Calculus - UNENE · 2020-03-19 · Calculus — Differentiation ©Wei-Chau Xie Physical Meanings of...

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Calculus

1

Calculus — Differentiation © Wei-Chau Xie

Derivatives

Some simple derivatives

d

dxxn = n xn−1

d

dxeax = a eax

d

dxln x = 1

x

d

dxsinax = a cosax

d

dxcosax = −a sinax

2


Important Rules for Evaluating Derivatives

[ f(x) + g(x)]′ = f ′(x) + g ′(x)

[c f(x)]′ = c f ′(x)

[ f(x) g(x)]′ = f ′(x) g(x) + f(x) g ′(x) Product Rule

[

f(x)

g(x)

]′= f ′(x) g(x) − f(x) g ′(x)

g2(x)Quotient Rule

y = y(u), u = u(x) =⇒dy

dx= dy

du· du

dxChain Rule

3


Example

Find the derivative of y = tan x.

y ′ = (tan x)′ =( sin x

cos x

)′

= (sin x)′ cos x − sin x (cos x)′

cos2 xQuotient Rule

( fg

)′= f ′g − f g ′

g2

= cos x · cos x − sin x · (− sin x)

cos2 x

= cos2 x + sin2 x

cos2 xsin2 x + cos2 x = 1

= 1

cos2 x= sec2 x

4


Example

Find the derivative of y = ln cos x.

y = ln cos x can be written as

y = ln u, u = cos x

Apply the Chain Rule

dy

dx= dy

du· du

dx

= 1

u· (− sin x) = 1

cos x· (− sin x)

= − tan x

5


Example

Find the derivative of y = ax.

y = ax can be written as

y = ax = eln ax = ex · ln a

∴ y = eu, u = x · ln a

Apply the Chain Rule

dy

dx= dy

du· du

dx

= eu · ln a = ax ln a

6


Higher-Order Derivatives

d2y

dx2= d

dx

(

dy

dx

)

d3y

dx3= d

dx

(

d2y

dx2

)

...

dny

dxn= d

dx

(

dn−1y

dxn−1

)

7


Physical Meanings of Derivatives

Consider a function y = y(x).

The first-order derivative y ′(x0) is the slope of the curve y(x) at (x0, y0).

The first-order derivative y ′(x0) is the rate of change of the function at x0.

The curvature of the curve is κ = y ′′

(1+ y ′2)3/2.

In many applications of physics and engineering, such as the equation of

bending in beams, the approximations to the fluid flow around surfaces,∣

∣ y ′∣∣≪1, and hence κ≈y ′′ .

Consider the motion of a particle (mass) with displacement x(t).

The first-order derivativedx

dt= v(t) is the velocity of the particle.

The second-order derivatived2x

dt2= a(t) is the acceleration of the particle.

8


Example

Find the second-order derivative of y = ex(sin x + cos x).

y ′ = [ex(sin x + cos x)]′

Product Rule ( f g)′ = f ′g + f g ′

= (ex)′(sin x + cos x) + ex (sin x + cos x)′

= ex (sin x + cos x) + ex (cos x − sin x)

= 2 ex cos x

y ′′ = [2 ex cos x]′


= 2 [(ex)′ cos x + ex (cos x)′]

= 2 [ex cos x + ex (− sin x)]

= 2 ex (cos x − sin x)

9


Example

Find the second-order derivative of y = ex2.

Rewrite y = eu, u = x2

∴dy

dx= dy

du· du

dxChain Rule

= eu · 2x = 2x ex2

∴d2y

dx2= [2x ex2

]′


= 2 [(x)′ ex2 + x (ex2)′]

= 2(

1 · ex2 + x · 2x ex2)

= 2 ex2(1 + 2x2)

10

Calculus — Taylor Series © Wei-Chau Xie

Taylor Series

In many applications, it is desirable to approximate complicated functions by

simple functions, such as polynomials.

Let f(x) and its derivatives f ′(x), f ′′(x), . . . , f (n)(x) exist and be

continuous in a closed interval a6x6b. Then for x0 in [a, b]

f(x) = f(x0) + f ′(x0)

1! (x−x0) + f ′′(x0)

2! (x−x0)2 + · · · + f (n)(x0)

n! (x−x0)n + · · ·

=∞∑

n=0

f (n)(x0)

n! (x−x0)n

Function f(x) can be approximated by a polynomial of degree n:

f(x) = f(x0) + f ′(x0)

1! (x−x0) + f ′′(x0)

2! (x−x0)2 + · · · + f (n)(x0)

n! (x−x0)n + Rn(x)

Rn = f (n+1)(ξ)

(n+1)! (x−x0)n+1, x0 6ξ 6x

11


MacLaurin Series

When x0 = 0, a Taylor series is also referred to as a MacLaurin series

f(x) = f(0) + f ′(0)

1! x + f ′′(0)

2! x2 + · · · + f (n)(0)

n! xn + · · · =∞∑

n=0

f (n)(0)

n! xn

Approximation of function f(x):

f(x) ≈ f(0) + f ′(0)

1! x + f ′′(0)

2! x2 + · · · + f (n)(0)

n! xn

The error in the approximation is

∣

∣Rn(x)∣

∣ = f (n+1)(ξ)

(n+1)! xn+1, 06ξ 6x

6M

(n+1)!∣

∣x∣

∣

n+1, M = max

∣

∣ f (n+1)(x)∣

∣

12


Example

Find the Taylor series at x0 = 0 (MacLaurin series) of f(x) = ex.

∵ f(x) = ex, f(0) = e0 = 1

f ′(x) = ex, f ′(0) = e0 = 1

f ′′(x) = ex, f ′′(0) = e0 = 1...

f (n)(x) = ex, f (n)(0) = e0 = 1

∴ ex = f(0) + f ′(0)

1! x + f ′′(0)

2! x2 + · · · + f (n)(0)

n! xn + · · ·

= 1 + x + x2

2! + x3

3! + · · · + xn

n! + · · ·

13


Example

Find the Taylor series at x0 = 0 (MacLaurin series) of f(x) = sin x.

∵ f(x) = sin x, f(0) = sin 0 = 0

f ′(x) = cos x, f ′(0) = cos 0 = 1

f ′′(x) = − sin x, f ′′(0) = − sin 0 = 0

f ′′′(x) = − cos x, f ′′′(0) = − cos 0 = −1

f (4)(x) = sin x, f (4)(0) = sin 0 = 0...

∴ f (2n)(0) = 0, f (2n+1)(0) = (−1)n

∴ sin x = f(0) + f ′(0)

1! x + f ′′(0)

2! x2 + · · · + f (n)(0)

n! xn + · · ·

= 0 + (−1)0

1! x + 0

2! x2 + (−1)1

3! x3 + · · · + (−1)n

(2n+1)! x2n+1 + · · ·

= x − x3

3! + x5

5! − x7

7! + · · · + (−1)n x2n+1

(2n+1)! + · · ·

14


sin x = x − x3

3! + x5

5! − x7

7! + · · · + (−1)n x2n+1

(2n+1)! + · · ·

–2

–1

0

1

2

3

0.5 1 321.5 2.5

sinx

x−x

x3

3!−

x7

7!+

x5

5!

−xx3

3!+

x5

5!

x

−xx3

3!

15

Calculus — Integration © Wei-Chau Xie

Integrals of Fundamental Functions

∫

xn dx = xn+1

n+1, n 6=−1

∫

1

xdx = ln

∣

∣x∣

∣, x 6=0

∫

ex dx = ex,

∫

eax dx = 1a

eax

∫

sin x dx = − cos x

∫

cos x dx = sin x

∫

tan x dx = − ln∣

∣cos x∣

∣

∫

cot x dx = ln∣

∣sin x∣

∣

∫

sec x dx = ln∣

∣sec x + tan x∣

∣

∫

csc x dx = ln∣

∣csc x − cot x∣

∣

16

Calculus — Integration - Change of Variables © Wei-Chau Xie

Techniques of Integration — Change of Variables

Example

Evaluate I =∫

x√

1 − x2 dx.

Let 1 − x2 = u, −2x dx = du =⇒ x dx = − 12

du

I =∫

√

1 − x2 · x dx =∫ √

u ·(

− 12

du)

= − 12

∫

u12 du = − 1

2· u

12+1

12 +1

+ C

∫

un du = un+1

n+1+ C

= − 13

u32 + C = − 1

3(1 − x2)

32 + C

17


Example

Evaluate I =∫

√

a2 − x2 dx, a>0.

Let x = a sin u =⇒ u = sin−1 x

a, dx = a cos u du sin2 x + cos2 x = 1

I =∫

√

a2 − a2 sin2 u · a cos u du, cos u =√

1 − sin2 u

= a2

∫

cos2 u du = a2

∫

1 + cos 2u

2du

= a2

2

(

u + 1

2sin 2u

)

+ C = a2

2u + a2

2sin u cos u + C

= a2

2sin−1 x

a+ a2

2· x

a·√

1 −( x

a

)2+ C

= a2

2sin−1 x

a+ 1

2x√

a2 − x2 + C

18


Some Useful Formulas

d

dx(ax + b) = a =⇒ dx = 1

ad(a x + b)

d

dx(xn+1) = (n+1) xn =⇒ xn dx = 1

n+1d(xn+1)

d

dx(ln x) = 1

x=⇒

1

xdx = d(ln x)

d

dx(eax) = a eax =⇒ ea x dx = 1

ad(ea x)

d

dx(sin x) = cos x =⇒ cos x dx = d(sin x)

d

dx(cos x) = − sin x =⇒ sin x dx = −d(cos x)

19


Example

Evaluate I =∫

tan x dx.

I =∫

tan x dx =∫

sin x

cos xdx

= −∫

1

cos xd(cos x), sin x dx = −d(cos x)

= − ln∣

∣cos x∣

∣ + C

∫

1

udu = ln

∣

∣u∣

∣, u = cos x

20


Example

Evaluate I =∫

2x ex2dx.

I =∫

ex2 · 2x dx =∫

ex2d(x2), xn dx = 1

n+1d(xn+1)

= ex2 + C

∫

eu du = eu, u = x2

21


Example

Evaluate I =∫

3x3

2 − x4dx.

I = 3

∫

1

2 − x4· x3 dx

= 3

∫

1

2 − x4· 1

4d(x3+1), xn dx = 1

n+1d(xn+1)

= 3

4

∫

1

2 − x4· 1

−1d( 2 − x4 ), dx = 1

ad(a x + b)

= − 3

4ln

∣

∣2 − x4∣

∣ + C,

∫

1

udu = ln

∣

∣u∣

∣, u = 2−x4

22


Example

Evaluate I =∫

sin3 x dx.

I =∫

sin2 x · sin x dx

= −∫

sin2 x d(cos x), sin x dx = −d(cos x)

= −∫

(1 − cos2 x) d(cos x), sin2 x + cos2 x = 1

= − cos x + 13

cos3 x + C,

∫

un du = un+1

n+1, u = cos x

23


Example

Evaluate I =∫

sin2 x cos3 x dx.

I =∫

sin2 x cos2 x · cos x dx

=∫

sin2 x cos2 x d(sin x), cos x dx = d(sin x)

=∫

sin2 x (1 − sin2 x) d(sin x), sin2 x + cos2 x = 1

=∫

(sin2 x − sin4 x) d(sin x)

= 13

sin3 x − 15

sin5 x + C,

∫

un du = un+1

n+1, u = sin x

24


Techniques of Integration — Integration by Parts

∫

f(x) dx =∫

u dv = u v −∫

v du

The important step is to split function f(x) into the product of two functions∫

f(x) dx =∫

u(x) · g(x) dx =∫

u(x) d[v(x)]

to identify functions u(x) and v(x).

Some rules for selecting functions u(x) and v(x):

1. It is easy to find v(x).

2.

∫

v du is easier to evaluate than

∫

u dv.

3. Functions, such as ln x, sin−1 x, cos−1 x, tan−1 x, . . . , whose derivatives

are of an easier form to integrate, should be left out as u(x).

25

Calculus — Integration - Integration by Parts © Wei-Chau Xie

Example

Evaluate I =∫

x2 cos x dx.

I =∫

x2 · cos x dx =∫

x2 d(sin x), cos x dx = d(sin x)

= x2 sin x −∫

sin x · 2x dx, Integration by parts

= x2 sin x − 2

∫

x · sin x dx

= x2 sin x + 2

∫

x d(cos x), sin x dx = −d(cos x)

= x2 sin x + 2(

x cos x −∫

cos x dx)

= x2 sin x + 2 x cos x − 2 sin x + C

26


Example

Evaluate I =∫

x ln x dx.

I =∫

ln x · x dx =∫

ln x · 12

d(x2), xn dx = 1

n+1d(xn+1)

= 12

(

x2 ln x −∫

x2 · 1

xdx

)

, Integration by parts

= 12

(

x2 ln x −∫

x dx)

= 12

(

x2 ln x − 12

x2)

+ C

= 14

x2(2 ln x − 1) + C

27


Example

Evaluate I =∫

ex sin x dx.

I =∫

sin x · ex dx =∫

sin x d(ex), ex dx = d(ex)

= ex sin x −∫

ex · cos x dx, Integration by parts

= ex sin x −∫

cos x d(ex)

= ex sin x −{

ex cos x −∫

ex · (− sin x) dx

}

∫

ex sin x dx = ex sin x − ex cos x −∫

ex sin x dx

∴

∫

ex sin x dx = 12

ex(sin x − cos x) + C

28


Techniques of Integration — Partial Fractions

Partial fraction decomposition is an essential step in

integrating rational fractions

solving ordinary differential equations using the method of Laplace transform

Consider a rational functionN(x)

D(x), where N(x) and D(x) are polynomials.

If the degree (power) of N(x) is higher or equal to that of Q(x), use long

division to simplify first.

Completely factorize the denominator D(x) into factors of the form

(αx+β)m and (ax2+bx+c)n,

where ax2+bx+c is an unfactorable quadratic.

29

Calculus — Integration - Partial Fractions © Wei-Chau Xie

For each factor of the form (αx+β)m, the partial fraction decomposition is

Am

(αx+β)m+

Am−1

(αx+β)m−1+ · · · + A1

(αx+β).

For each factor of the form (ax2+bx+c)n, the partial fraction decomposition

is n terms

Bn x+Cn

(ax2+bx+c)n+

Bn−1 x+Cn−1

(ax2+bx+c)n−1+ · · · + B1 x+C1

(ax2+bx+c).

Example

5x3+2x+7

2x8+7x7−10x5−6x4−x3= 5x3+2x+7

x3(2x−1)(x2+2x−1)2

= A3

x3+ A2

x2+ A1

x+ B

2x−1+ C2 x+D2

(x2+2x−1)2+ C1 x+D1

x2+2x−1

30


The Cover-Up Method

Suppose D(x) has a factor (x−a)m, the partial fraction decomposition is

N(x)

D(x)= N(x)

(x−a)m D1(x)= Am

(x−a)m+

Am−1

(x−a)m−1+ · · · + A1

(x−a)+ N1(x)

D1(x)

in which D1(x) does not have (x−a) as a factor.

Multiplying both sides of the equation by (x−a)m yields

N(x)

D1(x)= Am + Am−1(x−a) + · · · + A1(x−a)m−1 + N1(x)

D1(x)(x−a)m

Setting x = a =⇒ Am = N(x)

D1(x)

∣

∣

∣

∣

x=a

This result can be restated as follows:

To find Am, ‘‘cover-up’’ (remove) the term (x−a)m and set x = a:

Am = N(x)

×(x−a)m D1(x)

∣

∣

∣

∣

x=a

Works only for the highest power of repeated linear factor.

31


Example

Evaluate I =∫

4

(x−1)2(x+1)dx.

Using partial fractions:4

(x−1)2(x+1)= A2

(x−1)2+ A1

x−1+ B

x+1

To find A2, cover-up (x−1)2 and set x = 1 =⇒ A2 = 4

x+1

∣

∣

∣

∣

x=1

= 2

To find B, cover-up (x+1) and set x = −1 =⇒ B = 4

(x−1)2

∣

∣

∣

∣

x=−1

= 1

For A1, set x = 0 : 4

(0−1)2(0+1)= 2

(0−1)2+ A1

0−1+ 1

0+1=⇒ A1=−1

∴ I =∫ {

2

(x−1)2− 1

x−1+ 1

x+1

}

dx = − 2

x−1− ln

∣

∣x−1∣

∣ + ln∣

∣x+1∣

∣ + C

32


Example

Evaluate I =∫

8

(x−1)(x2+2x+5)dx.

Using partial fractions:

8

(x−1)(x2+2x+5)= A

x−1+ Bx+C

x2+2x+5

= A(x2+2x+5) + (Bx+C)(x−1)

(x−1)(x2+2x+5)= (A+B)x2+(2A−B+C)x+(5A−C)

(x−1)(x2+2x+5)

To find A, cover-up (x−1) and set x = 1:

A = 8

x2+2x+5

∣

∣

∣

∣

x=1

= 8

1+2+5= 1

33


Compare the coefficients of the numerators

x2 : A + B = 0 =⇒ B = −A = −1

x : 2A − B + C = 0 =⇒ C = B − 2A = −1 − 2·1 = −3

1 : 5A − C = 8 Use this equation as a check: 5·1−(−3)=8

∴8

(x−1)(x2+2x+5)= 1

x−1− x+3

x2+2x+5= 1

x−1− (x+1) + 2

(x+1)2+22

= 1

x−1− (x+1)

(x+1)2+22− 2

(x+1)2+22

Complete the Squares: a2 + 2·a·b + b2 = (a + b)2

3x2 + 2x + 5 = 3(

x2 + 23

x)

+ 5 = 3[

x2 + 2·x·13

+(

13

)2−

(

13

)2]

+ 5

= 3{[

x2 + 2·x·13

+(

13

)2]

+ 53

−(

13

)2}

= 3{(

x + 13

)2+ 14

9

}

= 3{(

x + 13

)2+

(

√143

)2}

34


I =∫ {

1

x−1− (x+1)

(x+1)2+22− 2

(x+1)2+22

}

dx

=∫

1

x−1d(x−1) − 1

2

∫

1

(x+1)2+22d[(x+1)2+22] − 2

∫

1

(x+1)2+22d(x+1)

= ln∣

∣x−1∣

∣ − 12

ln∣

∣(x+1)2 + 22∣

∣ − tan−1 x+1

2+ C

∫

1

xdx = ln

∣

∣x∣

∣,

∫

1

x2 + a2dx = 1

atan−1 x

a

35


Definite Integrals

∫

f(x) dx = F(x) =⇒∫ b

af(x) dx = F(x)

∣

∣

∣

b

x=a= F(b) − F(a)

The integral A =∫ b

af(x) dx is the area bounded by curve y = f(x) and the

x-axis between a and b.

x

y

y = f(x)

a b

A

36

Calculus — Integration - Definite Integrals © Wei-Chau Xie

Example

Find the area bounded by the sine curve y = sin x and the x-axis between 0

and π .

x0 π

y=sin x

A =∫ π

0sin x dx = − cos x

∣

∣

∣

π

0

= −(cos π − cos 0) = −[(−1) − 1] = 2

37


Example

Evaluate I =∫ 4

0

x+2√2x + 1

dx.

Let√

2x + 1 = u, x = u2−1

2, dx = u du

x = 0 =⇒ u =√

2·0+1 = 1; x = 4 =⇒ u =√

2·4+1 = 3

I =∫ 4

x=0

x + 2√2x+1

dx =∫ 3

u=1

u2−1

2+ 2

u· u du

= 12

∫ 3

1(u2 + 3) du = 1

2

[

13

u3 + 3u]3

1

= 12

[

(13·33 + 3·3

)

−(1

3·13 + 3·1

)

]

= 223

38


Example

Determine the area bounded by parabola y2 = 2x and line y = x−4.

x0

y 2=2x

y=x−4y

(8, 4)

(2, −2)

The intersections are given by

{

y2 = 2x

y = x−4

Substitute the second equation into the first

(x−4)2 = 2x =⇒ x2 − 10x + 16 = 0 =⇒ (x−2)(x−8) = 0

x = 2 =⇒ y = x−4 = −2, x = 8 =⇒ y = x−4 = 4

The points of intersection are (2, −2) and (8, 4).

39


x0

x=

y 2=2x

y=x−4

x=y+4

y

ydy

y 2

2(8, 4)

(2, −2)

Consider a horizontal strip of height dy as shown

dA =[

( y+4) − y2

2

]

dy

A =∫ 4

y=−2

(

y + 4 − y2

2

)

dy =[

12

y2 + 4 y − 16

y3]4

y=−2

=(

12·42 + 4·4 − 1

6·43

)

−[

12(−2)2 + 4(−2) − 1

6(−2)3

]

= 18

40

Calculus — Partial Derivatives © Wei-Chau Xie

Partial Derivatives

Consider a functiony = f(x1, x2, . . . , xn)

of n independent variables x1, x2, . . . , xn.

The partial derivatives are defined as derivatives when all but the variable of

interest are held fixed during the differentiation.

∂ y

∂xi

is the first-order derivative of y w.r.t. xi, when

x1, x2, . . . , xi−1, xi+1, . . . , xn are held fixed (or regarded as constants).

For ‘‘nice’’ functions (partial derivatives exist and are continuous), mixed

partial derivatives must be equal regardless of the order in which the

differentiation is performed.

For examples, consider f(x, y),

∂2f

∂x ∂ y= ∂2f

∂y ∂x,

∂3f

∂x2∂ y= ∂3f

∂y ∂x2, . . .

41


Example

Given z = x2 − y3 + 2x y2, find∂z

∂x,

∂z

∂ y,

∂2z

∂x2,

∂2z

∂ y2,

∂2z

∂x ∂ y.

∂z

∂x= ∂

∂x(x2 − y3 + 2 x y2) = 2x + 2 y2 y is fixed

∂z

∂ y= ∂

∂ y(x2 − y3 + 2 x y2) = −3 y2 + 4 x y x is fixed

∂2z

∂x2= ∂

∂x

(∂z

∂x

)

= ∂

∂x(2x + 2 y2) = 2 y is fixed

∂2z

∂ y2= ∂

∂ y

( ∂z

∂ y

)

= ∂

∂ y(−3 y2 + 4 x y) = −6 y + 4 x x is fixed

∂2z

∂x ∂y= ∂

∂x

( ∂z

∂ y

)

= ∂

∂x(−3 y2 + 4 x y) = 4 y y is fixed

42


Chain Rules for Partial Derivatives

Suppose z = f(u, v), u = g(x, y), v = h(x, y)

z

u v

y yx x

∂z

∂x= ∂z

∂u

∂u

∂x+ ∂z

∂v

∂v

∂x

∂z

∂ y= ∂z

∂u

∂u

∂ y+ ∂z

∂v

∂v

∂ y

43

Calculus — Chain Rules for Partial Derivatives © Wei-Chau Xie

Example

Find∂z

∂sif z = f(x, y), x = g(s, t), y = h(s, t),

in which z = x2 e y + y ln x, x = s2 cos t, y = 4t2 sin 2s.

By the Chain Rule:

∂z

∂s= ∂z

∂x

∂x

∂s+ ∂z

∂ y

∂ y

∂s

z

x y

t ts s

z = x2 e y + y ln x

∂z

∂x= ∂

∂x(x2 e y + y ln x) = 2x e y + y· 1

xy is fixed

∂z

∂ y= ∂

∂ y(x2 e y + y ln x) = x2 e y + ln x x is fixed

x = s2 cos t

∂x

∂s= ∂

∂s(s2 cos t) = 2s cos t t is fixed

44

Calculus — Chain Rules for Partial Derivatives © Wei-Chau Xie

y = 4t2 sin 2s

∂ y

∂s= ∂

∂s(4t2sin 2s) = 8t2 cos 2s t is fixed

∴∂z

∂s= ∂z

∂x

∂x

∂s+ ∂z

∂ y

∂ y

∂s

=(

2x e y + y

x

)

(2s cos t) + (x2 e y + ln x) (8t2 cos 2s)

45


Implicit Function

Equation F(x, y) = 0 defines y as an implicit function of x: y = y(x).

Differentiate F[x, y(x)] = F(x, y) = 0 with respect to x:

dF

dx= ∂F

∂x+ ∂F

∂ y

dy

dx= 0

F

y

x

x

∴dy

dx= −

∂F

∂x

∂F

∂y

, provided∂F

∂y6= 0

46

Calculus — Partial Derivatives of Implicit Functions © Wei-Chau Xie

Example

F(x, y) = yex + sin(x y) + 4 = 0, finddy

dx.

∂F

∂x= ∂

∂x[ yex + sin(x y) + 4] = y ex + cos(x y)·y y is fixed.

∂F

∂y= ∂

∂y[ yex + sin(x y) + 4] = ex + cos(x y)·x x is fixed.

dy

dx= −

∂F

∂x

∂F

∂y

= −y [ex + cos(x y)]ex + x cos(x y)

47

Calculus — Multiple Integrals © Wei-Chau Xie

Double Integrals

Let f(x, y) be defined in a closed region R of the x y-plane.

x

1Ak (xk, yk)

y

R

Divide R into n small regions of area 1Ak, k = 1, 2, . . . , n.

Let (xk, yk) be some point in 1Ak.

∫∫

R

f(x, y) dA = limn→∞

n∑

k=1

f(xk, yk) 1Ak

in which max1Ak → 0 when n → ∞.

48


Double Iterated Integrals

Convert double integrals to double iterated integrals.

Use horizontal lines to divide the region into horizontal strips.

x

x=x1(y) x=x2(y)

y

R

y

c

d

dy

dAy+dy

∫∫

R

f(x, y) dA =∫ d

y=c

{ ∫ x2( y)

x=x1( y)f(x, y) dx

}

dy

49

Calculus — Double Iterated Integrals © Wei-Chau Xie

Use vertical lines to divide the region into vertical strips.

y= y2(x)

dx

dA

x

y

xa bx+dx

R

y= y1(x)

∫∫

R

f(x, y) dA =∫ b

x=a

{ ∫ y2(x)

y= y1(x)

f(x, y) dy

}

dx

50

Calculus — Double Integrals © Wei-Chau Xie

Some Physical Meanings of I =∫∫

R

f(x, y) dA

If f(x, y)= 1 in R, then

I =∫∫

R

f(x, y) dA =∫∫

R

1 · dA = Area of region R

If f(x, y)=ρ(x, y)= density, I is the mass of a plate with unit thickness,

cross section R, and variable density ρ(x, y).

If z = f(x, y), I is the volume of a cylinder with flat bottom R and height

f(x, y).

x

z

y

z = f(x,y)

R

51


Example

Evaluate

∫∫

R

x2 y dA, where R is bounded by y =√

x+4, y = 0, x+ y = 2.

Method 1: Take horizontal strips

x

y

1

y= x+4x

y=0

x+y=2

x=2–yx=y2–4

2

–4 –3 –2 –1 0 1 2

R

∫∫

R

x2 y dA =∫ 2

y=0

[ ∫ 2−y

x=y2−4x2 y dx

]

dy =∫ 2

y=0y[

13

x3]2−y

x=y2−4dy

= 13

∫ 2

y=0(72 y − 12 y2 − 42 y3 − y4 + 12 y5 − y7) dy

= 13

[

36 y2 − 4 y3 − 212

y4 − 15

y5 + 2 y6 − 18

y8]2

y=0= 56

5

52


Method 2: Take vertical strips

x

y

1

y= x+4x

y= x+4x

y=0 y=0

x+y=2

y=2–x

2

–4 –3 –2 –1 0 1 2

R1 R2

∫∫

R

x2 y dA =∫ 0

x=−4

[ ∫

√x+4

y=0x2 y dy

]

dx +∫ 2

x=0

[ ∫ 2−x

y=0x2 y dy

]

dx

=∫ 0

x=−4x2

[

12

y2]

√x+4

y=0dx +

∫ 2

x=0x2

[

12

y2]2−x

y=0dx

= 12

[ ∫ 0

x=−4(4x2 + x3) dx +

∫ 2

x=0(4x2 − 4x3 + x4) dx

]

= 12

{

[

43

x3 + 14

x4]0

x=−4+

[

43

x3 − x4 + 15

x5]2

x=0

}

= 565

53


Triple Integrals

Let V be a closed volume bounded by a piecewise smooth surface S.

Divide V into n small volumes 1Vk, k = 1, 2, . . . , n.

Let (xk, yk, zk) be some point in 1Vk.

x

z

S

y

1Vk

V

(xk, yk, zk)

∫∫∫

Vf(x, y, z) dV = lim

n→∞

n∑

k=1

f(xk, yk, zk) 1Vk

in which max1Vk→0 when n→∞.

54


Triple Iterated Integrals

Triple integrals are evaluated using triple iterated integrals.

x

z

yz=z1(x,y)

z=z2(x,y)

Sxy

∫∫∫

Vf(x, y, z) dV =

∫∫

Sxy

{ ∫ z2(x, y)

z=z1(x, y)f(x, y, z) dz

}

dA

A double integral over Sxy is obtained after evaluating the inner integral.

55

Calculus — Triple Iterated Integrals © Wei-Chau Xie

Example

Find the volume bounded by the surfaces x+ y+z = 4, y = 3z, x = 0, y = 0.

x

z

yy=3z

x+y+z =4

y+z =4

x=4–y–z

x=0

4

4

4

z

Syz

y

y=3z

4

430

V =∫∫∫

VdV =

∫∫

Syz

{ ∫ 4−y−z

x=0dx

}

dA =∫∫

Syz

x∣

∣

∣

4−y−z

x=0dA

=∫∫

Syz

(4 − y − z) dA

56

Calculus — Triple Iterated Integrals © Wei-Chau Xie

z

Syz

y

y=3z

z =4–y

4

430z=

y

3

V =∫∫

Syz

(4 − y − z) dA =∫ 3

y=0

{ ∫ 4−y

z= y3

(4 − y − z) dz

}

dy

=∫ 3

y=0

[

4z − yz − 12

z2]4−y

z= y3

dy =∫ 3

y=0

(

89

y2 − 163

y + 8)

dy

=[

827

y3 − 83

y2 + 8 y]3

y=0= 8

57

Calculus — Vector Fields © Wei-Chau Xie

Scalar Field

If f(x, y, z) is a scalar function defined at every point (x, y, z) in a region of

space, then the function is a scalar field.

Only one quantity (magnitude) is required to describe a scalar field.

For example, temperature field T(x, y, z) is a scalar field.

Vector Field

If F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is a vector function de-

fined at every point (x, y, z) in a region of space, then it is a vector field.

Two quantities are required to define a vector field: the scalar field of its

magnitude and the vector field of its direction.

For example, a force field F(x, y, z) and the velocity field V(x, y, z) of a fluid

are vector fields.

58


Gradient of a Scalar Field

Denote

∇ =( ∂

∂x,

∂

∂y,

∂

∂z

)

= i∂

∂x+ j

∂

∂y+ k

∂

∂z

Gradient of a scalar field f(x, y, z) is defined as the vector field

grad f = ∇f =( ∂ f

∂x,

∂ f

∂y,

∂ f

∂z

)

= i∂ f

∂x+ j

∂ f

∂y+ k

∂ f

∂z

The gradient at a point is a vector pointing in the direction of the steepest

slope or grade at that point, i.e. the direction of the gradient is the direction

of the max rate of change of f.

The steepness of the slope at that point is given by the magnitude of the

gradient, i.e. the magnitude of gradient gives the max rate of change of f.

59


Normal Vectors of a Surface

If f(x, y, z) = 0 defines a surface S and f has continuous first partial derivatives,

then at any point P of S, the vector ∇f is normal to the surface S.

∴ Normal vector N = ∇f =⇒ Unit normal vector n = ∇f∣

∣∇f∣

∣

x

z

yf(x, y, z)=0

SP

N=∇f

60


Divergence of a Vector Field

Divergence of a vector field F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is

defined as the scalar field

div F = ∇·F =( ∂

∂x,

∂

∂y,

∂

∂z

)·(P, Q, R) = ∂P

∂x+ ∂Q

∂y+ ∂R

∂z

Divergence measures the extend to which a vector field flow behaves like a

source or a sink at a given point.

If the divergence is nonzero at some point, then there must be a source (positive

divergence) or a sink (negative divergence) at the point.

A vector field with constant zero divergence is called incompressible or

solenoidal. In this case, no net flow can occur across any closed surface.

61


Curl of a Vector Field

Curl of a vector field F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is defined

as the vector field

curl F = ∇×F =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

i j k

∂

∂x

∂

∂y

∂

∂z

P Q R

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

=(∂R

∂y− ∂Q

∂z

)

i +(∂P

∂z− ∂R

∂x

)

j +(∂Q

∂x− ∂P

∂y

)

k

Curl describes the infinitesimal rotation of a 3-dimensional vector filed.

The direction of the curl is the axis of rotation, as determined by the right-hand

rule. The magnitude of the curl is the magnitude of rotation.

A vector field whose curl is zero is called irrotational.

62


Laplacian

Laplacian of a scalar function f(x, y, z) is the divergence of the gradient of the

function

∇2f = ∇·∇f =( ∂

∂x,

∂

∂y,

∂

∂z

)·( ∂ f

∂x,

∂ f

∂y,

∂ f

∂z

)

= ∂

∂x

(∂ f

∂x

)

+ ∂

∂y

(∂ f

∂y

)

+ ∂

∂z

(∂ f

∂z

)

∴ ∇2f = ∇·∇f = ∂2f

∂x2+ ∂2f

∂y2+ ∂2f

∂z2

63

Calculus — Surface Integrals © Wei-Chau Xie

Surface Integral Involving Vector Fields

Let f(x, y, z) = F(x, y, z)·n, where n is the unit normal vector to surface S.

Surface integral

∫∫

SF(x, y, z)·n dS is called the flux of vector field F(x, y, z).

The Divergence Theorem (Gauss)

Let S be a closed surface and n the unit outward normal to S. V is the region

enclosed by S. Then

∫∫

⊂⊃SF·n dS =

∫∫∫

V∇·F dV

If F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k, then

∫∫

⊂⊃SF·n dS =

∫∫

⊂⊃S(P i + Q j + R k)·n dS =

∫∫∫

V

(∂P

∂x+ ∂Q

∂y+ ∂R

∂z

)

dV

64


Example

Evaluate the surface integral∫∫

⊂⊃S(2x yz i − y2z j + xz k)·n dS

where S is the surface enclosing the volume bounded by x= y, x+ y=2, z=0,

z=2, and x=0, and n is the unit inward normal to S.

Apply the Divergence Theorem

I =∫∫

⊂⊃S(2xyz i − y2z j + xz k)·n dS

= −∫∫∫

V

{

∂(2xyz)

∂x+ ∂(−y2z)

∂y+ ∂(xz)

∂z

}

dV

= −∫∫∫

V(2 yz − 2 yz + x) dV

= −∫∫∫

Vx dV

65


x

x

z

y

y=x

y=x

x+y=2

y=2–x

z=2

z=0

2

2

2

Sxy

y

2

1

20

z=0Sxy

Intersection of x = y and x+y = 2 =⇒ x = 1, y = 1.

I = −∫∫∫

Vx dV = −

∫∫

Sxy

{ ∫ 2

z=0x dz

}

dA = −∫∫

Sxy

xz∣

∣

∣

2

z=0dA

= −∫ 1

x=0

{ ∫ 2−x

y=x2x dy

}

dx = −∫ 1

x=02xy

∣

∣

∣

2−x

y=xdx

= −∫ 1

x=04(x−x2) dx = −4

[

12

x2 − 13

x3]1

x=0= −2

3

66

Calculus - UNENE · 2020-03-19 · Calculus — Differentiation ©Wei-Chau Xie Physical Meanings of...

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