notesinteg_06

7/28/2019 notesinteg_06

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2 Double Integrals and Line Integrals

2.1 Double Integrals

2.1.1 Introduction

ba

dc

f(x, y)dxdy

is a simple example of a double integral.The integral

dc f(x, y)dx is known as the inner integral and is a

function of y.The integral

b

a dy is known as the outer integral.

Example 2.1.1 :- Find 1

0

1

1x2ydxdy.

Solution Inner integral is1

1 x2ydx =

1

3x3y

11 =

2

3y

Outer integral is

1

0

2

3ydy = 2

3 1

2y2

1

0= 1

3

So1011 x2ydxdy = 13 .

Note :- A double integral can be interpreted as the volume underthe function f(x, y) in the area defined by c x d, a y b.

Note :- The order of integration may be changed. For exampleba

dc f(x, y)dxdy =

dc

ba f(x, y)dydx.

2.1.2 The Region of Integration

The limits on the inner integral may be functions of the variable forthe outer integral. For example1

x=0

xy=0

f(x, y)dydx

represents f(x, y) being integrated over the triangle defined by y = 0,x = 1, y = x. The extremes on x are x = 0 to x = 1 and, for each value of x,the values of y vary between y = 0 to y = x.


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Example 2.1.3 :- Find

1

x=0

x

y=0

(1 xy) dy dx

Solution 1

x=0

xy=0

(1 xy) dy dx =

1

x=0

y 1

2xy2

xy=0

dx

=

1

x=0

x 1

2x3 (0 0)

dx

=

12

x2 18

x810

=1

2 1

8 0 = 3

8

Sometimes, the region of integration is such that the integral has tobe split into several parts. For example, to integrate f(x, y) = x + 1 overthe rhombus befined by (2,0), (0,1), (-2,0) and (0,-1) involves evaluating

0

2 1

2x+1

1

2x

1(x + 1) dy dx +

2

0 1

2x+1

1

2x

1

(x + 1) dy dx

Sometimes reversing the order of integration allows an integral to beevaluated. For example

1

0

cos1

x

0x (cos y)4 dydx is a tricky integral

but is equivilent to/20

cos2 y

0(cos y)4 xdxdy which can be done much

more easily.

2.1.3 Change of Variable

Sometimes an integral can be made easier by a change of variables.Either the function or the limits may be made easier. If f(x, y) can

be expressed as g(u, v) then f(x, y) dx dy =

g (u, v) |J| du dv

. In this case, |J| is the magnitude ofJ i.e. the magnitude ofx

u

y

v x

v

y

u

and is known as the Jacobian. Care must be taken over the limits.


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For the rhombus of the previous section, if u = x2

+ y and v = x2 y,

then the integral comes to

1

u=1

1

v=1

(u + v + 1) 1 du dv

. In this case, the u + v + 1 corresponds to the x + 1 while the 1 isthe Jacobean J.

When coverting from rectangular (x, y) to polar (r, ) coordinates, theJacobean J is equal to r.


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2.2 Line Integrals

2.2.1 Introduction

It is possible to carry out integration along a line or a curve. Such an integral

may be written as c

F(x, y)dw

There are three main features influencing this integralF(x,y) This is the function to be integrated

e.g. F(x, y) = x2 + 4y2

C This is the curve along which integration takes

place. e.g.y = x2 or x = t 1; y = t2

dw This states the variable that the integration takesplace with respect to. Four main cases aredxdyds Here s is the coordinate along the

curve.

ds may be written as ds =

(dx)2

+ (dy)2

OR ds =

1 +

dydx

2dx

F1dx + F2dy This is a combination of thefirst two cases.

The line integral C

f(x, y) ds

represents the area beneath the function f(x, y) but above the curve C. Theintegrals

C

f(x, y) dx

and C

f(x, y) dy

represent the projections of this area onto the planes above the x and y axes.


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2.2.2 Techniques and Examples

The technique with a line integral is to express all quantities in an integral interms of a single variable. Often, if the integral is with respect to x or

y, the curve C and the function F may be expressed in terms ofthe relevant variable.

Example 2.2.1

Find c

x (1 + 4y) dx

where C is the curve y = x2, starting from x = 0, y = 0 and ending at

x = 1, y = 1.

Solution As we are interested only in points along C, y may be replaced byx2. The limits on x will be 0 to 1. So the integral becomes

1

x=0

x + 4x3

dx =

x2

2+ x4

10

=1

2+ 1

(0) = 32

If the integral is to be carried out with respect to s, it often helpsto convert everything to be in terms of x.

Example 2.2.3

Find c

x (1 + 4y) ds

where C is the curve y = x2, starting from x = 0, y = 0 and ending atx = 1, y = 1.

Solution All variables may be converted to xy = x2

ds =

1 +

dydx

2dx =

1 + (2x)2dx =

1 + 4x2dx

So, the integral is1

x=0 x

1 + 4x2

1 + 4x2dx

=1

x=0 x

1 + 4x23/2

dx

This can be evaluated using the transformation U = 1 + 4x2 so dU = 8xdxi.e. dx = dU

8. When x = 0, U = 1 and when x = 1, U = 5.


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Integral = 18

5

U=1 U3/2dU = 1

8

2

5

U5/2

51

= 120

55/2 1 = 2.745

When the curve is given in terms of a parameter, it is often bestto convert all variables to that coordinate

Example 2.2.6

Find c

x2ydx 2xydy

where C is the curve defined by x = t2 + 1, y = t3 tbetween t = 0 and t = 1.

Solution

From the equations for x and y, dx = 2tdt, dy =

3t2 1 dtSo integral is

1

t=0

t2 + 1

2 t3 t 2tdt 2 t3 t t2 + 1 3t2 1 dt

=

1

t=0

t4 + 2t2 + 1

2t4 2t2 2 t5 t 3t2 1 dt

=

1

0

2t8 + 2t6 2t4 2t2 6t7 + 2t5 + 6t3 2t dt

=

2

9t9 +

2

7t7 2

5t5 2

3t3 3

4t8 +

1

3t6 +

3

2t4 t2

10

=2

9+

2

7 2

5 2

3 3

4+

1

3+

3

2 1

= 5991260

2.3 The curve of integration

2.3.1 Compound Curves

Sometimes, the curve C can have different forms in different places.In such cases, the integral can be expressed as the sum of differentintegrals

Example 2.3.1


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Find C

xydx

where C is the curve y = x2 from (0,0) to (1,1) followed by y = 2 xfrom (1,1) to (2,0).

Solution Integral is1

x=0

xx2dx +

2

x=1

x (2 x) dx =

1

0

x3dx +

2

1

2x x2 dx

=

x4

4

10

+

x2 x

3

3

21

=

1

4 +

4 8

3 1

1

3

=1

4+ 3 7

3=

11

12

2.3.2 Different Routes of Integration

There are some functions F(x,y) for which the integral does not dependon the route of integration but merely on the start and finish points.

Example 2.3.2

Evaluate c

2x sin ydx + x2 cos ydy

for

i) C is the straight line y = 2

x from x = 0 to x = 1ii) C is the curve y = sin1 x from x = 0 to x = 1iii) C is the line y = 0 from x = 0 to x = 1 followed by the line

x = 1 from y = 0 to y = 2

All routes go from (0, 0) to (1, 2

)

Solution In all three cases, the integral equals 1. In fact, by allroutes between (0, 0) and (1,

2), the integral is 1.

Example 2.2.3

Evaluate C

x3ydx + xy2dy

where


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i) C is the line y = x2 from x = 0 to x = 1ii) C is the curve y = x2 from x = 0 to x = 1

Solution

Case i) Integral = 920Case ii) Integral = 19

42

In this example, the integral is dependent on the route of integrationand not just the start and end points.

2.3.3 Closed Curves

Sometimes, the curve of integration will return to its point of origin.In such a case, the symbol

rather than

is used for integration.

Example 2.2.11

Find C

x2ydx + xdy

where C is the route once around the unit circle moving anti-clockwise,

starting and ending at (1, 0).

Solution C is represented by x = cos t, y = sin t with the limitson t being 0 to 2.

x = cos t : dx = sin tdty = sin t : dy = cos tdtThe integral is2

t=0

cos2 t sin t ( sin tdt) + cos t cos tdt =

2

0

cos2 t sin2 t + cos2 t dt

=

3

4

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