8 Integrable functions

8.1 Riemann integral

8.1.1 An example

Suppose you want to find the area under the graph of the function () = 2 between = −1 and = 2 (see

Figure 12).

y

−1 −0.5 0.25 0.75 1 2

f(x) = x2

x

Figure 12: Approximating the area under the graph of by .

To approximate this area, you could could proceed like this. Divide the interval [−1 2] into pieces (smallersubintervals), for example considering the division points

−1 −05 0 025 075 1Next, note that the sum of the areas of the rectangles below the graph of the function is smaller than the area

you are looking for, and the sum of the areas of the rectangles above the graph of the function is larger than

the area , that is

To compute , note that the first rectangle has base length 05 and height given by min∈[−1−05] () =(−05)2 = 025 (the height of the rectangle has to equal the minimum value of on the interval [−1−05] if youwant this rectangle to lie below the graph of the function and the approximation of the area to be the best best

approximation).

Doing the same for the remaining rectangles, you obtain

= 05 · (025)2 + 05 · (0)2 + 025 · (0)2 + 05 · (025)2 + 025 · (075)2 = 0297Similarly, to compute , note that in this case the area of the first rectangle is the length of the base (05)

times the height, which in this case is max∈[−1−05] () = (−1)2 = 1 (the height of the rectangle has to equal

the maximum value of on the interval [−1−05] if you want the graph of the function to lie below this rectangleand the approximation of the area to be the best best approximation).

You obtain

= 05 · (−1)2 + 05 · (−05)2 + 025 · (025)2 + 05 · (075)2 + 025 · (1)2 = 1718From the above calculation you obtain the crude estimate for the area under the graph of

0297 1718

Clearly, the above approximation of the area is not great, but if you consider more division points (very close

to each other), the method above will produce quite good results.

This is the main ideea in defining the Riemann integral

61

8.1.2 Definition of the Riemann integral

For a given function : [ ]→ R and a division

∆ : = 0 1 −1 =

of the interval [ ], we define:

- the lower Riemann sum of corresponding to the division ∆ by

(∆) =

X=1

( − −1) inf∈[−1]

() (40)

- the upper Riemann sum of corresponding to the division ∆ by

(∆) =

X=1

( − −1) sup∈[−1]

() (41)

- the Riemann sum of corresponding to the division ∆ and to the intermediate points ∈ [−1 ] by

(∆ ) =

X=1

( − −1) () (42)

Remark 8.1 If is a continuous function, by Weierstrass’s boundedness theorem, attains its maximum and

minimum value on each interval [−1 ], so the infimum (inf) and supremum (sup) in the above definitions are

just the minimum (min) and maximum (max) values of on these intervals.

Some properties of the lower/upper/Riemann sums are contained in the following.

Proposition 8.2 Let : [ ]→ R be a bounded function.

a) For any division ∆ of [ ] and any intermediate points we have

(∆) ≤ (∆ ) ≤ (∆) (43)

b) For any divisions ∆0 and ∆00 of [ ] we have

(∆) ≤ (∆) (44)

c) Refining the partition, the lower Riemann sum increase and the upper Riemann sum decrease. That is, if

∆0 ⊂ ∆00 (all the points of ∆0 are also points of ∆00), then

(∆0) ≤ (∆00) and (∆0) ≥ (∆00)

Proof. a) Follows immediately using

inf∈[−1]

() ≤ () ≤ sup∈[−1]

()

and summing over .

b) The left side is less than the area under the graph of , and the right side is greater than these, and the

inequality follows.

c) Follows using the inequalities

inf∈[0−10]

() ≤ inf∈[00−100 ]

() and sup∈[0−10]

() ≥ sup∈[00−100 ]

()

for any intervals£00−1

00

¤ ⊂ £0−1 0¤.With this preamble, we can now introduce the definition of Riemann integral of a function, as follows.

62

Definition 8.3 We say that the function : [ ]→ R is integrable on the interval [ ] if there exists a number such that for any 0 there exists a division ∆ such that

− (∆) (∆) + (45)

The number is called the Riemann integral of on [ ] and is denoted byZ

() = (46)

Remark 8.4 In other words, the above definition says that the function is integrable if the upper and lower

Riemann sums get arbitrarily close.

Example 8.5 Consider the function : [0 1]→ R defined by () = 2.

To see if is integrable, consider the partition

∆ : 0 1

2

− 1

1

(so =, = 0 1 ).

We have

(∆) =

X=1

( − −1) inf∈[ −1

] ()

=

X=1

µ

− − 1

¶inf

∈[ −1 ]2

=1

X=1

µ− 1

¶2=

1

3

³12 + 22 + + (− 1)2

´=

1

3(− 1) (2− 1)

6

=(− 1) (2− 1)

62

Similarly,

(∆) =

X=1

( − −1) sup∈[ −1

] ()

=

X=1

µ

− − 1

¶sup

∈[ −1 ]2

=1

X=1

µ

¶2=

1

3

¡12 + 22 + + 2

¢=

1

3 (+ 1) (2+ 1)

6

=(+ 1) (2+ 1)

62

Since the limits

lim→∞

(∆) = lim→∞

(− 1) (2− 1)62

=2

6=1

3

and

lim→∞

(∆) = lim→∞

(+ 1) (2+ 1)

62=2

6=1

3

63

are equal, for any 0 we can find a partition ∆ such that

1

3− (∆) (∆)

1

3+

so by the definition it follows that () = 2 is integrable on [0 1] andZ 1

0

2 =1

3

8.1.3 Properties of the Riemann integral

In practice, the definition (as in the previous example) is rarely used to show that a function is integrable. Instead,

we can use the following.

Proposition 8.6 If : [ ]→ R is continuous on the interval [ ], then is integrable on [ ].

Proof. To be written.

Some properties of the Riemann integral are contained in the following.

Theorem 8.7 (Properties of the Riemann integral) Let : [ ]→ R.

a) If ∈ [ ] and is integrable on [ ] and on [ ], then is integrable on [ ] andZ

() =

Z

() +

Z

()

b) If is integrable on [ ] and ∈ R is a constant, then is integrable on [ ] andZ

() =

Z

()

c) If and are integrable on [ ] then + is integrable on [ ] andZ

() + () =

Z

() +

Z

()

d) If and are integrable on [ ] and () ≤ () for every ∈ [ ], thenZ

() ≤Z

()

In particular, ¯¯Z

()

¯¯ ≤

Z

| ()|

Proof. Follows from the corresponding properties of lower/upper Riemann sum.

For example, property b) follows using the fact that

(∆) = (∆) and (∆) = (∆)

Another important property of the Riemann integral is the following.

Theorem 8.8 (Mean value theorem for Riemann integral) If : [ ] → R is continuous on [ ], then

there exists ∈ [ ] such that1

−

Z

() = () (47)

64

Proof. Since is continuous on [ ], by Weierstrass boundedness theorem (Theorem 4.9) it follows that is

bounded on [ ] and attains its bounds, so there exists ∈ [ ] such that

() ≤ () ≤ ( ) ∈ [ ]

By part d) of the previous theorem, it follows thatZ

() ≤Z

() ≤Z

( )

or equivalent (noteR = (− ) for any constant ∈ R)

() ≤ 1

−

Z

() ≤ ( )

The number 1−

R () (called the mean value of on [ ]) lies between the values () and ( ) of

. Since is continuous, by the intermediate value theorem it follows that there exists between and such

that1

−

Z

() = ()

concluding the proof.

8.1.4 Antiderivatives

In order to find an easier way for computing Riemann integrals, we introduce the notion of antiderivative.

Definition 8.9 A differentiable function is called an antiderivative or indefinite integral of a function on

an interval [ ] if 0 () = () for every ∈ [ ].

Example 8.10 Consider the function () = 2. Then () = 3

3is an antiderivative of on any interval , for

0 () =µ3

3

¶0=32

3= 2 = () ∈

Also, note that by adding an arbitrary constant C ∈ R to () gives another derivative of :µ3

3+ C

¶0= 2 = () ∈

So 3

3+ C (for any C ∈ R) are antiderivatives of () = 2.

The next theorem shows that once we have found one antiderivative of a function, we can find all its

antiderivatives by just adding constants to it.

Theorem 8.11 If is an antiderivative of on an interval , the any other antiderivative of on is of the

form + C for a constant C ∈ R.

Proof. Let 1 be another antiderivative of .

We have

(1 ()− ())0= 01 ()− 0 () = ()− () = 0 ∈ [ ]

Since the derivative of the function 1 ()− () is identically zero on [ ], by Proposition 6.18 it follows that

the function is constant on [ ], so there exists ∈ R such that

1 ()− () = C ∈ [ ]

or equivalent

1 () = () + C ∈ [ ] concluding the proof.

65

We will denote the set of antiderivatives of on an interval byZ () = () + C

where is one (any) antiderivative of . Note that the interval does not appear explicitly in this notation, and

it should be clear from the context of the problem.

To find an antiderivative of a continuous functions we may use the following.

Theorem 8.12 If : [ ]→ R is continuous on [ ], then

() =

Z

()

is an antiderivative of on [ ], that is is differentiable on [ ] and 0 () = () for every ∈ [ ].

Proof. Consider 0 ∈ [ ] arbitrarily fixed.For an arbitrarily ∈ [ ]− {0}, using Theorem 8.7 a), we have:

()− (0)

− 0=

1

− 0

µZ 0

() −Z

()

¶=

1

− 0

Z 0

()

By the mean value theorem (Theorem 8.8), there exists between 0 and such that

1

− 0

Z 0

() = ()

so combining with the above we obtain ()− (0)

− 0= ()

Since the function is continuous on [ ] (so in particular is continuous at 0) and since → 0 as → 0,

we obtain

lim→0

() = (0)

This shows that the limit

lim→0

()− (0)

− 0= lim

→0 () = (0)

exists and equals (0), so is differentiable at 0 and 0 (0) = (0), for any 0 ∈ [ ].The above theorem can be generalized as follows.

Corollary 8.13 (Differentiation wrt integration limits) If : [ ] → R is continuous and : [ ] →[ ] are differentiable functions, the function

() =

Z ()

()

() ∈ [ ]

is differentiable on [ ] and we have

0 () = ( ()) · 0 ()− ( ()) · () ∈ [ ]

The main relation between antiderivatives and the Riemann integral is contained in the following.

Theorem 8.14 (Fundamental theorem of Calculus / Leibniz-Netwon formula) If : [ ] → R is con-

tinuous on [ ] and is an antiderivative of on [ ], thenZ

() = ()| = ()− () (48)

66

Proof. By the previous theorem it follows that the functionZ

() ∈ [ ]

is an antiderivative of . Since is also an antiderivative of , it follows that they differ by a constant: there exists

C ∈ R such that () =

Z

() + C ∈ [ ]

We have

()− () =

ÃZ

() + C!−µZ

() + C¶=

Z

() + C − 0− C =Z

()

concluding the proof.

Example 8.15 The function () = 2 is continuous on [0 1] and an antiderivative of is () = 3

3, so using

the above theorem we can easily compute the integral of on [0 1] (compare with the previous calculation of the

integral) like this: Z 1

0

2 =3

3

¯10

=13

3− 0

3

3=1

3

As another example, consider the following.

Example 8.16 An antiderivative of () = 2+ 3 cos is () = 2 + 3 sin (check!), so the integral of over

the interval [ 2] isZ 2

2+ 3 cos = 2 + 3 sin¯2=³(2)

2+ 3 sin (2)

´− ¡2 + 3 sin¢ = 32

8.1.5 Techniques of integration

There are mainly three methods for computing an integralR () : using a formula, using a substitution (in

fact this method just simplifies the integral, so that it becomes “easier”), or integration by parts.

• Using a formula

To be able to compute integrals, you should start by learning the antiderivatives of some common functions (see

the table of antiderivatives at the end of the notes, for example).

Example 8.17 Using the fact that the antiverivative of sin is − cos and the Leibniz-Newton formula, we haveZ 2

0

sin = − cos|20 = − cos 2− (− cos 0) = 0 + 1 = 1

• Using a substitution

If you have to compute an integral, maybe a substitution will help you to simplify (and hence compute) the

integral. There are mainly two possibilities.

The first is to try to substitute = () for a convenient function () ( is the “new” variable).

Since = (), = 0 (), which can be written in differential form = 0 () . The substitution formula

is in this case Z () =

Z ( ()) 0 ()

If the resulting integral in is easier (can be computed), it means you have the “right” substitution. If not...

try again (or use integration by parts).

67

Example 8.18 ComputeR√+ 1.

Consider the substitution = − 1, so = . The integral becomesZ√+ 1 =

Z(− 1)√− 1 + 1

=

Z√−√

=

Z32 − 12

=52

52− 32

32+ C

=2

5(+ 1)

52 − 23(+ 1)

32+ C

A second possibility is to consider a substitution of the form () = . Since = (), = 0 () which can

be written in differential form = 0 () . The substitution formula is in this caseZ ( ()) 0 () =

Z ()

If the resulting integral in is easier (can be computed), it means you have the “right” substitution. If not...

try again (or use integration by parts).

Example 8.19 ComputeR2

3

.

Note that if you consider 3 = , then 32 = , so 2 can be replaced by 13. The given integral can be

computed as follows Z2

3

=1

3

Z =

1

3 + C = 1

3

3

+ C

Example 8.20 ComputeR¡2 + 3

¢4.

Note that if you consider 2 + 3 = then 2 = , or = 12. The given integral can be computed as

follows Z¡2 + 3

¢4 =

1

2

Z4 =

1

2

5

5+ C =

¡2 + 3

¢510

+ C

• Using integration by parts

The integration by parts formula is the followingZ0 = −

Z0

Choosing 0 in the following order works in most cases:

1. Exponential ( 22 3, etc)

2. Sine or cosine (sin (2) cos sin, etc)

3. Power (2 32 5, etc)

4. 1 (the constant function 1)

Example 8.21 ComputeR2.

Since there is an exponential inside the integral, this should be chosen as 0. So 0 = 2 (hence =R2 =

122) and = (so 0 = 0 = 1).

68

Using the integration by parts formula we obtainZ2 =

Z µ1

22¶0·

=1

22 · −

Z1

22 · 1

=2

2− 12

Z2

=2

2− 12·

2

2+ C

=2

2− 2

4+ C

Example 8.22 ComputeRln.

The integral does not contain exponentials, sine or cosine, powers. So choose 0 = 1 (so =R1 = ) and

= ln (so 0 = (ln)0 = 1).

Using the integration by parts formula we obtainZln =

Z()

0ln = ln−

Z · 1

= ln−

Z1 = ln− + C

8.2 The line integral

The line integral is a natural generalization of the definite Riemann integralR () . In the integral

R ()

we integrate the function () from = to = along the line segment [ ] from to . Replacing the line

segment [ ] by a general curve in the plane or in the space, we obtain the line (or curve) integralR ·

defined below.

We begin by defining the notion of curve (either in the plane R2 or in the space R3).

Definition 8.23 A curve is the oriented image (graph) of a continuous function : [ ] → R3, () =( () () ()), ∈ [ ].The function = () is called a parametrization of the curve , and () is called the initial/starting

point of the curve and () is called the terminal/endpoint of the curve . If () = () be say that the

curve is a closed curve.

If is has a derivative 0 () = (0 () 0 () 0 ()) which is continuous at all points ∈ ( ) and 0 () 6= (0 0 0),we say that is a smooth curve.

Remark 8.24 Some useful examples of curves/parametrizations are given below:

1. A parametrization of the line segment from (1 1 1) to (2 2 2) is given by : [0 1]→ R3, where

() = (1− )+

= (1− ) (1 1 1) + (2 2 2)

= ((1− )1 + 2 (1− ) 1 + 2 (1− ) 1 + 2) ∈ [0 1]

2. A parametrization of an arc of circle of center (0 0) and radius 1, between angles and is : [ ]→ R2,where

() = (cos sin ) ∈ [ ]

More generally, if the circle has radius and center (0 0), the corresponding parametrization is

() = (0 + cos 0 + sin ) ∈ [ ]

3. A parametrization which is very useful in practice (exercises) is that of the graph of a given continuous function

: [ ]→ R2. In this case, the parametrization is given by : [ ]→ R2, where

() = ( ()) ∈ [ ]

69

For example, the parametrization of a part of a parabola, given by the graph of the function : [−2 3]→ R2, () = 2 − 5+ 7 is given by

() =¡ 2 − 5+

¢ ∈ [−2 3]

(to remember this easily, just set = ∈ [−2 3] - the parameter, and then = () = () = 2 − 5+ 7, so () = ( () ()) =

¡ 2 − 5+

¢).

4. If : : [ ] → R3 is a given curve, then the reversed curve − obtaining by reversing the initial andstarting points of the curve , has the parametrization − : [ ]→ R3, where

− () = (+ − ) ∈ [ ]

With this preparation, we can now give the definition of the line integral, as follows:

Definition 8.25 Given a continuous function = (1 2 3) defined on smooth curve : = () = ( () () ()),

∈ [ ], we define the line integral of over the curve byZ

· =Z

( ()) · 0 ()

Remark 8.26 Given two vectors = (1 2 3) and = (1 2 3) in R3, · denotes the scalar product/dotproduct of the vectors and , defined by

· = 11 + 22 + 33

The line integralR · defined above is therefore equal toZ

1+ 2 + 3

=

Z

(1 ( () () ()) 2 ( () () ()) 3 ( () () ())) · (0 () 0 () 0 ())

=

Z

1 ( () () ())0 () + 2 ( () () ())

0 () + 3 ( () () ()) 0 ()

This formula says that in order to compute the line integralR · we integrate the dot product of the vectors

( ()) = (1 ( ()) 2 ( ()) 3 ( ())) and 0 () = (0 () 0 () 0 ()) over the interval [ ].

Remark 8.27 It is not immediate from the definition that the line integralR · is independent of the

parametrization. That is, if 1 : [ ] → R3 and 2 : [ ] → R3 are two parametrizations of the same orientedcurve , then Z

(1 ()) · 01 () =Z

(2 ()) · 02 ()

However, this follows by the change of variables in the definite integral, as follows: if 1 and 2 both describe

the same smooth curve , then it can be shown that 1 () = 2 ( ()), for some bijective function : [ ]→ [ ].

With the change of variables = (), we obtainZ

(2 ()) · 02 () =

Z =−1()

=−1() (2 ( ())) · 02 ( ())0 ()

=

Z

(1 ()) · 01 ()

=

Z

(1 ()) · 01 ()

Example 8.28 Compute the line integralR · where ( ) = ¡ 2 sin ¢ and is the arc of the circle

of center (0 0) and radius 1 located in the first quadrant (i.e. ≥ 0).A parametrization of the given curve is : [0 2]→ R3, where

() = (cos sin 0) ∈h0

2

i

70

We have 0 () =¡(cos )

0 (sin )

0 (0)

0¢= (− sin cos 0), and therefore the given integral isZ

· =

Z 2

0

( ()) · 0 ()

=

Z 2

0

¡cos sin2 sin 0

¢ · (− sin cos 0) =

Z 2

0

− sin cos + sin2 cos

= −sin2

2+sin3

3

¯=2=0

=

µ−12+1

3

¶−µ−02+0

3

¶= −1

6

Remark 8.29 (Physical interpretation of the line integral) In Physics, the work done by a constant force

acting on a particle moving in a straight line, is defined as = Force ·distance. In general, given a force (not

necessarily constant) and a particle moving on a curve (not necessarily a line segment), we can compute the

work done by dividing the curve by points () into small parts , = 1 , such that each curve is

approximately a straight line segment, and the force is approximately constant on .

It follows that the work done in moving the particle along the curve is approximately = ( ()) ·( ()− (−1)) ≈ ( ()) · 0 ()∆ (note that the tangent vector 0 () is a good approximation of the curve by a straight line if the points () are sufficiently close to each other and the curve is smooth), and therefore

we obtain

≈1 + + =

X=1

=

X=1

( ()) · ()∆

To make this formula exact, we pass to the limit with →∞, and we obtain

= lim→∞

X=1

( ()) · ()∆ =Z

( ()) · 0 ()

which shows that the line integralR · gives the work done by a force in moving a particle along the curve

. This is the reason for which the line integralR · is sometimes also called the work integral.

Since the line integral is defined by a definite (Riemann) integral, it also has the corresponding properties. More

precisely, we have:

Theorem 8.30 (Properties of the line integral) If are continuous functions defined on the smooth curve

, then

1.R

¡ +

¢ · = R · + R

· (additivity)

2.R

¡¢ · =

R · , for any constant ∈ R (homogeneity)

3. If = 1 ∪2 is formed by union of two curves, thenR1∪2 · =

R1

· + R2

· (additivity withrespect to the curve)

4.R− · = −

R ·

Proof. Follow from the corresponding properties of the definite integral: the properties 1) and 2) follow from the

properties Z

( () + ()) =

Z

() +

Z

()

and Z

() =

Z

()

71

of the definite integral.

The property 3) follows from the propertyZ

() =

Z

() +

Z

()

for any , and the property 4) follows from the propertyZ

() = −Z

()

For the definite integral, we saw (the Leibniz-Newton formula) that if 0 () = () for ∈ [ ], thenZ

() = ()− ()

We will see that with the appropriate definitions, a similar formula holds for the line integral. First, we introduce

the concept of independence of path of a line integral, as follows:

Definition 8.31 We say that the line integralR · is independent of the path of integration in the

domain , if for any two points ∈ and any two curves 1 and 2 contained in with starting point

and endpoint , the corresponding line integrals are equal:Z1

· =Z2

·

We have:

Theorem 8.32 If = (1 2 3) is continuous in a domain ⊂ R3, then the line integralR · is independent

of path in if and only if there exists a function : → R such that

= grad =

µ

¶in

In this case, the line integralR · can be computed as follows:Z

· = ()− ()

where is the endpoint and is the starting point of the curve .

Proof. Assume that there exists the function : → R such that = grad , that is 1 = 2 =

and

3 =

Let : = () = ( () () ()) : [ ] → R3 be a curve with starting point = () and endpoint

= (). and If the curve We haveZ

· =

Z

( ()) · 0 ()

=

Z

1 ( ())0 () + 2 ( ())

0 () + 3 ( ()) 0 ()

=

Z

( ())0 () +

( ()) 0 () +

( ()) 0 ()

=

Z

( () () ())

= ( ())|==

= ( ())− ( ())

= ()− ()

72

and therefore the line integralR · does not depend on the path of integration (depends only on its endpoint

and starting point).

Conversely, if the line integralR · does not depend on the path of integration, it can be shown that the

function : → R defined by

( ) =

Z

·

where is an arbitrary curve contained in with starting point (0 0 0) ∈ arbitrarily fixed and endpoint

( ) ∈ , is differentiable in and satisfies the condition = grad .

To see this, we choose a curve = 1 ∪ 2 such that the last part of the curve (the part close to the endpoint( )) is a line segment 2 = [(− ) ( )] parallel to the - axis. We obtain:

( ) =

Z

·

=

Z1

· +Z2

·

= (− ) +

Z2

·

and therefore

( )− (− ) =

Z2

·

Using the parametrization 2 : : [0 1]→ R3 given by

() = (1− ) (− ) + ( )

= (− )

we obtain

( )− (− ) =

Z2

·

=

Z 1

0

1 (+ − ) · (+ − )0+ 2 (+ − ) ()

0+

1 (+ − ) ()00 + 2 (+ − ) · ()0 + 1 (+ − ) · ()0

=

Z 1

0

1 (+ − ) · + 2 (+ − ) · 0 + 1 (+ − ) · 0

=

Z 1

0

1 (+ − )

and therefore

lim→0

( )− (− )

− (− )= lim

→0

Z 1

0

1 (+ − )

= 1 ( )

which shows that has partial derivative with respect to

( ) = 1 ( )

A similar proof shows that = 2 and

= 3, and therefore = (1 2 3) =

³

´= grad ,

concluding the proof.

Example 8.33 Consider the line integralR · , where = (+ + cos ) and is a curve from =

(0 0 0) to = (1 2 2)

73

It can be checked that = grad³2

2+ 2

2+ sin

´, and therefore from the above theorem it follows that

Z

· =2

2+

2

2+ sin

¯(122)(000)

=

µ1

2+22

2+ sin

2

¶−µ0

2+ 02 + sin 0

¶=

1

2+ 2 + 1

=7

2

In general, it is not very easy to see whether a given function is the gradient of a certain function . The

following theorem gives us a simple criterion for deciding whether = grad :

Theorem 8.34 Let be a simply connected domain and = (1 2 3) have continuous partial derivatives.

ThenR · is independent of path in if and only if curl = 0, where

curl =

¯¯

1 2 3

¯¯ = µ3

− 2

−3

+

1

2

− 1

¶

Proof. AssumeR · is independent of path. By the previous theorem, = grad , or 1 =

, 2 =

and

3 =for some differentiable function : → R.

We have

3

− 2

=

µ

¶−

µ

¶=

2

− 2

= 0

by Schwarz’s theorem (since has continuous partial derivatives and = grad =³

´, has continuous

second order partial derivatives, and therefore 2

= 2

).

Similar proof shows −3+ 1

= 0 and 2

− 1

= 0, and therefore curl = (0 0 0) = 0.

We will not give here the proof of the converse (it requires Stokes’s theorem).

8.3 The line integral with respect to the arc length

We will first introduce the notion of arc length of a curve. Consider a smooth curve, with parametrization

: [ ] → R3, () = ( () () ()). To approximate the length of the curve , we consider a partition

∆ : = 0 1 = and we approximate the length of the curve by the length of the polygonal line

passing through the points = () = ( () () ()), = 0 1 . We have

Length () ≈X=1

|| ()− (−1)||

=

X=1

q( ()− (−1))

2+ ( ()− (−1))

2+ ( ()− (−1))

2

Applying Lagrange theorem to the function : [−1 ] → R it follows that there exists a point ∗ ∈ [−1 ]

74

such that ()− (−1) = 0 (∗ ) ( − −1), and similarly for the functions () and (). We obtain

Length () ≈X=1

q(0 (∗ ) ( − −1))

2+ (0 (∗∗ ) ( − −1))

2+ (0 (∗∗∗ ) ( − −1))

2

=

X=1

q(0 (∗ ))

2+ (0 (∗∗ ))

2+ (0 (∗∗∗ ))

2( − −1)

=

X=1

q(0 (∗ ))

2+ (0 (∗∗ ))

2+ (0 (∗∗∗ ))

2∆

To make the above approximation exact, we pass to the limit with ∆ = − −1 → 0. Since is a smooth

curve, by the definition of the Riemann integral it follows that the limit of the above sum exists and we have

lim∆&0

X=1

q(0 (∗ ))

2+ (0 (∗∗ ))

2+ (0 (∗∗∗ ))

2∆ =

Z

q(0 ())2 + (0 ())2 + (0 ())2

We are thus lead to the following:

Definition 8.35 Given a smooth curve with parametrization : [ ]→ R, () = ( () () ()), we definethe length of the curve by

Length () =

Z

q(0 ())2 + (0 ())2 + (0 ())2

Remark 8.36 The length of the curve defined above depends on the parametrization = (). Since there are

many possible paramtrizations of a given smooth curve , we should check that the above definition is independent

on the choice of the parametrization. However, this follows by the independence of the line integral with respect to

the arc length (see below).

Example 8.37 Let’s use the above formula to compute the length of the circle of radius centered at the origin.

A parametrization of is given by : [0 2]→ R, () = ( cos sin 0).We have 0 () = (− sin cos 0), and using the definition above we obtain

Length () =

Z 2

0

q(− sin )2 + ( cos )2 + 02

=

Z 2

0

= |20= 2

which coincides with the formula known from geometry (i.e. the length of a circle of radius is 2).

To make it easier to remember the next definition, we consider a small part of the curve in the previous

discussion, say between the points () and (+∆). By the above definition, its length ∆ is given by

∆ =

Z +∆

q(0 ())2 + (0 ())2 + (0 ())2

and by the mean value theorem we obtain

∆ =

q(0 (∗))2 + (0 (∗))2 + (0 (∗))2∆

where ∗ is a intermediate point between and +∆.

Passing to the limit with ∆→ 0, it follows that

lim∆→0

∆

∆= lim∆→0

q(0 (∗))2 + (0 (∗))2 + (0 (∗))2 =

q(0 ())2 + (0 ())2 + (0 ())2

75

or

=

q(0 ())2 + (0 ())2 + (0 ())2

This formula shows that the length = () of an infinitesimal arc of the curve (called the arc length of the

curve ) has length

=

q(0 ())2 + (0 ())2 + (0 ())2

and justifies the following definition:

Definition 8.38 Given a continuous function = ( ) defined on a smooth curve : = () = ( () () ()),

∈ [ ], we define the line integral of with respect to the arc length of the curve byZ

=

Z

( ()) ||0 ()|| =Z

( ())

q(0 ())2 + (0 ())2 + (0 ())2

Remark 8.39 It is not immediate from the definition that the line integralR is independent of the para-

metrization. That is, if 1 : [ ] → R3 and 2 : [ ] → R3 are two parametrizations of the same oriented curve, then Z

(1 ()) ||01 ()|| =Z

(2 ()) ||02 ()||

However, this follows by the change of variables in the definite integral, as follows: if 1 and 2 both describe

the same smooth curve , then it can be shown that 1 () = 2 ( ()), for some bijective function : [ ]→ [ ].

With the change of variables = (), we obtainZ

(2 ()) ||02 ()|| =

Z −1()

−1() (2 ( ())) ||02 ( ())||0 ()

=

Z

(2 ( ())) ||02 ( ())0 ()||

=

Z

(1 ())¯¯(2 ( ()))

0¯¯

=

Z

(1 ()) · ||01 ()||

Example 8.40 Compute the line integralR+2− where is the arc of the circle of center (0 0) and radius

1 located in the first quadrant (i.e. ≥ 0) oriented counterclockwise.A parametrization of the given curve is : [0 2]→ R3, where

() = (cos sin 0) ∈h0

2

i

We have 0 () =¡(cos )

0 (sin )

0 (0)

0¢= (− sin cos 0), and therefore the given integral isZ

+ 2 − =

Z 2

0

(cos + 2 sin − 0)q(− sin )2 + (cos )2 + 02

=

Z 2

0

cos + 2 sin

= sin − 2 cos |=2=0

= (1− 2 · 0)− (0− 2 · 1)= 3

Since the line integral with respect to the arc length is defined by a definite (Riemann) integral, it also has the

corresponding properties. More precisely, we have:

Theorem 8.41 (Properties of the line integral wrt arc length) If are continuous functions defined on

the smooth curve , then

76

1.R( + ) =

R+

R (additivity)

2.R() =

R, for any constant ∈ R (homogeneity)

3. If = 1 ∪2 is formed by union of two curves, thenR1∪2 =

R1

+R2

(additivity with respect

to the curve)

4.R− =

R

Proof. Consider : [ ]→ R, () = ( () () ()) a parametrization of the curve .By definition, we haveZ

( + ) =

Z

( ( ()) + ( ()))

q(0 ())2 + (0 ())2 + (0 ())2

=

Z

( ())

q(0 ())2 + (0 ())2 + (0 ())2+Z

( ())

q(0 ())2 + (0 ())2 + (0 ())2

=

Z

+

Z

and similarly Z

() =

Z

( ( ()))

q(0 ())2 + (0 ())2 + (0 ())2

=

Z

( ())

q(0 ())2 + (0 ())2 + (0 ())2+

=

Z

using the additivity and homogeneity of the Riemann integral.

To prove 3), since = 1 ∪ 2, there exists a point ∈ ( ) such that 1 : [ ] → R, 1 () = () is a

parametrization of 1 and 2 : [ ]→ R, 2 () = () is a parametrization of 2.

From the definition of the line integral with respect to the arc length, we obtainZ1∪2

=

Z

( ()) ||0 ()||

=

Z

( ()) ||0 ()|| +Z

( ()) ||0 ()||

=

Z

(1 ()) ||01 ()|| +Z

(2 ()) ||02 ()||

=

Z1

+

Z2

To prove part 4), note that a parametrization for the reversed curve − is 1 : [ ]→ R, 1 () = (), and that

we have 01 () =

( (+ − )) = −0 (+ − ). Using the definition and using the substitution = + −

(note that = −), we obtainZ−

=

Z

(1 ()) ||01 ()||

=

Z

( (+ − )) ||− (+ − )||

= −Z

( ()) || ()||

=

Z

( ()) || ()||

=

Z

77

which concludes the proof.

There are 3 applications of the line integral with respect to the arc length, as follows:

1. Length of a curve

It is immediate from the two definitions of above that the length of a curve can be computed using the line

integral:

Length () =

Z

1

(just compare the first definition above with the second one, in which ≡ 1)2. Mass of a curve with mass density = ( )

The mass of a curve with variable mass density (for example a wire which is not uniform, having parts that

are heavier/lighter) can be computed using the line integral as follows:

Mass () =

Z

To see why this formula is so, divide the curve into small parts by points (), = 1 , where

: [ ]→ R is a parametrization of and = 0 1 = is a partition of [ ].

If the partition is fine enough and the mass density is continuous, the part of the curve between points

(−1) and () has approximately constant density, hence its mass is approximately ( ()) (density)

times the length of this arc. Summing up the masses of all such arcs, we obtain

Mass () ≈X=1

( ())

q(0 (∗ ))

2+ (0 (∗∗ ))

2+ (0 (∗∗∗ ))

2

To make the above approximation exact, we pass to the limit with the norm of the partition tending to zero,

and we obtain

Mass () =

Z

( ())

Z

( ())

q(0 ())2 + (0 ())2 + (0 ())2

which by definition equalsR.

3. The center of mass ( ) of a given curve with mass density = ( ) is given by

=

RR

=

RR

=

RR

The reason for the above formulae comes from Physiscs (the sum of moments with respect to the center of

mass of all forces involved equals zero), but we will not discuss it here.

Example 8.42 Find the length, mass and center of mass of the curve with mass density ( ) = 3, where

is the arc of parabola = 2 lying in the first quadrant.

A paramterization of the curve is : [0 1]→ R, () =¡ 2 0

¢.

From the above formulae we obtain:

- Length of

Length () =

Z

1

=

Z 1

0

q(1)

2+ (2)

2+ 02

= 4

Z 1

0

r1

4+ 2

78

Using the formulaR √

2 + 2 = 2

√2 + 2 + 2

2ln¡+√2 + 2

¢+ , we obtain

Length () = 4

Ã

2

r1

4+ 2 +

14

2ln

Ã+

r1

4+ 2

!!¯¯1

0

= 4

Ã1

2

√5

2+1

8ln

Ã1 +

√5

2

!!

=√5 +

1

2ln

Ã1 +

√5

2

!

- Mass of

Mass () =

Z

=

Z 1

0

3

q(1)

2+ (2)

2+ 02

= 3

Z 1

0

p1 + 42

Using the substitution = 1 + 42 (note that = 8) we obtain

Mass () = 3

Z 1

0

p1 + 42

= 3

Z 5

1

√1

8

=3

8

Z 5

1

12

=3

8

32

32

¯51

=5√5− 14

- Center of mass of : ( )

For example, the coordinate of the center of mass is

=

RR

=

R 102 · 3√1 + 42

5√5−14

=12

5√5− 1

Z 1

0

3p1 + 42

79

With the substitution = 1 + 42 (note that = 8), we obtain

=12

5√5− 1

Z 5

1

− 14

√1

8

=3

8¡5√5− 1¢

Z 5

1

32 − 12

=3

8¡5√5− 1¢

µ52

52− 32

32

¶¯51

=3

8¡5√5− 1¢

µ552

52− 5

32

32

¶− 3

8¡5√5− 1¢

µ1

52− 1

32

¶=

15√5

8¡5√5− 1¢

µ2− 2

3

¶− 3

8¡5√5− 1¢

µ− 415

¶=

5√5

2¡5√5− 1¢ + 1

10¡5√5− 1¢

=25√5 + 1

10¡5√5− 1¢

and similarly for and .

8.4 The double integral

The double integral can be viewed as an extension of the definite integralR () , as follows. Consider a function

of two variables, = ( ) : → R defined on a domain ⊂ R2. We defined the integral by dividing the interval[ ] into subintervals [−1 ], = 1 , of length ∆ = − −1 and taking limits of the sums of the form

X=1

(∗ )∆

where ∗ ∈ [−1 ] are arbitrarily chosen points in the interval [−1 ], = 1 .Proceeding similarly, we can divide the given region into · rectangles with sides parallel to the and

axes (see Figure 13), and we can form the sum

X=1

X=1

¡∗

∗

¢∆∆

where¡∗

∗

¢is an arbitrary point in the rectangle , = 1 and = 1 .

Figure 13: The domain dividided into rectangles .

80

Figure 14: The domain between the graph of two functions () ≤ ≤ (), ∈ [ ].

If the function is continuous on and is a “nice” domain (its boundary is a smooth curve), it can be shown

that the limit as ∆,∆ → 0 of the above sum exists and is finite, and we denote the limit byZ Z

( ) = lim∆∆→0

X=1

X=1

¡∗

∗

¢∆∆ (49)

and and call it the double integral of over the domain .

The double integral has similar properties with the definite integral, as follows:

Theorem 8.43 Let : → R be continuous functions defined on the domain ⊂ R2. We have:

1.R R

( ( ) + ( )) =

R R ( ) +

R R ( ) (additivity)

2.R R

( ) =

R R ( ) for any constant ∈ R (homogeneity)

3. If = 1 ∪2, thenR R

1∪2 ( ) =

R R1

( ) +R R

2 ( )

4. There exists (0 0) ∈ such thatR R

( ) = (0 0) ·Area ()

8.4.1 Evaluation of the double integral

There are essentially two different ways for computing the double integral: as an iterate (succesive) integral or by a

change of variables (see also Green’s theorem for an alternate possibility for the evaluation of the double integral).

Double integral as an iterate integral Assume that the domain is the region between the graphs of two

functions : [ ]→ R of the variable (see Figure 14), that is:

= {( ) : () ≤ ≤ () ∈ [ ]}

Computing the limit in (49) as an iterate integral, we obtainZ Z

( ) = lim∆∆→0

X=1

X=1

¡∗

∗

¢∆∆

= lim∆→0

X=1

⎛⎝ lim∆→0

X=1

¡∗

∗

¢∆

⎞⎠∆=

Z

ÃZ ()

()

( )

!

81

Figure 15: The domain between the graph of two functions () ≤ ≤ (), ∈ [ ].

and therefore in this case we can compute the double integral by first integrating the function first with respect

to (between the limit () and ()) and then integrating the resulting function with respect to (between the

limits and ): Z Z

( ) =

Z

ÃZ ()

()

( )

! (50)

A similar formula holds if we reverse the roles of the variables and , that is if the domain can be described

as

= {( ) : () ≤ ≤ () ∈ [ ]} for some functions : [ ]→ R of the variable (see Figure 15).In this case, we can compute the double integral by first integrating with respect to (between () and ())

and then integrating the resulting function with respect to the variable (between and ):Z Z

( ) =

Z

ÃZ ()

()

( )

! (51)

Example 8.44 Compute the double integralR R

where is the upper half of the unit disk 2 + 2 1.

The given domain is of the form in Figure 14 (sketch a picture!), where () = 0 and () =√1− 2,

∈ [−1 1], and also of the form in Figure 14, where () = −p1− 2 and () =

p1− 2, ∈ [0 1].

Using the first representation above, we haveZ Z

=

Z 1

−1

ÃZ √1−20

!

=

Z 1

−1

2

2

¯=√1−2=0

=

Z 1

−1

1− 2

2

=

2− 3

6

¯=1=−1

=

µ1

2− 16

¶−µ−12− −16

¶=

2

3

82

Using the second representation above, we haveZ Z

=

Z 1

0

ÃZ √1−2−√1−2

!

=

Z 1

0

|=√1−2

=−√1−2

=

Z 1

0

2p1− 2

= −23

¡1− 2

¢32 ¯=1=0

=2

3

obtaining the same result.

Change of variables in the double integral Recall that for the definite integralR () , using the change

of variable = () : [ ]→ [ ], we obtainZ

() =

Z

( ())0 ()

Similarly, changing the variables in the double integralR R

( ) by½

= ( )

= ( ) ( ) ∈ ∗ (52)

we obtain the following change of variables formulaZ Z

( ) =

Z Z∗

( ( ) ( ))

¯ ( )

( )

¯ (53)

The formula says that changing the variables from ( ) ∈ → ( ) ∈ ∗, we have to change accordingly the

domain of integration (from to ∗), and the area element by¯()

()

¯, where

( )

( )=

¯

¯is the Jacobian of the change of variables (52) - the determinant of the matrix of partial derivatives of the “old”

variables with respect to the “new” variables (note that the corresponding “Jacobian” for the definite

integral is just = 0 ()).

Remark 8.45 To see why this formula is so, we go back to the definition (49), and we see that the area element

is the limiting area ∆∆ of a rectangle, and similarly for and ∆∆.

Approximating = ( ) and = ( ) by a Taylor’s formula of order 1 near the point ( ), we obtain

a linear transformation ½ ( ) ≈ ( ) +

(− ) +

( − )

( ) ≈ ( ) +(− ) +

( − )

where all the partial derivatives are evaluated at the point ( ).

It can be shown that under a linear transformation, the area element element changes by a factor equal to the

absolute value of the Jacobian, that is

∆∆ =

¯ ( )

( )

¯∆∆

which by initial remark shows that

=

¯ ( )

( )

¯

83

Example 8.46 Compute the integralR R

2+2 where is the disk 2 + 2 9.

By using the change of variables (polar coordinates):½ = cos

= sin ∈ [0 3] and ∈ [0 2]

the corresponding Jacobian is

( )

( )=

¯

¯=

¯cos sin

− sin cos

¯= cos2 + sin2 =

and therefore we obtain Z Z

2+2 =

Z Z[03]×[02]

2 cos2 +2 sin2

Computing the las integral as an iterated intgral, we haveZ Z

2+2 =

Z 2

0

µZ 3

0

2

¶

=

Z 2

0

1

2

2

¯=3=0

=

Z 2

0

9

2

=9

2

¯=2=0

= 9

8.4.2 Green’s Theorem

Green’s theorem gives us a formula which relates the double integral and the line integral, introduced in the previous

section.

Theorem 8.47 (Green’s theorem) Let ⊂ R2 be a bounded domain, having as boundary a smooth curve

(oriented such that the domain is on the left of the curve ). If = (1 2) : → R is continuous on and 12have continuous partial derivatives on 7 , thenZ Z

µ2

− 1

¶ =

Z

(1 2) ·

Remark 8.48 Note that the integrand on the left is just the third component of the curl , where = (1 2 3).

More generally, a similar formula holds (Stokes’s theorem), which states thatZ Z

curl¡¢ · = Z

where is an oriented surface with normal vector and is the boundary of the surface , oriented accordingly

(such that the surface is “to the left” of ).

Proof. It is sufficient to prove the theorem in the case when the domain is of the form in Figure 14 or in Figure

15 (we can reduce the proof to this case by dividing the domain into domains of this form and then use the

property 3) in Theorem 8.43).

Assume therefore that the domain is of the form in Figure 14, that is = {( ) : () ≤ ≤ () ∈ [ ]} 7This means that 12 have partial derivatives in a domain ∗ containing .

84

Computing the double integral as an itrated integral, we haveZ Z

1

( ) =

Z

ÃZ ()

()

1

( )

!

=

Z

1 ( )|=()=()

=

Z

1 ( ())− 1 ( ())

=

Z

1 ( ()) −Z

1 ( ())

The boundary curve of can be written as the union of two curves

= ∪ −

where is the “lower” curve given by = (), with parametrization

: () = ( ()) ∈ [ ]

and is the “upper” curve given by = (), with parametrization

: () = ( ()) ∈ [ ]

We have Z

(1 0) · =

Z

(1 0) · +Z−

(1 0) ·

=

Z

(1 0) · −Z

(1 0) ·

=

Z

(1 ( ())) · (1 0 ()) −Z

(1 ( ())) · (1 0 ())

=

Z

1 ( ()) −Z

1 ( ())

= −Z Z

1

A similar proof shows that we also haveZ

(0 2) · =Z Z

2

and therefore adding the last two equalities, we obtainZ

(1 2) · =Z Z

2

− 1

concluding the proof.

8.4.3 Applications of the double integral

There are several applications of the double integral, among which we will reffer to the computation of volumes,

areas and center of mass.

Given a continuous function = ( ) : → R with ≥ 0 in , the volume of the region beneath the graphof the function and above the domain (see Figure ??) is given by

Volume =

Z Z

( )

85

Remark 8.49 To see why this is so, remember that the double integral is the limit of the sum in (49), in which

( )∆∆ gives an approximation of the parallelipiped with base the rectangle and height ( ). Sum-

ming up over and taking the limit, this formula (that is, the double integral) gives the area under the graph of

the function = ( ).

Example 8.50 We can compute the volume of a sphere of radius (i.e. 2 + 2 + 2 2) by using the double

integral as follows: the volume of the sphere is by symmery twice the volume under the graph of the function

( ) =p2 − 2 − 2 above the domain : 2 + 2 2 (sketch a picture!).

Therefore, we obtain

Volume = 2

Z Z

p2 − 2 − 2

To compute this integral, we can either use an iterate integral, or a substitution. Choosing the substitution

(change to polar coordinates) ½ = cos

= sin ∈ [0 ] and ∈ [0 2]

the corresponding Jacobian is

( )

( )=

¯

¯=

¯cos sin

− sin cos

¯= cos2 + sin2 =

and therefore we obtain

Volume = 2

Z Z[0]×[02]

p2 − 2 cos2 − 2 sin2

= 2

Z 2

0

ÃZ

0

p2 − 2

!

= 2

Z 2

0

−13

¡2 − 2

¢32 ¯==0

= 2

Z 2

0

1

3

¡2¢32

=2

33

¯=2=0

=4

33

Taking ( ) ≡ 1 and remembering that a cylinder with a height of 1 has volume equal to the area of its base,from the previous formula we obtain a formula for computing the area of a given domain :

Area () =

Z Z

1

Example 8.51 We can compute the area of the disk : 2 + 2 2 by using the double integral as follows:

Area =

Z Z

1

Writing the domain as the domain between the curves () = −√2 − 2 and () =√2 − 2, with

∈ [−], we can compute the double integral as an iterated integral, as folows:

Area =

Z

−

ÃZ √2−2

−√2−21

!

=

Z

−2p2 − 2

To compute the integral, we can either use the formulaR √

2 − 2 = 2

√2 − 2 + 2

2arcsin

+ or we can

use a substitution, say = sin , with ∈ [−2 2], and after computation we obtainArea = 2

86

Interpreting the function ( ) ≥ 0 as the density of mass at the point ( ) ∈ , and proceeding similarly

with the previous remark, we see that the mass of a domain with mass density ( ) is given by

Mass () =

Z Z

( )

The center of mass of can be computed by the following formulae:

=

R R ( ) R R

( )

and =

R R ( ) R R

( )

9 Exercises (incomplete)

1. Compute the integralR 10 using the definition.

2. Differentiate with respect to the given integrals.

(a)R 1√2 + 1

(b)R 42 (+ 1)

3

(c)R 1+21−2

11+2

(d)R sin1

1

+√

3. Show that for any ∈ (0 1) we have Z 1+

1−

− 1 (2− )

= 0

4. Compute the integralR by using the definition, considering the partition

∆ : 2 =

5. Redo the previous exercise in the particular caseR 212, then use it to show that

lim→∞

Ã1

(+ 1)2+

1

(+ 2)2+ +

1

(+ )2

!=1

2

6. Compute the indicated indefinite integrals.

a)Rsin cos 7 b)

R2 cos 3 c)

R2 ln

d)R

cos 21−sin 2 e)

R1√

16−3+2 f)R

12+4−4

g)R

2

3+32+3+1 h)

R2+3−1

4+3+2+ i)

Rln¡6¢

j)R

12−5+4 k)

R2 cos l)

R3 +

√

m)R

4+√+1

n)

R2√− 1 o)

R√2 + 4

p)R4+2 q)

R

2

r)Rln

s)Rln5

t)R

2+32+3−4 u)

Rarcsin√1−2

v)R (√+1)10√

x)

R5 y)

R−4

(2+4)(+1)

7. Compute the following definite integrals.

a)R 32¡1 + 22

¢4 b)

R 41

√+1√ c)

R 21

2+1(+1)4

d)R 40

cos(1+sin)2

e)R 10√2 + 1 f)

R 41

√√

g)R 40

cos2 2 h)R 20

√2− 2 cos i)

R 40

cos 2

j)R 130

√1−2

87

8. (More exercise to be added)

9.

10. Describe the given region as the region between the graphs of two functions:

) = {( ) : 0 ≤ ≤ 3−1 ≤ ≤ 1}) =

©( ) : ≥ 2 − 9 ≤ 2− 1ª

) =©( ) : 2 + 2 ≤ 2ª

) =n( ) : (− 1)2 + 2 ≤ 4 (+ 1)2 + 2 ≤ 4

o) =

½( ) : 2 + 2 ≤ 4 2 + 1

42 ≥ 1 ≥ 0

¾) : ∆, where (−1 0) (2 0) (0 3)

11. Write the double integralR R

( ) as an iterated integral for the domains given is in the previous

exercise.

12. Evaluate the given double integrals

)

ZZ

sin () ; = {( ) : ∈ [0 1] ∈ [0 ]}

)

ZZ

¡2 + 2

¢ : 4, where (1 1) (2−1) (2 1)

)

ZZ

2 = {( ) : 0 ≤ ≤ 2; 0 ≤ ≤ }

)

ZZ

2 =©( ) : 2 + 2 ≤ 2ª

)

ZZ

=© ≥ 2 − 9, ≤ 2− 1 ∈ [−2 4]ª

)

ZZ

=© ≥ 2 − 9, ≤ 2− 1 ∈ [−2 3]ª

13. Find the volume of the regions indicated below

) 2 + 2 + 2 ≤ 2

) ≥ 2 + 2 ≤ 3

)2

2+

2

2+

2

2≤ 1

14. Evaluate the given double integrals by using a change of variables

)

ZZ

=©( ) : 2 + 2 ≤ 4 ≤ √3 ≥ 0ª

)

ZZ

ln¡2 + 2

¢2 + 2

:©( ) : 1 ≤ 2 + 2 ≤ 4ª

)

ZZ

(+ ) :©( ) : 2 + 2 ≤ +

ª

)R R

(+ ) is the parallelogram with vertices (0 0) (4 0) (1 3) (5 3)

88

3 Calculate the area, the mass and center of mass of the domain , where the mass density is given by

( ) = :

) : {( ) : ∈ [−1 1] ∈ [0 2]}) : 4, where (1 1) (1 4) (4 1)

) :n( ) : (− 1)2 + ( − 1)2 ≤ 1

o) =

n( ) :

2

2+ 2

2≤ 1

o

89

AppendixAntiderivatives of some common function

Function Antiderivative Domain

+1

+1 ∈ R ( ∈ R∗ for 0)

sin − cos ∈ Rcos sin ∈ R1

cos2 tan ∈ R−©2 ±

2: ∈ Zª

− 1sin2

cot ∈ R− { : ∈ Z}1√1−2 arcsin ∈ (−1 1)− 1√

1−2 arccos ∈ (−1 1)1

1+2arctan ∈ R

− 11+2

arccot ∈ R ∈ R

ln ∈ R

1

ln || ∈ R∗

90