In these slides, we treat some exercises about cartesian and polar coordinates

1. Lebanese University - Faculty of Sciences
Section ¶
Chapter 2: Double Integration
Solved Problems
Dr. Kamel ATTAR
attar.kamel@gmail.com
F Wednesday 24/Mars/2021 F

2. 2Ú22
Exercises
Solutions
1 Exercises
Cartesian coordinates
Polar coordinates
2 Solutions
Cartesian coordinates
Polar coordinates
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

3. 3Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Cartesian coordinates
. Exercise 1.
Let D = {(x, y) ∈ R2
: x2
+ y2
≥ 1, y ≤ x2
+ 1, 0 ≤ y , 0 ≤ x ≤ 1}
1. Sketch D.
2. Using Cartesian coordinates, write the expression J =
ZZ
D
f(x, y)dxdy by two different ways.
Go to Solution
. Exercise 2. Let D = {(x, y) ∈ R2
: x2
+ y2
≥ 1, y ≤ x2
+ 2 , 0 ≤ y, −1 ≤ x ≤ 1}
1. Sketch D.
2. Using Cartesian coordinates, write the expression J =
ZZ
D
f(x, y)dxdy by two different ways.
Go to Solution
. Exercise 3. Let K =
Z 1
0
dx
Z 1
x
sin(y2
) dy
1. Write and sketch the domain of integration D.
2. Calculate K.
Go to Solution
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

4. 4Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
. Exercise 4. Let I =
Z 1
0
dx
Z x
x2
2xe2y3
−3y2
dy
1. Write and sketch the domain of integration D.
2. Write I in another form by changing the order of integration.
3. Calculate I.
Go to Solution
. Exercise 5.
1. Calculate I =
Z 1
0
dy
Z 2−y
√
1−y2
x dx.
2. Sketch the domain of integration of I.
3. Refind the value of I by changing the order of integration.
Go to Solution
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

5. 5Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Polar coordinates
. Exercise 6. Calculate the following integrals by using the polar coordinates.
1.
ZZ
D
x2
y dxdy , D =
n
(x, y); y ≥ 0 , x2
+ y2
− 2x ≤ 0
o
2.
ZZ
D
ln(x2
+ y2
) dxdy , D =
n
(x, y); x2
+ y2
≥ 1 , x2
+ y2
≤ 4
o
3.
ZZ
D
dxdy
(x2 + y2)2
, D =
n
(x, y); x ≥ 1 , x2
+ y2
− 2x ≤ 0
o
4.
ZZ
D
y dxdy , D =
n
(x, y) ∈ R2
+; x2
+ y2
≤ 1 , x2
+ y2
≥ 2y
o
Go to Solution
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

6. 6Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
. Exercise 7. Let ∆ be the domain defined by:
∆ = {(x, y) ∈ R2
: x2
+ y2
− 2y ≤ 0, y ≥ 1, x ≥ 0}
1. Draw ∆.
2. By using the polar coordinates, calculate I =
ZZ
∆
dxdy
(x2 + y2)2
Go to Solution
. Exercise 8. Let ∆ be the domain defined by:
∆ = {(x, y) ∈ R2
: x2
+ y2
− 2x ≤ 0, y ≥ 0, x ≥ 1} ,
1. Draw ∆.
2. By using the polar coordinates, calculate I =
ZZ
∆
dxdy
(x2 + y2)2
Go to Solution
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

7. 7Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 1.
Vertical Section
J =
ZZ
D
f(x, y)dxdy =
Z 1
0
Z x2
+1
√
1−x2
f(x, y) dy
!
dx
Horizontal Section
J =
ZZ
D
f(x, y)dxdy
=
Z 1
−1
Z 1
√
1−y2
f(x, y) dx
!
dy
+
Z 2
1
Z 1
√
y−1
f(x, y) dx
!
dy .
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

8. 8Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 2.
Vertical Section
J =
Z 1
−1
Z x2
+2
√
1−x2
f(x, y) dy
!
dx
Horizontal Section
J =
Z 1
0
Z −
√
1−y2
−1
f(x, y) dx +
Z 1
√
1−y2
f(x, y) dx
!
dy
+
Z 2
1
Z 1
−1
f(x, y) dx
!
dy
+
Z 3
2
Z −
√
y−2
−1
f(x, y) dx +
Z 1
√
y−2
f(x, y) dx
!
dy
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

9. 9Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 3.
1.
2. We change the order of integration and obtain
K =
Z 1
0
Z y
0
sin(y2
) dx
dy =
Z 1
0
y sin(y2
) dy = −
1
2
[cos(y2
)]
Z 1
0
= −
1
2
(cos 1−1) .
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

10. 10Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 4.
1. D = {(x, y) ∈ R2
: 0 ≤ x ≤ 1 et x2
≤ y ≤ x}
2.
0 ≤ y ≤ 1 et y ≤ x ≤
√
y
Thus
I =
Z 1
0
dy
Z y
√
y
2xe2y3
−3y2
dx
3.
I =
Z 1
0
e2y3
−3y2
dy
Z √
y
y
2x dx =
Z 1
0
e2y3
−3y2
dy
h
x2
i√
y
y
=
Z 1
0
(y − y2
)e2y3
−3y2
dy
= −
1
6
h
e2y3
−3y2
i1
0
= −
1
6
e−1
− 1
.
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

11. 11Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 5.
1.
I =
Z 1
0
dy
Z 2−y
√
1−y2
x dx =
Z 1
0
x2
2
2−y
√
1−y2
dy =
1
2
Z 1
0
(2y2
− 4y + 3) dy =
5
6
.
2. D =
n
(x, y) ∈ R
2
; 0 6 y 6 1 and
q
1 − y2 6 x 6 2 − y
o
.
3. Fixant x d’abord et cherchons la variation verticale de y on obtient que,
I Pour 0 6 x 6 1 on a y ∈ [
p
1 − x2; 1],
I Pour 1 6 x 6 2 on a y ∈ [0; 2 − x].
I =
ZZ
D
x dx dy =
Z 1
0
dx
Z 1
√
1−x2
x dy +
Z 2
1
dx
Z 2−x
0
x dy =
Z 1
0
x(1 −
p
1 − x2) dx +
Z 2
0
x(2 − x) dx
=
Z 1
0
(x − x
p
1 − x2) dx +
Z 2
1
(2x − x
2
) dx =
1
2
−
1
2
.
2
3
(1 − x
2
)
3
2
1
0
+ 3 −
7
3
=
5
6
.
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

12. 12Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 6.
1.
x2
+ y2
− 2x ≤ 0 =
⇒ (x − 1)2
+ y2
≤ 1
Using polar coordinates:
x = r cos θ , y = r sin θ , |J| = r .
We have
x2
+ y2
− 2x ≤ 0 =
⇒ r2
− 2r cos θ = 0 =
⇒ r = 2 cos θ
We obtain
0 ≤ θ ≤
π
2
et 0 ≤ r ≤ 2 cos θ
Finally
I =
Z π
2
0
Z 2 cos θ
0
r2
cos2
θr sin θ rdrdθ =
Z π
2
0
cos2
θ sin θdθ
Z 2 cos θ
0
r4
dr
=
32
5
Z π
2
0
cos7
θ sin θdθ = −
32
5
cos8 θ
8
π
2
0
=
4
5
.
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

13. 13Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
2.
Using polar coordinates
x = r cos θ , y = r sin θ , |J| = r .
We have
x2
+ y2
= 1 =
⇒ r = 1 et x2
+ y2
= 4 =
⇒ r = 2
I =
Z 2π
0
Z 2
1
ln(r2
)rdrdθ = 2
Z 2π
0
dθ
Z 2
1
ln(r)rdr u = ln r , v0
= r
= 2π [r2
ln r]2
1 −
Z 2
1
rdr
!
= 2π(4 ln 2 −
3
2
) . u0
=
1
r
, v =
r2
2
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

14. 14Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
3.
Using polar coordinates
x = r cos θ , y = r sin θ , |J| = r .
We have
x = 1 =
⇒ r1 =
1
cos θ
et x
2
+ y
2
− 2x = 0
=
⇒ r
2
− 2r cos θ = 0 =
⇒ r2 = 2 cos θ
Then
1
cos θ
≤ r ≤ 2 cos θ
To find the angle limits, we have to find the intersections points. Indeed
{x
2
+y
2
−2x = 0}∩{x = 1} =
⇒ −1+y
2
= 0 =
⇒ y = ±1 =
⇒ A = (1, 1) et B(−1, 1) .
D’où
−
π
4
≤ θ ≤
π
4
I =
Z π
4
− π
4
Z 2 cos θ
1/ cos θ
1
r4
rdrdθ =
Z π
4
− π
4
dθ
−
1
2r2
2 cos θ
1/ cos θ
=
Z π
4
− π
4
cos2
θ
2
−
1
8 cos2 θ
!
dθ =
π
8
.
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

15. 15Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
4.
We have
x
2
+ y
2
− 2y = 0 =
⇒ r
2
− 2r sin θ = 0
=
⇒ r1 = 2 sin θ
and
x
2
+ y
2
= 1 =
⇒ r2 = 1
Then
2 sin θ ≤ r ≤ 1 .
{x
2
+ y
2
− 2y = 0} ∩ {x
2
+ y
2
= 1} =
⇒ 1 − 2y = 0 =
⇒ y =
1
2
=
⇒ A =
√
3
2
,
1
2
!
.
Thus 0 ≤ θ ≤
π
6
I =
Z π
6
0
Z 1
2 sin θ
r sin θrdrdθ =
Z π
6
0
sin θdθ
r2
2
#1
2 sin θ
=
1
2
Z π
6
0
sin θ − 4 sin
3
θ
dθ
= −
√
3
4
+
1
2
− 2
Z π
6
0
sin
3
θdθ =
√
3
2
−
5
6
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

16. 16Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 7.
1.
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

17. 17Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
2. For x2
+ y2
− 2y = 0 we have r(r − 2 sin θ) = 0, which is implies that r = 2 sin θ
because r 6= 0.
For y = 1, we have r sin θ = 1 then r =
1
sin θ
Therefore
∆ =





π
4
≤ θ ≤
π
2
1
sin θ
≤ r ≤ 2 sin θ
and so
I =
Z π
2
π
4
Z 2 sin θ
1
sin θ
1
r4
r dr
!
dθ =
1
2
Z π
2
π
4
sin2
θ −
1
4 sin2
θ
dθ
Having in mind
sin2
θ =
1 − cos(2θ)
2
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

18. 18Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Thus
I =
1
4
Z π
2
π
4
dθ −
1
4
Z π
2
π
4
cos(2θ)dθ −
1
8
Z π
2
π
4
cos2 θ
sin2
θ
dθ
cos2 θ
.
Using the fact that
Z
u0
cos u = − sin u and
Z
1
tan2 θ
dθ
cos2 θ
=
Z
1
u2
du = −
1
u
= −
1
tan θ
we find
I =
1
4
h
θ
i π
2
π
4
−
1
8
h
sin(2θ)
iπ
2
π
4
−
1
8
Z π
2
π
4
1
tan2 θ
dθ
cos2 θ
.
=
π
16
+
1
8
+
1
8
h 1
tan θ
iπ
2
π
4
=
π
16
+
1
8
−
1
8
=
π
16
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

19. 19Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Solution 8.
1.
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

20. 20Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
2.
I =
Z π
4
0
Z 2 cos θ
1/ cos θ
1
r4
· r dr
!
dθ = −
1
2
Z π
4
0
h 1
r2
i2 cos θ
1/ cos θ
dθ
= −
1
2
Z π
4
0
1
4 cos2 θ
− cos2
θ
dθ = −
1
8
Z π
4
0
1
cos2 θ
dθ +
1
2
Z π
4
0
cos2
θ dθ
We know that
cos2
θ =
1 + cos(2θ)
2
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

21. 21Ú22
Exercises
Solutions
Cartesian coordinates
Polar coordinates
Thus
I =
1
4
Z π
4
0
dθ +
1
4
Z π
4
0
cos(2θ)dθ −
1
8
Z π
4
0
sin2
θ
cos2 θ
dθ
sin2
θ
.
Using the fact that
Z
u0
cos u = − sin u, (cot θ)
0
= −
1
sin2
θ
and
Z
1
cot2 θ
dθ
sin2
θ
= −
Z
1
u2
du =
1
u
=
1
cot θ
= tan θ
We get
I =
1
4
h
θ
i π
4
0
+
1
8
h
sin(2θ)
i π
4
0
−
1
8
Z π
4
0
1
cot2 θ
dθ
sin2
θ
.
=
π
16
+
1
8
−
1
8
h
tan θ
i π
4
0
=
π
16
+
1
8
−
1
8
=
π
16
Go Back
Dr. Kamel ATTAR | Chapter 2: Double Integration | Solved Problems

22. Thank you! Questions?