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Multiple Integrals
Advanced Calculus Lecture 1 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 4, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Double integrals 10.0
Consider a function f (x, y) , defined over
7.5
5.0
a rectangle D = [a, b] × [c, d ] . Suppose,
2.5
for simplicity, that f (x, y) ≥ 0 for every
0.0 0 10
4
3
2
1
y
2
(x, y) in D .
x
3 4 5 6 7
Double integrals (continue) 10.0
Let S be the solid bounded by the
7.5
xy -plane (for a ≤ x ≤ b and
5.0
2.5
c ≤ y ≤ d ), and the surface of the
0.0 0
function f (x, y) .The double integral of f (x, y) is the volume of the solid S . Dr. Lahcen Laayouni
Advanced Calculus
1
10 2 3 x
4 5 6 7
2 y
3
4
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Partitioning of the domain We partition D as fellows
yn=d
a = x0 < x1 < · · · < xm = b
y
j
Rij
yj−1
c = y0 < y1 < · · · < yn = d . The partition consists of mn rectangles
y
1
y0=c x0=a
Rij . (1 ≤ i ≤ m, 1 ≤ j ≤ n) . Riemann sum Let
(xij∗ , yij∗ )
x
x1
x
i−1
xm=b
i
10.0
be any arbitrary point in Rij ,
7.5
5.0
then the Riemann sum is defined by
2.5
R(f ) =
m X n X
f (xij∗ , yij∗ )∆Aij
0.0 0
,
1
10 2 3
i=1 j=1
x
4 5 6
∆Aij = ∆xi ∆yj = (xi − xi−1 )(yj − yj−1 ) Dr. Lahcen Laayouni
Advanced Calculus
2 y
3
4
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
The double integral over a rectangle The function f (x, y) is integrable over the rectangle D and has double integral ZZ I= f (x, y)dA . D
Iff the limit of the Riemann sum exists as n, m goes to ∞ , that is ZZ
D
f (x, y)dA =
lim
n,m→∞
m X n X
f (xij∗ , yij∗ )∆Aij .
i=1 j=1
Remark The term dA is called the area element and represents the limit of ∆A = ∆x∆y in the Riemann sum, written also dxdy or dydx . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Find an approximate value of RR 2 2 D (x + y )dA , where D = [0, 1] . Use a Riemann sum corresponding to the partition of D into four smaller squares with points selected at the center of each.
(3/4,3/4)
(1/4,3/4) 1/2
(1/4,1/4)
(3/4,1/4) 1/2
Solution The partition of D is formed by x = 1/2 and y = 1/2 , which divide D into four squares, each of area ∆A = 1/4 . The centers are (1/4, 1/4), (1/4, 3/4), (3/4, 1/4), and (3/4, 3/4) , so that RR 1 1 9 1 2 ≈ ( 14 + 16 ) 4 + ( 14 + 16 )4+ D (x + y )dA ( 34 +
Dr. Lahcen Laayouni
1 1 16 ) 4
+ ( 34 +
9 1 16 ) 4
Advanced Calculus
= 0.8125
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Double integral over general domains
1 0.9
If f (x, y) is defined and bounded on D ,
R
0.8 0.7 0.6
let ˆf be the extension of f if (x, y) in D ˆf (x, y) = f (x, y) 0 otherwise
D
0.5 0.4 0.3 0.2 0.1 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Double integral over general domains (continue) If D is a bounded domain, then it is contained is some rectangle R with sides parallel to the coordinates axes. If ˆf is integrable over R , then f is integrable over D and ZZ ZZ ˆf (x, y)dA . f (x, y)dA = D
Dr. Lahcen Laayouni
R
Advanced Calculus
1
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Properties of the double integral If f and g are integrable over D and if a and b are constants, ZZ 1f (x, y)dA = 0 if D has zero area. D
2-
ZZ
1dA = area of D .
D
3- if f (x, y) ≥ 0 then 4- if f (x, y) ≤ 0 then 5-
ZZ
ZZ
D
ZZ
D
f (x, y)dA = volume of S . f (x, y)dA = −volume of S .
(af (x, y) + bg(x, y))dA = a
D Dr. Lahcen Laayouni
ZZ
f (x, y)dA + b D Advanced Calculus
ZZ
D
g(x, y)dA .
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Examples Let R be the rectangle a ≤ x ≤ b, c ≤ y ≤ d , then ZZ ZZ 2dA = 2 dA = 2 area of R = 2(b − a)(d − c) . R
Let I =
R
ZZ
(sin(y) + y 5 + 10)dA . Using property (5),
x 2 +y 2 ≤1
I
= =
ZZ
ZZ
sin(y)dA + x 2 +y 2 ≤1
5
y dA +
x 2 +y 2 ≤1
ZZ
10dA x 2 +y 2 ≤1
I1 + I2 + I3 .
The domain D is a symmetric disk of radius 1 . Since sin(y) is an odd function of y , then ZZ ZZ sin(y)dA = − x 2 +y 2 ≤1,y ≥0
x 2 +y 2 ≤1,y ≤0
Dr. Lahcen Laayouni
sin(y)dA ⇒ I1 = 0 .
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Examples (continue) Similarly, y 5 is an odd function and D is a symmetric domain about x -axis, thus I2 = 0 . SinceZZ I3 =
10dA = 10
x 2 +y 2 ≤1
ZZ
dA = 10 × (area of D) = 10π .
x 2 +y 2 ≤1
I = I1 + I2 + I3 = 0 + 0 + 10π = 10π . Additivity of domains If D1 , D2 , . . . , Dk are nonoverlapping domains on each of which f is integrable, then f is integrable on D = D1 ∪ D2 ∪ · · · ∪ Dk and ZZ Z k Z X f (x, y)dA = f (x, y)dA . D
j=1
Dr. Lahcen Laayouni
Dj
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Let D be the disk x 2 + y 2 ≤ 1 , then the ZZ q integral I = 1 − x 2 − y 2 dA is the D
volume of a hemisphere of radius 1 , thus
1.0
0.5 z
0.0 −1.0
0
−0.5
I=
Volume of the sphere 2
0.0
=
2π 3
x
.
Dr. Lahcen Laayouni
0.5 1.0
Advanced Calculus
−1
y
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
y-simple domain The domain D in the xy-plane is said to be y-simple if it is bounded by two vertical lines x = a and x = b , and two continuous graphs y = c(x) and y = d (x) . x-simple domain The domain D in the xy-plane is said to be x-simple if it is bounded by two horizontal lines y = c and y = d , and two continuous graphs x = a(y) and x = b(y) . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Iteration of double integrals If f (x, y) is continuous on the bounded y-simple domain D given by a ≤ x ≤ b and c(x) ≤ y ≤ d (x) , then ZZ
f (x, y)dA =
D
Z
b
dx a
d(x)
Z
f (x, y)dy .
c(x)
Similarly, if f (x, y) is continuous on the x-simple domain D given by c ≤ y ≤ d and a(y) ≤ x ≤ b(y) , then ZZ
D
f (x, y)dA =
Z
d
dy c
Dr. Lahcen Laayouni
Z
b(y )
f (x, y)dx .
a(y )
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Find V the volume of the solid lying above the square Q defined by 0 ≤ x ≤ 1 and 1 ≤ y ≤ 2 and below the plane z = 1 − x + y . Solution Since the square is both y-simple and x-simple then we can iterate in either direction. For instance if we integrate first w.r.t x , then ZZ V = (1 − x + y)dA Q
=
Z
=
Z
2
dy 1
1
(1 − x + y)dx =
0
2 1
Z
Z
2 1
1 dy (x − x 2 /2 + xy) 0
2 (1/2 + y)dy = (y/2 + y 2 /2) = 2 1
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Evaluate I =
ZZ
xy 2 dA over the triangle T with vertices T
(0, 0), (1, 0), and (1, 1) . Solution Here the triangle is x-simple and y-simple. Integrating first w.r.t. y , we obtain Z 1 Z x Z 1 x 2 I= dx xy dy = dx (xy 3 /3) 0
Z
1
0
0
0
0
1 x 4 /3dx = x 5 /15 0 = 1/15
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Solution (continue) Now integrating first w.r.t. x , we obtain Z 1 Z 1 Z 1 1 2 I= dy xy dx = dy (x 2 /2y 2 ) 0
1/2
Z
y
1
0
y
(y 2 − y 4 )dy
0
1 = 1/2 (y 3 /3 − y 5 /5) 0 = 1/15
Dr. Lahcen Laayouni
Advanced Calculus
Advanced Calculus Lecture 1 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 4, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Double integrals 10.0
Consider a function f (x, y) , defined over
7.5
5.0
a rectangle D = [a, b] × [c, d ] . Suppose,
2.5
for simplicity, that f (x, y) ≥ 0 for every
0.0 0 10
4
3
2
1
y
2
(x, y) in D .
x
3 4 5 6 7
Double integrals (continue) 10.0
Let S be the solid bounded by the
7.5
xy -plane (for a ≤ x ≤ b and
5.0
2.5
c ≤ y ≤ d ), and the surface of the
0.0 0
function f (x, y) .The double integral of f (x, y) is the volume of the solid S . Dr. Lahcen Laayouni
Advanced Calculus
1
10 2 3 x
4 5 6 7
2 y
3
4
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Partitioning of the domain We partition D as fellows
yn=d
a = x0 < x1 < · · · < xm = b
y
j
Rij
yj−1
c = y0 < y1 < · · · < yn = d . The partition consists of mn rectangles
y
1
y0=c x0=a
Rij . (1 ≤ i ≤ m, 1 ≤ j ≤ n) . Riemann sum Let
(xij∗ , yij∗ )
x
x1
x
i−1
xm=b
i
10.0
be any arbitrary point in Rij ,
7.5
5.0
then the Riemann sum is defined by
2.5
R(f ) =
m X n X
f (xij∗ , yij∗ )∆Aij
0.0 0
,
1
10 2 3
i=1 j=1
x
4 5 6
∆Aij = ∆xi ∆yj = (xi − xi−1 )(yj − yj−1 ) Dr. Lahcen Laayouni
Advanced Calculus
2 y
3
4
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
The double integral over a rectangle The function f (x, y) is integrable over the rectangle D and has double integral ZZ I= f (x, y)dA . D
Iff the limit of the Riemann sum exists as n, m goes to ∞ , that is ZZ
D
f (x, y)dA =
lim
n,m→∞
m X n X
f (xij∗ , yij∗ )∆Aij .
i=1 j=1
Remark The term dA is called the area element and represents the limit of ∆A = ∆x∆y in the Riemann sum, written also dxdy or dydx . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Find an approximate value of RR 2 2 D (x + y )dA , where D = [0, 1] . Use a Riemann sum corresponding to the partition of D into four smaller squares with points selected at the center of each.
(3/4,3/4)
(1/4,3/4) 1/2
(1/4,1/4)
(3/4,1/4) 1/2
Solution The partition of D is formed by x = 1/2 and y = 1/2 , which divide D into four squares, each of area ∆A = 1/4 . The centers are (1/4, 1/4), (1/4, 3/4), (3/4, 1/4), and (3/4, 3/4) , so that RR 1 1 9 1 2 ≈ ( 14 + 16 ) 4 + ( 14 + 16 )4+ D (x + y )dA ( 34 +
Dr. Lahcen Laayouni
1 1 16 ) 4
+ ( 34 +
9 1 16 ) 4
Advanced Calculus
= 0.8125
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Double integral over general domains
1 0.9
If f (x, y) is defined and bounded on D ,
R
0.8 0.7 0.6
let ˆf be the extension of f if (x, y) in D ˆf (x, y) = f (x, y) 0 otherwise
D
0.5 0.4 0.3 0.2 0.1 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Double integral over general domains (continue) If D is a bounded domain, then it is contained is some rectangle R with sides parallel to the coordinates axes. If ˆf is integrable over R , then f is integrable over D and ZZ ZZ ˆf (x, y)dA . f (x, y)dA = D
Dr. Lahcen Laayouni
R
Advanced Calculus
1
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Properties of the double integral If f and g are integrable over D and if a and b are constants, ZZ 1f (x, y)dA = 0 if D has zero area. D
2-
ZZ
1dA = area of D .
D
3- if f (x, y) ≥ 0 then 4- if f (x, y) ≤ 0 then 5-
ZZ
ZZ
D
ZZ
D
f (x, y)dA = volume of S . f (x, y)dA = −volume of S .
(af (x, y) + bg(x, y))dA = a
D Dr. Lahcen Laayouni
ZZ
f (x, y)dA + b D Advanced Calculus
ZZ
D
g(x, y)dA .
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Examples Let R be the rectangle a ≤ x ≤ b, c ≤ y ≤ d , then ZZ ZZ 2dA = 2 dA = 2 area of R = 2(b − a)(d − c) . R
Let I =
R
ZZ
(sin(y) + y 5 + 10)dA . Using property (5),
x 2 +y 2 ≤1
I
= =
ZZ
ZZ
sin(y)dA + x 2 +y 2 ≤1
5
y dA +
x 2 +y 2 ≤1
ZZ
10dA x 2 +y 2 ≤1
I1 + I2 + I3 .
The domain D is a symmetric disk of radius 1 . Since sin(y) is an odd function of y , then ZZ ZZ sin(y)dA = − x 2 +y 2 ≤1,y ≥0
x 2 +y 2 ≤1,y ≤0
Dr. Lahcen Laayouni
sin(y)dA ⇒ I1 = 0 .
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Examples (continue) Similarly, y 5 is an odd function and D is a symmetric domain about x -axis, thus I2 = 0 . SinceZZ I3 =
10dA = 10
x 2 +y 2 ≤1
ZZ
dA = 10 × (area of D) = 10π .
x 2 +y 2 ≤1
I = I1 + I2 + I3 = 0 + 0 + 10π = 10π . Additivity of domains If D1 , D2 , . . . , Dk are nonoverlapping domains on each of which f is integrable, then f is integrable on D = D1 ∪ D2 ∪ · · · ∪ Dk and ZZ Z k Z X f (x, y)dA = f (x, y)dA . D
j=1
Dr. Lahcen Laayouni
Dj
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Let D be the disk x 2 + y 2 ≤ 1 , then the ZZ q integral I = 1 − x 2 − y 2 dA is the D
volume of a hemisphere of radius 1 , thus
1.0
0.5 z
0.0 −1.0
0
−0.5
I=
Volume of the sphere 2
0.0
=
2π 3
x
.
Dr. Lahcen Laayouni
0.5 1.0
Advanced Calculus
−1
y
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
y-simple domain The domain D in the xy-plane is said to be y-simple if it is bounded by two vertical lines x = a and x = b , and two continuous graphs y = c(x) and y = d (x) . x-simple domain The domain D in the xy-plane is said to be x-simple if it is bounded by two horizontal lines y = c and y = d , and two continuous graphs x = a(y) and x = b(y) . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Iteration of double integrals If f (x, y) is continuous on the bounded y-simple domain D given by a ≤ x ≤ b and c(x) ≤ y ≤ d (x) , then ZZ
f (x, y)dA =
D
Z
b
dx a
d(x)
Z
f (x, y)dy .
c(x)
Similarly, if f (x, y) is continuous on the x-simple domain D given by c ≤ y ≤ d and a(y) ≤ x ≤ b(y) , then ZZ
D
f (x, y)dA =
Z
d
dy c
Dr. Lahcen Laayouni
Z
b(y )
f (x, y)dx .
a(y )
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Find V the volume of the solid lying above the square Q defined by 0 ≤ x ≤ 1 and 1 ≤ y ≤ 2 and below the plane z = 1 − x + y . Solution Since the square is both y-simple and x-simple then we can iterate in either direction. For instance if we integrate first w.r.t x , then ZZ V = (1 − x + y)dA Q
=
Z
=
Z
2
dy 1
1
(1 − x + y)dx =
0
2 1
Z
Z
2 1
1 dy (x − x 2 /2 + xy) 0
2 (1/2 + y)dy = (y/2 + y 2 /2) = 2 1
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Example Evaluate I =
ZZ
xy 2 dA over the triangle T with vertices T
(0, 0), (1, 0), and (1, 1) . Solution Here the triangle is x-simple and y-simple. Integrating first w.r.t. y , we obtain Z 1 Z x Z 1 x 2 I= dx xy dy = dx (xy 3 /3) 0
Z
1
0
0
0
0
1 x 4 /3dx = x 5 /15 0 = 1/15
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Double integrals Iteration of double integrals
Multiple Integrals
Solution (continue) Now integrating first w.r.t. x , we obtain Z 1 Z 1 Z 1 1 2 I= dy xy dx = dy (x 2 /2y 2 ) 0
1/2
Z
y
1
0
y
(y 2 − y 4 )dy
0
1 = 1/2 (y 3 /3 − y 5 /5) 0 = 1/15
Dr. Lahcen Laayouni
Advanced Calculus
