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Double Integrals and Green’s TheoremThis unit is based on Sections 9.10 through 9.12 , Chapter 9.All assigned readings and exercises are from the textbookObjectives:Make certain that you can define, and use in context, the terms,

concepts and formulas listed below:• evaluate double integrals in Cartesian and polar• Use double integrals to evaluate area, volume, center of mass,

moment of inertia,…etc.• use Green’s theorem to convert a line integral along a boundary

of a into a double integral, and to convert a double integral to a line integral along the boundary of a region

• use Green’s theorem to evaluate line integrals, and to determine work, area and moment of inertia.

Reading: Read Section 9.10 - 9.12, pages 505-524.Exercises: Complete problems

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The integral over the region R may be calculated by 2 successive integrations:

∫∫R dxdyyxf ),(

2. Similarly, for integrating first over x and then over y

∫∫ ∫ ∫ ⎥⎦⎤

⎢⎣⎡=

R

b

a

xg

xgdxdyyxfdxdyyxf

)(

)(

2

1

),(),(

Double integrals may be written in the form

∫∫ ∫ ∫ ⎥⎦⎤

⎢⎣⎡=

R

d

c

yh

yhdydxyxfdxdyyxf

)(

)(

2

1

),(),(

1. We may integrate first over y, keeping x constant with proper limits of integration and then the whole expression is integrated w.r.t. x

Double integrals in Cartesian coord.

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Engineering Application of double integralsArea: set f(x,y) = 1: ∫∫∫∫ ==

RRdxdydAA

Volume: for f(x,y) > 0

Center of mass for a lamina with density ρ(x,y)

Moment of inertia about the x-axis (Ix) and the y-axis (Iy):

∫∫=R

dxdyyxfV ),(

mM

dxdyyx

dxdyyxyy

mM

dxdyyx

dxdyyxxx x

R

Ry

R

R ====∫∫∫∫

∫∫∫∫

),(

),(~ ;),(

),(~ρ

ρ

ρ

ρ

∫∫∫∫ ==RyRx dxdyyxxIdxdyyxyI ),( ;),( 22 ρρ

∫∫=R

dxdyyxm ),(ρ

R

Mass for a lamina with density ρ(x,y)

Mx & My are the moments of lamina about x & y axes.

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• Example 1 (P9-10.14)

• Example 2 (P9-10.16)

• Example 3 (P9-10.26)

Find the volume bounded by:

z=4-y2, z=0, x=0, x=3, 1st octant

Complete the solution

Complete the solution

Complete the solution

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• Example 4 (P9-10.43)

• Example 5 (P9-10.56)

Find the center of mass for the shape bounded by:y=x, x+y=6, y=0, with mass density ρ(x,y) = 2y

mM

dxdyyx

dxdyyxyy

mM

dxdyyx

dxdyyxxx x

R

Ry

R

R ====∫∫∫∫

∫∫∫∫

),(

),(~ ;),(

),(~ρ

ρ

ρ

ρ

Find the moment of inertia about the x-axis:y=x2, y=√x, with mass density ρ(x,y) = x2

∫∫=Rx dxdyyxyI ),(2ρ

Complete the solution

Complete the solution

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Double Integrals in Polar Coordinates

The integral over the region R may be calculated by 2 successiveintegrations using two different orders:

The element of area in polar coordinates is given by

222);sin();cos( yxrryrxddrrdydxdA

+===

==

θθ

θwith

∫ ∫

∫∫ ∫ ∫∫∫

⎥⎦⎤

⎢⎣⎡=

⎥⎦⎤

⎢⎣⎡==

b

a

rh

rh

R

g

gR

rdrdru

drrdrurdrdrrfdxdyyxf

θθ

θθθθθβ

α

θ

θ

)(

)(

)(

)(

2

1

2

1

),(

),()sin,cos(),(

r-first

θ-first

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• Example 1 (P9-11.5)Find the volume bounded by; r = 5 cos 3θ, z = 0 and z = 3

Use symmetry, i.e., integrate over half of the area

3D

Projection in the xy-plane

Sketch the projected area.

Complete the solution

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• Example 2 (P9-11.26)

Find the integral by changing to polar coordinates

Write in the integral polar coordinates

Sketch the projected area.

Complete the solution

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• Example 3 (P9-11.34)

Find the integral over R

Since the region is circular, change to polar coordinates

Complete the solution

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• Example 4 (P9-11.17)

Find the moment of inertia of the lamina bounded by r = a, with ρ = k, around the x - axis

Since the region is circular, change to polar coordinates

∫∫=Rx dxdyyxyI ),(2ρ

Start with:

y = r sin θ, ρ = k, dxdy = r dr dθ

Complete the solution

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• Example 4 (P9-11.13): Find the center of mass of the laminabounded by y = √3 x, y=0, x=3, ρ = r2

In evaluating these integrals, you need to express:

Change to polar coordinates

Start with:

x = r cos θ, y = r sin θ and dxdy = r dr dθ

mM

dxdyyx

dxdyyxyy

mM

dxdyyx

dxdyyxxx x

R

Ry

R

R ====∫∫∫∫

∫∫∫∫

),(

),(~ ;),(

),(~ρ

ρ

ρ

ρ

r cosθ =3

θ=π/3

My =Mx =

==Complete the solution

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Green’s Theorem

Most of the functions that engineers deal with satisfy these conditions.

There is an important relationship between line and double integrals expressed in terms of Green’s theorem in the plane.

If the functions P(x,y) and Q(x,y) and their derivatives are finite and continuousfunctions in a region R and on its boundary, the closed curve C, then

[ ] ∫∫∫ ⎥⎦

⎤⎢⎣

⎡∂∂

−∂∂

=+RC

dxdyyP

xQQdyPdx

The direction of integration along C is such that the region R is always to the left

The theorem is a special case of two important theorems: Divergence and Stokes Theorems (to be discussed)

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Example (P9-12.1)

[ ] ∫∫∫ ⎥⎦

⎤⎢⎣

⎡∂∂

−∂∂

=+RC

dxdyyP

xQQdyPdx

Verify Green’s theorem: P = x - y, & Q = xy. C is the triangle: (0,0), (1,0), (1,3)

x=1

y=3x

y=o

R C

LHS =

RHS =

Complete the solution

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Green’s theorem in region with holesThe theorem could be extended to a region bounded by 2 or more smooth simple closed curves. In such a case: C = C1 U C2.

[ ]∫ +C

QdyPdx• Example (P9-12.24) Evaluate the line integral:

[ ]

1−==

⎥⎦

⎤⎢⎣

⎡∂∂

−∂∂

=+ ∫∫∫yx

RC

PoQ

dxdyyP

xQQdyPdx

& Complete the solution

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[ ]∫ +C

QdyPdxUse Green’s Theorem to evaluate the line integral:

C is the triangle: (1,2), (2,2), (2,4)

y=2x

y=2

x=2

R C

[ ]

22 33

22

yQxyQ

xPxyP

dxdyyP

xQQdyPdx

x

y

RC

=⇒=

=⇒=

⎥⎦

⎤⎢⎣

⎡∂∂

−∂∂

=+ ∫∫∫

Example (P9-12.9)

Complete the solution