Chapter 6.1: Double integrals

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Riemann’s definition of the double integral

Suppose that f : D ⊂ R2 → R is continuous nonnegativefunction. D = [ a,b ]× [ c,d ].We want to compute the volume of the body T :

T = {(x , y , z) ∈ R3; x ∈ [ a,b ], y ∈ [ c,d ], 0 ≤ z ≤ f (x , y)}.Definition: Let f (x , y) be a bounded function on D. Then underthe double (Riemann) integral of the function f over the set Dwe understand number∫

D

∫f (x , y) dxdy = lim

n→∞m→∞

n∑i=1

m∑j=1

f (xi , yj)4xi4yj ,

If the limit is finite and does not depend on the choice of thedivision a = x0 ≤ x1 ≤ · · · ≤ xn = b, c = y0 ≤ y1 ≤ · · · ≤ yn = band the choice of the points xi ∈ [ xi−1, xi ] a yj ∈ [ yj−1, yj ],i=1,...,n, j=1,...,m.

Theorem: If the function f is continuous on D = [ a,b ]× [ c,d ],then there exists double integral

∫D

∫f (x , y) dxdy .

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Computation of the double integral over the rectangle

Fubini’s Theorem for the rectangle integration set:Let f (x , y) be a continuous function on D = [ a,b ]× [ c,d ].Then ∫

D

∫f (x , y) dxdy =

b∫a

d∫c

f (x , y)dy

dx

=

d∫c

b∫a

f (x , y)dx

dy

Theorem: Let g(x) be a continuous function on [ a,b ] and h(y)be a continuous function on [ c,d ]. Then

∫D

∫f (x , y) dxdy =

b∫a

g(x)dx

d∫c

h(y)dy

.

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Double integral over the standard set

Theorem: Let f (x , y) be a bounded function on the rectangleO = [ a,b ]× [ c,d ] and continuous on O except for a finitenumber of points or except for a finite number of arc. Then∫

O

∫f (x , y) dxdy exists.

We introduce the term standard set D:1 D is bounded, D ⊂ R2

2 the boundary H(D) is created by the finite number ofsimple closed curves and points.

Define the function

g(x , y) =

{f (x , y) for (x , y) ∈ D0 for (x , y) ∈ [ a,b ]× [ c,d ] \ D

T = {(x , y , z); (x , y) ∈ D, 0 ≤ z ≤ f (x , y)} is a body.The volume of the body T is

V =

∫∫[ a,b ]×[ c,d ]

g(x , y) dxdy

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Double integral over the standard set - 2

Definition: Let the function f (x , y) be bounded on the standardset D ⊂ [ a,b ]× [ c,d ]. Put

g(x , y) =

{f (x , y) for (x , y) ∈ D0 for (x , y) ∈ [ a,b ]× [ c,d ] \ D

,

then under the double integral of the function f over the set Dwe understand the number∫

D

∫f (x , y) dxdy =

∫∫[ a,b ]×[ c,d ]

g(x , y) dxdy ,

if the right-hand-side integral exists. The set D is called theintegration domain.

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Double integral over the standard set - 3

Theorem: Let D be the standard set. Consider K ⊂ D, whereK is created by the finite number of arcs and points. Let f and gbe the functions that are bounded on D and continuous andequal on the set D \K . Then there exists double integrals of thefunctions f and g over the set D and are equal, i.e.∫

D

∫f (x , y) dxdy =

∫D

∫g(x , y) dxdy .

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Properties of the double integrals

Theorem: Let all following integrals exists. Then1∫

D

∫α f (x , y)+β g(x , y) dxdy = α

∫D

∫f (x , y) dxdy+β

∫D

∫g(x , y) dxdy ,

α, β ∈ R,2 ∫

D

∫f (x , y) dxdy =

n∑i=1

∫Di

∫f (x , y) dxdy .

where D =n⋃

i=1Di ,

3 ∫D

∫f (x , y) dxdy ≥ 0,

for a nonnegative function f (x , y).

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Standard set 1st and 2nd kind

Definition:We say that D is the standard set of the 1st kind if

D = {(x , y) ∈ R2; x ∈ [ a,b ], ϕ1(x) ≤ y ≤ ϕ2(x)},

where ϕ1(x), ϕ2(x) are continuous functions on [ a,b ] andϕ1(x) ≤ ϕ2(x) ∀ x ∈ [ a,b ].

We say that D is the standard set of the 2nd kind if

D = {(x , y) ∈ R2; y ∈ [ c,d ], ψ1(y) ≤ x ≤ ψ2(y)},

where ψ1(y), ψ2(y) are continuous functions on [ c,d ] andψ1(y) ≤ ψ2(y) ∀ x ∈ [ c,d ].

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Computation of the double integral

Fubini Theorem: Let∫D

∫f (x , y) dxdy exists.

1 Let D be the set of the 1st kind,

D = {(x , y) ∈ R2; x ∈ [ a,b ], ϕ1(x) ≤ y ≤ ϕ2(x)}.

Then ∫D

∫f (x , y) dxdy =

b∫a

ϕ2(x)∫ϕ1(x)

f (x , y)dy

dx .

2 Let D be the set of the 2nd kind,

D = {(x , y) ∈ R2; y ∈ [ c,d ], ψ1(y) ≤ x ≤ ψ2(y)}.

Then ∫D

∫f (x , y) dxdy =

d∫c

ψ2(y)∫ψ1(y)

f (x , y)dx

dy .

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Planar area of the set

Remark: Double integral can be used for the computation of theplanar area of the set D ⊂ R2. The planar area is (numerically)equal to the volume of the body of the height 1, i.e.

PD =

∫D

∫1 dxdy .

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Substitution method for double integral

Consider mapping−→Φ : H ⊂ R2 → R2,−→

Φ (u, v) = (ϕ(u, v), ψ(u, v)) = (x , y). Equations x = ϕ(u, v)

y = ψ(u, v)are called transformation equations.

Definition: Mapping−→Φ is called regular if

1−→Φ is injective H, i.e. ∀ (u1, v1), (u2, v2) ∈ R2 holds

(u1, v1) 6= (u2, v2) ⇒−→Φ (u1, v1) 6=

−→Φ (u2, v2).

2−→Φ is continuously differentiable on H.

3

J(u, v) = det

( ∂ϕ∂u (u, v) ∂ϕ

∂v (u, v)

ψϕ∂u (u, v) ∂ψ

∂v (u, v)

)6= 0

for ∀ (u, v) ∈ H.

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Substitution method for double integral 2.

Theorem: Assume that the integral∫D

∫f (x , y) dxdy , where D is

the standard set, exists. Let−→Φ = (ϕ, φ) be a regular mapping

such that it images standard set H onto the standard set D, i.e.−→Φ (H) = D. Then∫

D

∫f (x , y) dxdy =

∫H

∫f (ϕ(u, v), ψ(u, v)) |J(u, v)|dudv

Very often we use the transformation to the polar coordinates.

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Polar coordinates

Transformation equations for polar coordinates are

x = r cos ty = r sin t , r ∈ (0,∞), t ∈ [ 0,2π)

(r , t) 7→−→Φ (r , t) = (ϕ(r , t), ψ(r , t)) = (x , y)

J(r , t) = det

(cos t −r sin tsin t r cos t

)= r cos2 t + r sin2 t = r 6= 0.

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Improper integrals

Definition: Consider the set D ⊂ R2 such that1 D is unbounded2 the boundary of D consists of the finite number of the

segments and rays.Let Dn = D ∩ {(x , y) ∈ R2, x2 + y2 ≤ n2}.Let f (x , y) a function bounded on D.Assume that for ∀n ∈ N the integral∫

Dn

∫f (x , y) dxdy

exists.Then under the improper double integral we understand∫

D

∫f (x , y) dxdy = lim

n→∞

∫Dn

∫f (x , y) dxdy ,

if the right-hand-side limit exists.14/21

Improper integrals - 2

Theorem: If the function f (x , y) is continuous and bounded onD = [ a,∞)× [ c,∞) then∫D

∫f (x , y) dxdy =

∫ ∞a

(∫ ∞c

f (x , y)dy)

dx =

∫ ∞c

(∫ ∞a

f (x , y)dx)

dy

it the right-hand-side integrals exist.

Remark: AnalogouslyD = [ a,b ]× [ c,∞) D = (−∞,∞)× [ c,∞)D = [ a,∞)× [ c,d ] D = (−∞,∞)× [ c,d ]...

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Laplace integral

∫ ∞0

e−ax2dx a > 0.

Consider that a = 1, i.e. we want to compute the integral

I =

∞∫0

e−x2dx . It holds that

∫D

∫e−(x

2+y2)dxdy =

(∫ ∞0

e−x2dx)(∫ ∞

0e−y2

dy)

= I2,

where D = [ 0,∞)× [ 0,∞).

We obtain I = 12√π and after substitution t =

√ax it holds∫ ∞

0e−ax2

dx =12

√π

a.

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Appendix: Triple integral

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Triple integral

Analogously as for double integral we have∫∫D

∫f (x , z, y)dxdydz,

where D ⊂ R3 is an integration domain.

If f (x , y , z) is the density of the body D at point (x , y , z),thenthe triple integral over the set D represents the weight ofthe body D.

If D = [ a,b ]× [ c,d ]× [ e, f ] then∫∫D

∫f (x , z, y)dxdydz =

∫ b

a

(∫ d

c

(∫ f

ef (x , y , z)dz

)dy

)dx .

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Triple integral - 2

D ⊂ R3:

∫∫D

∫f (x , z, y)dxdydz =

∫ b

a

(∫D(x)

∫f (x , y , z)dydz

)dx =

=

∫ b

a

(∫ ϕ2(x)

ϕ1(x)

(∫ ψ2(x ,y)

ψ1(x ,y)f (x , y , z)dz

)dy

)dx .

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Substitution

Theorem: Let H ⊂ R3 and the mapping−→Φ : H → R3 is

regular, i.e.

1−→Φ is injective and

−→Φ (H) = D

2−→Φ is continuously differentiable on H

3 det J(u, v ,w) 6= 0 pro ∀ (u, v ,w) ∈ H,

then∫∫D

∫f (x , y , z)dxdydz =

∫∫H

∫f (−→Φ (u, v ,w))|det J(u, v ,w)|dudvdw

Remark: The most common is the substitution to the sphericalcoordinates or to the cylindrical coordinates.

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Spherical and cylindrical cooredinates

Transformation equations for spherical coordinates are

x = r cos t sin uy = r sin t sin uz = r cos u, r ∈ (0,∞), t ∈ [ 0,2π), u ∈ [ 0, π)

(r , t ,u) 7→−→Φ (r , t ,u) = (ϕ1(r , t ,u), ϕ2(r , t ,u), ϕ3(r , t ,u)) = (x , y , z)

detJ(r , t ,u) = −r2 sin u.

Transformation equations for cylindrical coordinates are

x = r cos ty = r sin tz = z, r ∈ (0,∞), t ∈ [ 0,2π), z ∈ R

(r , t , z) 7→−→Φ (r , t , z) = (ϕ1(r , t , z), ϕ2(r , t , z), ϕ3(r , t , z)) = (x , y , z)

detJ(r , t , z) = r .

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