Double Integral : Lecture 5

‫ﻣﺪﺭﺱ ﺍﺯﻫﺎﺭ ﻣﺎﻟﻚ‬

Double Integral

Definition: let R be closed region in the (x, y )- plane. If f is a function

of two variables that is define on the region R, then the double integrals
on R is written by

n
Lim
n →∞
∑ f (x
r =1
r , y r )∆A r = ∫∫ f (x, y) dA
∆A r → 0 R

dydx ‫ﺍﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻤﻨﺤﻨﻴﺎﺕ ﺑﻬﺬﻩ ﺍﻟﺼﻴﻐﺔ ﻳﺆﺧﺬ ﺍﻟﻤﻘﻄﻊ ﺷﺎﻗﻮﻟﻲ‬

y2
b y2 R
∫∫ f (x, y) dA = ∫ ∫ f (x, y) dy dx
R a y1
y1

a b

dxdy ‫ﺍﻣﺎ ﺍﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻤﻨﺤﻨﻴﺎﺕ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﻟﻲ ﻳﺆﺧﺬ ﺍﻟﻤﻘﻄﻊ ﺍﻓﻘﻴﺎ‬

x1 x2
d
d x2
R
∫∫ f (x, y) dA = ∫ ∫ f (x, y) dxdy
R c x1
c

1
Double Integral : Lecture 5
‫ﻣﺪﺭﺱ ﺍﺯﻫﺎﺭ ﻣﺎﻟﻚ‬

Examples:
U

3 2
1) Evaluate ∫ ∫ (1 + 8xy )dydx
0 1

i) sketch: since dxdy ⇒ vertical

y =1 , y=2
ii)
3 2 3
y2 2
∫ ∫ (1 + 8xy )dydx = ∫ (y + 8x
0 1 0
) dx
2 1
3
= ∫ {1 + 12x} dx
0

x2 3
= ( x + 12 )
2 0
= (3 + 6(9)) - (0) = (3 + 54) = 57

∫∫ (2 x − y
2
2) Evaluate )dA over the triangular R enclosed by
R
y =1− x , y =1+ x , y=3

i) sketch:
y =1− x , y =1+ x
if x = 0 ⇒ y = 1 , if x = 0 ⇒ y = 1
if y = 0 ⇒ x = 1 , if y = 0 ⇒ x = −1
⇒ (0,1) & (1,0) , ⇒ (0,1) & (-1,0)

3 y −1
244
3 y −1

∫∫ (2 x − y )dA = ∫ ∫ (2 x − y )dxdy = ∫ ( x − y x) dy = 18 −
2 2 2 2

R 1 1− y 1 1− y 6

2
Double Integral : Lecture 5
‫ﻣﺪﺭﺱ ﺍﺯﻫﺎﺭ ﻣﺎﻟﻚ‬

2 1

∫∫
2
3) Evaluate ex dx dy
0 y
2
Reverse the order of integration

Since dxdy horizontal

y
x= ⇒ y = 2x
2
x =1
for y from 0 → 2
2 1 1 2x 1 2x

∫∫ dx dy = ∫ ∫ e x dy dx = ∫ e x y dx
2 2 2
ex
0 y 0 0 0 0
2

= e -1

ππ
sin x
4) Evaluate ∫∫
0 y
x
dx dy

From left x= y
y=π
From right x = π
value of y , from 0 ⇒ x
reverse the order

x=π
π π π x
sin x sin x
⇒ ∫∫
0 y
x
dx dy = ∫ ∫
0 0
x
dy dx

π π
sin x x sin x
=∫ ⋅ y dx = ∫ ⋅ x dx
0
x 0 0
x
= 2

5)
2 2 2 y

∫ ∫ 2y sin xy dydx = ∫ ∫ 2 y 2 sin xy dxdy

2

0 x 0 0

[ ]
2 2
= ∫ [− 2 y cos xy]0 dy = ∫ − 2 y cos y 2 + 2 y dy
y

0 0

[
= - sin y + y 2
]
2 2
0 = 4 − sin4

3
Double Integral : Lecture 5
‫ﻣﺪﺭﺱ ﺍﺯﻫﺎﺭ ﻣﺎﻟﻚ‬
0 x+2

6) Write an equivalent double of integration reversed ∫∫ (x 2 + y 2 )dy dx

−1 − x

0 x +2

∫∫ (x 2 + y 2 )dy dx (-1,1)
−1 − x
1 0 2 0
= ∫ ∫ (x 2 + y 2 )dxdy + ∫ ∫ (x
2
+ y 2 )dxdy
0 −y 1 y −2

7) Draw the region bounded by y=ex, y=sinx, x=π, x=-π P P

and evaluate its area.

π ex
A= ∫π ∫ dydx
− sin x
π
ex
= ∫ y sin x
−π

= eπ − e −π

8) Find the area bounded by y=-x, y=-3x and x=y+4.

Solution:

(2,-2)
−2 y + 4 0 −y

A= ∫ ∫
−3 − y
dxdy + ∫ ∫ dxdy
−2 − y
3 3 (1,-3)
−2 0
y+4 −y
= ∫x
−3
−4
3
dy + ∫ x −4 dy = 2
−2 3

4
Double Integral : Lecture 5
‫ﻣﺪﺭﺱ ﺍﺯﻫﺎﺭ ﻣﺎﻟﻚ‬

Problems
U

1 1 ln 2 1
x
∫∫ ∫ ∫
2
1) dydx 2) xy e y x
dydx
0 0
(xy + 1) 2 0 0

π x2 ln 8 ln y
1 y
3) ∫∫
π 0
x
cos dydx
x
4) ∫ ∫1 0
e x + y dxdy
2

2 2x 1π
x
5) ∫∫
1 x
y
dydx 6) ∫∫
0 0
y cos xy dxdy

π
4 2 2 sin θ

∫∫ ∫ ∫ r cos θ drdθ
3
7) e x dxdy 8)
0 y 0 0

9) Evaluate ∫∫ dA
R
, R: 1st quadrant bounded by y 2 = x & x 2 = y
P P

10) Evaluate ∫∫ xydA

R
, R: the region bounded by y 2 = x & y = x

1
−
∫∫ x(1 + y ) 2 dA , R: the region in the 1 quadrant enclosed by:
2 st
11) Evaluate P P

y = x2 , y = 4 , x = 0

∫∫ sin (y ) dA , R: the region bounded by y = x , y = 2 & x = 0

3
12) Evaluate
R

1 1
13) ∫∫
0 y
x 2 e xy dxdy

References
U

1- calculus & Analytic Geometry (Thomas).

2- Calculus (Haward Anton).
3- Advanced Mathematics for Engineering Studies (‫ ﺭﻳﺎﺽ ﺍﺣﻤﺪ ﻋﺰﺕ‬.‫)ﺃ‬