g,
RR RR
(ii) αf is integrable, and R αf = 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,
RR RR
(v) If f ≤ g , then R ≤ R RR g , R R
(vi) |f | is integrable, and | R | ≤ R |f |.
Proof:
(v)
R R For any partition P of RR,R U(P, f ) ≤ U(P, g ), and so
R f = U(f ) ≤ U(g ) = R g.
(vi) U(P, |f |) − L(P, |f |) ≤ U(P,
R R f ) − L(P,
R R f ) forReach
R P.
Also, −|f | ≤ f ≤ |f | =⇒ − R |f | ≤ Rf ≤ R |f |.

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