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1
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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