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Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2008

WEN-CHINGLIEN Calculus (I)

(2)

16.3 Evaluating Double Integrals

(1)

Theorem

Let z =f(x,y)be a continuous function on the region R (i) If there exist continuous function x =g1(y)and x =g2(y)for which

R ={(x,y)|g1(y)≤x ≤g2(y), x ≤y ≤d}

then

d g (y) !

(3)

16.3 Evaluating Double Integrals

(1)

Theorem

Let z =f(x,y)be a continuous function on the region R (i) If there exist continuous function x =g1(y)and x =g2(y)for which

R ={(x,y)|g1(y)≤x ≤g2(y), x ≤y ≤d}

then Z Z

R

f(x,y)dA= Z d

c

Z g2(y)

g1(y)

f(x,y)dx

! dy

WEN-CHINGLIEN Calculus (I)

(4)

16.3 Evaluating Double Integrals

(1)

Theorem

Let z =f(x,y)be a continuous function on the region R (i) If there exist continuous function x =g1(y)and x =g2(y)for which

R ={(x,y)|g1(y)≤x ≤g2(y), x ≤y ≤d}

then

d g (y) !

(5)

16.3 Evaluating Double Integrals

(1)

Theorem

Let z =f(x,y)be a continuous function on the region R (i) If there exist continuous function x =g1(y)and x =g2(y)for which

R ={(x,y)|g1(y)≤x ≤g2(y), x ≤y ≤d}

then Z Z

R

f(x,y)dA= Z d

c

Z g2(y)

g1(y)

f(x,y)dx

! dy

WEN-CHINGLIEN Calculus (I)

(6)

16.3 Evaluating Double Integrals

(1)

Theorem

Let z =f(x,y)be a continuous function on the region R (i) If there exist continuous function x =g1(y)and x =g2(y)for which

R ={(x,y)|g1(y)≤x ≤g2(y), x ≤y ≤d}

then

d g (y) !

(7)

16.3 Evaluating Double Integrals

(1)

Theorem

Let z =f(x,y)be a continuous function on the region R (i) If there exist continuous function x =g1(y)and x =g2(y)for which

R ={(x,y)|g1(y)≤x ≤g2(y), x ≤y ≤d}

then Z Z

R

f(x,y)dA= Z d

c

Z g2(y)

g1(y)

f(x,y)dx

! dy

WEN-CHINGLIEN Calculus (I)

(8)

16.3 Evaluating Double Integrals

(1)

Theorem

Let z =f(x,y)be a continuous function on the region R (i) If there exist continuous function x =g1(y)and x =g2(y)for which

R ={(x,y)|g1(y)≤x ≤g2(y), x ≤y ≤d}

then

d g (y) !

(9)

Theorem

(ii) If there exist continuous functions y =h1(x)and y =h2(x)for which

R ={(x,y)|a≤x ≤b, h1(x)≤y ≤h2(x)}

then Z Z

R

f(x,y)dA= Z b

a

Z h2(x)

h1(x)

f(x,y)dy

! dx

WEN-CHINGLIEN Calculus (I)

(10)

Theorem

(ii) If there exist continuous functions y =h1(x)and y =h2(x)for which

R ={(x,y)|a≤x ≤b, h1(x)≤y ≤h2(x)}

then Z Z

R

f(x,y)dA= Z b

a

Z h2(x)

h1(x)

f(x,y)dy

! dx

(11)

Theorem

(ii) If there exist continuous functions y =h1(x)and y =h2(x)for which

R ={(x,y)|a≤x ≤b, h1(x)≤y ≤h2(x)}

then Z Z

R

f(x,y)dA= Z b

a

Z h2(x)

h1(x)

f(x,y)dy

! dx

WEN-CHINGLIEN Calculus (I)

(12)

Theorem

(ii) If there exist continuous functions y =h1(x)and y =h2(x)for which

R ={(x,y)|a≤x ≤b, h1(x)≤y ≤h2(x)}

then Z Z

R

f(x,y)dA= Z b

a

Z h2(x)

h1(x)

f(x,y)dy

! dx

(13)

Theorem

(ii) If there exist continuous functions y =h1(x)and y =h2(x)for which

R ={(x,y)|a≤x ≤b, h1(x)≤y ≤h2(x)}

then Z Z

R

f(x,y)dA= Z b

a

Z h2(x)

h1(x)

f(x,y)dy

! dx

WEN-CHINGLIEN Calculus (I)

(14)

Theorem

(ii) If there exist continuous functions y =h1(x)and y =h2(x)for which

R ={(x,y)|a≤x ≤b, h1(x)≤y ≤h2(x)}

then Z Z

R

f(x,y)dA= Z b

a

Z h2(x)

h1(x)

f(x,y)dy

! dx

(15)

Ex 2.

Z 4

0

Z y

1

(x +y)dx

! dy

Ex 3.

Z 4

2

Z 3

1

(2x +y)dy

dx

Ex 4.

Z Z

R

2xydA

WEN-CHINGLIEN Calculus (I)

(16)

Ex 2.

Z 4

0

Z y

1

(x +y)dx

! dy

Ex 3.

Z 4

2

Z 3

1

(2x +y)dy

dx

Ex 4.

Z Z

R

2xydA

(17)

Ex 2.

Z 4

0

Z y

1

(x +y)dx

! dy

Ex 3.

Z 4

2

Z 3

1

(2x +y)dy

dx

Ex 4.

Z Z

R

2xydA

WEN-CHINGLIEN Calculus (I)

(18)

Ex 2.

Z 4

0

Z y

1

(x +y)dx

! dy

Ex 3.

Z 4

2

Z 3

1

(2x +y)dy

dx

Ex 4.

Z Z

R

2xydA

(19)

Example 1:

Z Z

√x−y dxdy

WEN-CHINGLIEN Calculus (I)

(20)

Example 1:

Z Z

√x−y dxdy

(21)

Example 2:

Use the double integral to calculate the area of the regionΩenclosed by

y =x2 and x+y =2

WEN-CHINGLIEN Calculus (I)

(22)

Example 2:

Use the double integral to calculate the area of the regionΩenclosed by

y =x2 and x+y =2

(23)

Example 2:

Use the double integral to calculate the area of the regionΩenclosed by

y =x2 and x+y =2

WEN-CHINGLIEN Calculus (I)

(24)

Example 2:

Use the double integral to calculate the area of the regionΩenclosed by

y =x2 and x+y =2

(25)

Example 3:

Let R be rectanglea≤x ≤b,c ≤y ≤d show that if f is continuous on[a,b], andg is continuous on[c,d],

then Z Z

R

f(x)·g(y)dxdy = Z b

a

f(x)dx

Z d

c

g(y)dy

WEN-CHINGLIEN Calculus (I)

(26)

Example 3:

Let R be rectanglea≤x ≤b,c ≤y ≤d show that if f is continuous on[a,b], andg is continuous on[c,d],

then Z Z

R

f(x)·g(y)dxdy = Z b

a

f(x)dx

Z d

c

g(y)dy

(27)

Example 3:

Let R be rectanglea≤x ≤b,c ≤y ≤d show that if f is continuous on[a,b], andg is continuous on[c,d],

then Z Z

R

f(x)·g(y)dxdy = Z b

a

f(x)dx

Z d

c

g(y)dy

WEN-CHINGLIEN Calculus (I)

(28)

Example 3:

Let R be rectanglea≤x ≤b,c ≤y ≤d show that if f is continuous on[a,b], andg is continuous on[c,d],

then Z Z

R

f(x)·g(y)dxdy = Z b

a

f(x)dx

Z d

c

g(y)dy

(29)

Example 3:

Let R be rectanglea≤x ≤b,c ≤y ≤d show that if f is continuous on[a,b], andg is continuous on[c,d],

then Z Z

R

f(x)·g(y)dxdy = Z b

a

f(x)dx

Z d

c

g(y)dy

WEN-CHINGLIEN Calculus (I)

(30)

Example 4:

Z Z

√xy dxdy, Ω :0≤y ≤1, y2≤x ≤y

(31)

Example 4:

Z Z

√xy dxdy, Ω :0≤y ≤1, y2≤x ≤y

WEN-CHINGLIEN Calculus (I)

(32)

(2)Interchange the order of integration

For example, the iterated integral

Z 1

0

Z 1

y2

yex2dxdy

(33)

(2)Interchange the order of integration

For example, the iterated integral

Z 1

0

Z 1

y2

yex2dxdy

WEN-CHINGLIEN Calculus (I)

(34)

(2)Interchange the order of integration

For example, the iterated integral

Z 1

0

Z 1

y2

yex2dxdy

(35)

Solution:

reverse the order of integration

consider Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy

WEN-CHINGLIEN Calculus (I)

(36)

Solution:

reverse the order of integration

consider Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy

(37)

Solution:

reverse the order of integration

consider Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy

WEN-CHINGLIEN Calculus (I)

(38)

Step 1:

Identify the regionQ for which the iterated integral can be written as the double integral

Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy = Z Z

Q

f(x,y)dA

(39)

Step 1:

Identify the regionQ for which the iterated integral can be written as the double integral

Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy = Z Z

Q

f(x,y)dA

WEN-CHINGLIEN Calculus (I)

(40)

Step 1:

Identify the regionQ for which the iterated integral can be written as the double integral

Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy = Z Z

Q

f(x,y)dA

(41)

Step 1:

Identify the regionQ for which the iterated integral can be written as the double integral

Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy = Z Z

Q

f(x,y)dA

WEN-CHINGLIEN Calculus (I)

(42)

Step 2:

Find constantsaandband continuous functions g1and g2 so that the region Q can be expressed as

Q={(x,y)|a≤x ≤b, g1(x)≤y ≤g2(x)}

(43)

Step 2:

Find constantsaandband continuous functions g1and g2 so that the region Q can be expressed as

Q={(x,y)|a≤x ≤b, g1(x)≤y ≤g2(x)}

WEN-CHINGLIEN Calculus (I)

(44)

Step 2:

Find constantsaandband continuous functions g1and g2 so that the region Q can be expressed as

Q={(x,y)|a≤x ≤b, g1(x)≤y ≤g2(x)}

(45)

Step 2:

Find constantsaandband continuous functions g1and g2 so that the region Q can be expressed as

Q={(x,y)|a≤x ≤b, g1(x)≤y ≤g2(x)}

WEN-CHINGLIEN Calculus (I)

(46)

Step 3:

Rewrite the iterated integral as Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy = Z Z

Q

f(x,y)dA

= Z b

a

Z g2(x)

g1(x)

f(x,y)dydx

(47)

Step 3:

Rewrite the iterated integral as Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy = Z Z

Q

f(x,y)dA

= Z b

a

Z g2(x)

g1(x)

f(x,y)dydx

WEN-CHINGLIEN Calculus (I)

(48)

Step 3:

Rewrite the iterated integral as Z d

c

Z h2(y)

h1(y)

f(x,y)dxdy = Z Z

Q

f(x,y)dA

= Z b

a

Z g2(x)

g1(x)

f(x,y)dydx

(49)

Example 1:

Z 1

0

Z 1

y2

yex2dxdy

Example 2:

Z 1

0

Z

1−x

0

xy2dydx

WEN-CHINGLIEN Calculus (I)

(50)

Example 1:

Z 1

0

Z 1

y2

yex2dxdy

Example 2:

Z 1

0

Z

1−x

0

xy2dydx

(51)

Example 1:

Z 1

0

Z 1

y2

yex2dxdy

Example 2:

Z 1

0

Z

1−x

0

xy2dydx

WEN-CHINGLIEN Calculus (I)

(52)

Example 1:

Z 1

0

Z 1

y2

yex2dxdy

Example 2:

Z 1

0

Z

1−x

0

xy2dydx

(53)

Example 3:

1 Z 2

0

Z x2

0

(x −2y)dydx

2 Z 1

0

Z 1

x2

xey2dydx

WEN-CHINGLIEN Calculus (I)

(54)

Example 3:

1 Z 2

0

Z x2

0

(x −2y)dydx

2 Z 1

0

Z 1

x2

xey2dydx

(55)

Example 3:

1 Z 2

0

Z x2

0

(x −2y)dydx

2 Z 1

0

Z 1

x2

xey2dydx

WEN-CHINGLIEN Calculus (I)

(56)

Thank you.

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