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 23, 2022)
Lecture 30

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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
αf is integrable, and R αf = α R f for all α ∈ R,
(ii)

(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 Rf ≤ g,
RR R RR
(vi) |f | is integrable, and | R f | ≤ R |f |.
Proof:

R R For any partition P of RR,R U(P, f ) ≤ U(P, g ), and so

(v)
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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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 on an interval [a, b], we used a powerful result known as the
Fundamental Theorem of Calculus (Part II). If we find a function F on
[a, b] whose derivative is equal to f on [a, b], then
Z b Z b
f (x) dx = F 0 (x) dx = F (b) − F (a).
a 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 in which the problem is reduced to a repeated
evaluation of Riemann integrals.

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Theorem (Fubini Theorem on a Rectangle)
Let R := [a, b] × [c, d], and let f : R → R be integrable. Let I denote the
double integral of f on R.
Rd
(i) If for each fixed x ∈ [a, b], the Riemann integral c f (x, y ) dy exists,
Rb Rd
then the iterated integral a ( c f (x, y ) dy ) dx exists and is equal to
I.
Rb
(ii) If for each fixed y ∈ [c, d], the Riemann integral a f (x, y ) dx exists,
Rd Rb
then the iterated integral c ( 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
Z b Z d  Z d Z b 
f (x, y ) dy dx = I = f (x, y ) dx dy .
a c c a

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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
Z b Z d  Z b  Z d 
φ(x)ψ(y ) dy dx = φ(x) dx ψ(y ) dy .
a c a c

In particular, if r , s ∈ R with r ≥ 0 and s ≥ 0, then

 r +1
− ar +1
  s+1
− c s+1
Z Z 
r s b d
x y d(x, y ) = ,
[a,b]×[c,d] r +1 s +1

provided 0 < a < b and 0 < c < d.

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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
Rd
method. For x ∈ [a, b], we find the area A(x) := c f (x, y ) dy , of the
cross-section of D perpendicular to the
R b x-axis, or alternatively, for
y ∈ [c, d], we find the area B(x) := a f (x, y ) dx, of the cross-section of
D perpendicular to the y -axis. (See Figure.)
Z b Z d
Vol(D) = A(x) dx = B(y ) dy .
a c
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
Z 1 Z 1 1
1 1
 Z
2
(x + y ) dx dy = (x + y )3 dy
0 0 3 0 0
1 1
Z
7
= ((1 + y )3 − y 3 ) dy = .
3 0 6
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(ii) Let R := [0, 1] × [0, 1], f (0, 0) := 0, and for (x, y ) 6= (0, 0), let
3
f (x, y ) := xy (x 2 − y 2 )/(x 2 + y 2 ) . For x ∈ [0, 1], x 6= 0,
1
xy (x 2 − y 2 )
Z Z
x
A(x) := f (x, y ) dy = dy = .
0
2
(x + y )2 3
2(1 + x 2 )2
R1
(Substitute x 2 + y 2 = u.) Also, A(0) = 0 0 dy = 0. Hence
Z 1 Z 1  Z 1 Z 1
x 1
f (x, y ) dy dx = A(x) dx = dx = .
0 0 0 0 2(1 + x 2 )2 8

R 1 R 1 1 
By interchanging x and y , 0 dy = − .
0 f (x, y ) dx
8
Thus the two iterated integrals exist, but they are not equal.
Note that since f (1/n, 1/2n) = 24n2 /125 for all n ∈ N, the function f is
not bounded on R, and so it is not integrable on R. Thus Fubini’s
theorem is not applicable.

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Remark
Around 1910, Fichtenholz and Lichtenstein proved the following result.
Theorem
Let f : [a, b] × [c, d] → R be a bounded function. If for each fixed
Rd
x ∈ [a, b], the Riemann integral c f (x, y ) dy exists, and if for each fixed
Rb
y ∈ [c, d], the Riemann integral a f (x, y ) dx exists, then the iterated
Rb Rd Rd Rb
integrals a ( c f (x, y ) dy ) dx and c ( a f (x, y ) dx) dy exist and are
equal.

The above result is stronger than the Fubini theorem since it does not
assume that the function f : [a, b] × [c, d] → R is integrable; it only
assumes that f is bounded on [a, b] × [c, d].

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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
n X
X k
S(P, f ) := f (si , tj )(xi − xi−1 )(yj − yj−1 ),
i=1 j=1

where (si , tj ) ∈ [xi−1 , xi ] × [yj−1 , yj ] for i = 1, . . . , n and j = 1, . . . , k, 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].

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We 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  > 0. Then there is δ > 0 such that
U(P, f ) − L(P, f ) <  for every partition P satisfying µ(P) < δ.

Corollary
If f is integrable on R and (Pn ) is a sequence of partitions of R with
µ(Pn ) → 0, then U(Pn , f ) − L(Pn , f ) → 0. Further, if S(PnR, fR) is a double
Riemann sum corresponding to Pn and f , then S(Pn , f ) → Rf.

Proof: Given  > 0, find δ > 0 as in the theorem. Then there is n0 ∈ N

such that µ(Pn ) < δ for all n ≥ n0 , and so U(Pn , f ) − L(Pn , f ) < . Thus
U(Pn , f ) − L(Pn , f )R →
R 0. Since L(Pn , f ) ≤ S(Pn , f ) ≤ U(Pn , f ), and
L(Pn , f ) ≤ L(f
RR ) = R f = U(f ) ≤ U(Pn , f ), we see that
|S(Pn , f ) − R f | ≤ U(Pn , f ) − L(Pn , f ) → 0. 
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Caution:
If we define

µ(P) := max{(xi − xi−1 )(yj − yj−1 ) : 1 ≤ i ≤ n; 1 ≤ j ≤ k},

RR
then we may not obtain S(Pn , f ) → R f as µ(Pn ) → 0 for every
integrable function f on R. An example of a function of this kind is a bit
involved.
The preceding corollary allows us to find limits of certain sequences. For
example, let
n n
1 XX
sn := 4 (i + j)2 for n ∈ N.
n
i=1 j=1
n X
n  2
1 X i j
Then sn := 4 + = S(Pn , f ), where f (x, y ) := (x + y )2
n n n
i=1 j=1
for (x, y ) ∈ [0, 1] × [0, 1], and µ(Pn ) = 1/n for n ∈ N. Hence
R1R1
sn → 0 0 (x + y )2 d(x, y ) = 7/6.

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Double Integral over a Bounded Set
Let D be a bounded subset of R2 , and let f : D → R be a bounded
function. Consider a rectangle R := [a, b] × [c, d] such that D ⊂ R, and
define f ∗ : R → R by
(
f (x, y ) if (x, y ) ∈ D,
f ∗ (x, y ) :=
0 otherwise.

We say that f is integrable over D if f ∗ is integrable on R, and in this

case, the double integral of f over D is defined to be the double integral of
f ∗ on R, that is,
Z Z Z Z
f (x, y ) d(x, y ) := f ∗ (x, y ) d(x, y ).
D R

By the domain additivity of double integrals on rectangles, the integrability

of f over D and the value of its double integral are independent of the
choice of a rectangle R containing D and the corresponding extension f ∗
of f to R.
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Let D be a bounded subset of R2 ,and let f : D → R be Rintegrable
R over D.
We may also denote the double integral of f over D by Df.
If f is nonnegative, then the volume of the solid under the surface given by
z = f (x, y ) and above the region D is defined to be the double integral of
f over D. Thus Z Z
Vol(Ef ) := f (x, y ) d(x, y ),
D

where Ef := {(x, y , z) ∈ R3
: (x, y ) ∈ D and 0 ≤ z ≤ f (x, y )}. Next, if
g : D → R is integrable, and f ≤ g on D, then
Z Z
(g (x, y ) − f (x, y )) d(x, y )
D

is defined to be the volume between the surfaces given by z = f (x , y )

and z = g (x , y ).

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