The document discusses double integrals over polar equations and cylindrical coordinates. It explains that a double integral over a region D defined in polar coordinates by r1(θ) and r2(θ), where r1 < r2 and A < θ < B, can be evaluated as an integral of an integrand function f(r,θ) over the region D in terms of drdθ. It also introduces the cylindrical coordinate system, which uses polar coordinates for points in the xy-plane and z as the third coordinate, and provides examples of converting between rectangular and cylindrical coordinates and sketching curves defined by constant r and θ values.

1. Double Integrals Over Polar Equations

2. Double Integrals Over Polar Equations
* Double Integrals Over Regions Defined
in Polar Coordinates
* The Cylindrical Coordinates

3. Given a domain D defined by r1() and r2() where
r1 < r2 and A <  < B as shown in the figure,
D = {(r, )| r1 < r < r2, A <  < B}.
Double Integrals Over Polar Equations
Y
Z
A
B
X
D
r2()
r1()

4. Double Integrals Over Polar Equations
Y
Z
A
B
X
Partition the xy-plane
with Δr and Δ.
D
r2()
r1()
Given a domain D defined by r1() and r2() where
r1 < r2 and A <  < B as shown in the figure,
D = {(r, )| r1 < r < r2, A <  < B}.

5. …
Double Integrals Over Polar Equations
Partition the xy-plane
with Δr and Δ.
Given a domain D defined by r1() and r2() where
r1 < r2 and A <  < B as shown in the figure,
D = {(r, )| r1 < r < r2, A <  < B}.
Y
Z
A
B
X
D
r2()
r1()

6. …
Δr
Double Integrals Over Polar Equations
Partition the xy-plane
with Δr and Δ.
Given a domain D defined by r1() and r2() where
r1 < r2 and A <  < B as shown in the figure,
D = {(r, )| r1 < r < r2, A <  < B}.
Y
Z
A
B
X
D
r2()
r1()

7. …
Δ
Δr
Double Integrals Over Polar Equations
Partition the xy-plane
with Δr and Δ.
Given a domain D defined by r1() and r2() where
r1 < r2 and A <  < B as shown in the figure,
D = {(r, )| r1 < r < r2, A <  < B}.
Y
Z
A
B
X
D
r2()
r1()

8. arbitrary ΔrΔ–tile as shown and let (r*, *) be an
arbitrary point in the tile and Δ A be area of the tile,
then ΔA ≈ r*ΔrΔ.
…
Δ
Δr
(r*, *)
r*
Double Integrals Over Polar Equations
Partition the xy-plane
with Δr and Δ. Select an
Given a domain D defined by r1() and r2() where
r1 < r2 and A <  < B as shown in the figure,
D = {(r, )| r1 < r < r2, A <  < B}.
Y
Z
A
B
X
D
r2()
r1()

9. arbitrary ΔrΔ–tile as shown and let (r*, *) be an
arbitrary point in the tile and Δ A be area of the tile,
then ΔA ≈ r*ΔrΔ.
(HW. Why? In fact ΔA = r*ΔrΔ if r*is the center.)
…
Δ
Δr
(r*, *)
r*
Double Integrals Over Polar Equations
Partition the xy-plane
with Δr and Δ. Select an
Given a domain D defined by r1() and r2() where
r1 < r2 and A <  < B as shown in the figure,
D = {(r, )| r1 < r < r2, A <  < B}.
Y
Z
A
B
X
D
r2()
r1()

10. Double Integrals Over Polar Equations
Let z = f(r, ) ≥ 0 be a function over D.
Y
Z
A
B
X
z = f(r, )
D
r2()
r1()

11. The volume of the column over
one tile having z as the cover is
ΔV ≈ f(r*, *) r*ΔrΔ.
Double Integrals Over Polar Equations
Y
A
B
X
Let z = f(r, ) ≥ 0 be a function over D.
D
Z
z = f(r, )
r2()
r1()

12. The volume of the column over
one tile having z as the cover is
ΔV ≈ f(r*, *) r*ΔrΔ.
lim Σ(ΔV)
Δr,Δ0
Double Integrals Over Polar Equations
Y
A
B
X
Let z = f(r, ) ≥ 0 be a function over D.
D
Z
Hence the volume V over D is
the sum of all
z = f(r, )
r2()
r1()

13. The volume of the column over
one tile having z as the cover is
ΔV ≈ f(r*, *) r*ΔrΔ.
lim Σ(ΔV) = lim Σf(r*, *)r*ΔrΔ
= ∫∫f(r, )dA.
Δr,Δ0 Δr,Δ0
ΔA
Double Integrals Over Polar Equations
Y
A
B
X
Let z = f(r, ) ≥ 0 be a function over D.
D
D
Z
Hence the volume V over D is
the sum of all
z = f(r, )
r2()
r1()

14. The volume of the column over
one tile having z as the cover is
ΔV ≈ f(r*, *) r*ΔrΔ.
lim Σ(ΔV) = lim Σf(r*, *)r*ΔrΔ
= ∫∫f(r, )dA.
Δr,Δ0 Δr,Δ0
∫ ∫ f(r, ) rdrd, which is V,
r=r1()
r2()
=A
B
ΔA
Double Integrals Over Polar Equations
Y
A
B
X
Let z = f(r, ) ≥ 0 be a function over D.
D
D
Z
Hence the volume V over D is
the sum of all
z = f(r, )
in term of drd, we have
r2()
r1()
∫∫f(r, )dA =
D

15. The volume of the column over
one tile having z as the cover is
ΔV ≈ f(r*, *) r*ΔrΔ.
lim Σ(ΔV) = lim Σf(r*, *)r*ΔrΔ
= ∫∫f(r, )dA.
Δr,Δ0 Δr,Δ0
∫ ∫ f(r, ) rdrd, which is V,
r=r1()
r2()
=A
B
ΔA
Double Integrals Over Polar Equations
Y
A
B
X
Let z = f(r, ) ≥ 0 be a function over D.
D
D
Z
Hence the volume V over D is
the sum of all
z = f(r, )
in term of drd, we have
= volume of the solid over D = {A <  <B; r1 < r < r2}.
r2()
r1()
∫∫f(r, )dA =
D

16. Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

17. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

18. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
x
y
z
Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

19. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
4
x
y
z
(3, 120o, 4)
Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

20. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 3
Hence the point is (–3/2, 3, 4)
Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

21. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 3
Hence the point is (–3/2, 3, 4)
b. Convert (3, –3, 1) into to
cylindrical coordinate.
Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

22. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 3
Hence the point is (–3/2, 3, 4)
b. Convert (3, –3, 1) into to
cylindrical coordinate.
Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

23. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 3
Hence the point is (–3/2, 3, 4)
b. Convert (3, –3, 1) into to
cylindrical coordinate.
 = 315o, r = 9 + 9 = 18
Hence the point is (18, 315o, 1)
the cylindrical coordinate.
Cylindrical Coordinates
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

24. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 3
Hence the point is (–3/2, 3, 4)
b. Convert (3, –3, 1) into to
cylindrical coordinate.
(18, 315o, 0)
 = 315o, r = 9 + 9 = 18
Hence the point is (18, 315o, 1)
the cylindrical coordinate. x
y
Cylindrical Coordinates
z
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

25. Example A. a. Plot the point (3,120o, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
3
120o
4
x
y
z
(3, 120o, 4)
x = 3cos(120o) = –3/2
y = 3sin(120o) = 3
Hence the point is (–3/2, 3, 4)
b. Convert (3, –3, 1) into to
cylindrical coordinate.
(18, 315o, 0)
 = 315o, r = 9 + 9 = 18
Hence the point is (18, 315o, 1)
the cylindrical coordinate. x
y
Cylindrical Coordinates
z
(18, 315o, 1) = (3, –3, 1)
1
The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd coordinate.

26. Cylindrical Coordinates
The constant equations
r = k describes cylinders
of radius k, thus the name
"cylindrical coordinate".

27. The constant equations
r = k describes cylinders
of radius k, thus the name
"cylindrical coordinate".
Example B. a. Sketch r = 2
2
Cylindrical Coordinates
x
y
z

28. The constant equations
r = k describes cylinders
of radius k, thus the name
"cylindrical coordinate".
x
y
Example B. a. Sketch r = 2
2
The constant equations
 = k describes the
vertical plane through the
origin, at the angle k with
x-axis.
b. Sketch θ = 3π/4
3π/4
Cylindrical Coordinates
z
x
y
z

29. The constant equations
r = k describes cylinders
of radius k, thus the name
"cylindrical coordinate".
x
y
Example B. a. Sketch r = 2
2
The constant equations
 = k describes the
vertical plane through the
origin, at the angle k with
x-axis.
b. Sketch θ = 3π/4
3π/4
Cylindrical Coordinates
z
x
y
z

30. Cylindrical Coordinates
Graphs may be given in the cylindrical coordinate
using polar coordinates for points in the domain in the
xy–plane and z = f(r, ). Here are some examples of
cylindrical graphs.

31. Cylindrical Coordinates
Graphs may be given in the cylindrical coordinate
using polar coordinates for points in the domain in the
xy–plane and z = f(r, ). Here are some examples of
cylindrical graphs.
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}
→ 0 ≤ z ≤ 1

32. Cylindrical Coordinates
Graphs may be given in the cylindrical coordinate
using polar coordinates for points in the domain in the
xy–plane and z = f(r, ). Here are some examples of
cylindrical graphs.
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}
→ 0 ≤ z ≤ 1
y
z = sin(r)
x (π, 0, 1)

33. Cylindrical Coordinates
Graphs may be given in the cylindrical coordinate
using polar coordinates for points in the domain in the
xy–plane and z = f(r, ). Here are some examples of
cylindrical graphs.
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}
→ 0 ≤ z ≤ 1
y
z = sin(r)
b. z =  sin(r),
D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}
→ 0 ≤ z ≤ 2 π
x (π, 0, 1)

34. Cylindrical Coordinates
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}
→ 0 ≤ z ≤ 1
x
y
z = sin(r)
x y
z = sin(r)
b. z =  sin(r),
D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}
→ 0 ≤ z ≤ 2 π
(π/2, 2π, 2π)p
(π, 0, 1)
Graphs may be given in the cylindrical coordinate
using polar coordinates for points in the domain in the
xy–plane and z = f(r, ). Here are some examples of
cylindrical graphs.

35. Double Integrals Over Polar Equations
D D
Let’s set up the volume calculation for both solids.
I II

36. Double Integrals Over Polar Equations
The base D
D D
r = π
Let’s set up the volume calculation for both solids.
We note that D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}.
I II

37. ∫ ∫ f(r, ) rdrd
r=r1()
r2()
=A
B
Double Integrals Over Polar Equations
Vol(I) =
The base D
D D
r = π
Let’s set up the volume calculation for both solids.
We note that D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}. Using the
formula
I II
V =
Vol(II) =
, their volumes are

38. ∫ ∫ f(r, ) rdrd
r=r1()
r2()
=A
B
Double Integrals Over Polar Equations
Vol(I) =
The base D
D D
r = π
Let’s set up the volume calculation for both solids.
We note that D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}. Using the
formula
I II
V =
∫ ∫ sin(r) r drd,
r=0
π
=0
2π
Vol(II) =
, their volumes are

39. ∫ ∫ f(r, ) rdrd
r=r1()
r2()
=A
B
Double Integrals Over Polar Equations
Vol(I) =
The base D
D D
r = π
Let’s set up the volume calculation for both solids.
We note that D = {0 ≤ r ≤ π, 0 ≤  ≤ 2π}. Using the
formula
I II
V =
∫ ∫ sin(r) r drd,
r=0
π
=0
2π
Vol(II) = ∫ ∫  sin(r) r drd. (HW: Finish the problems.)
r=0
π
=0
2π
, their volumes are

40. Example D. Let z = f(r, ) = cos() over the domain
D = {(r, )| 0 < r < sin(), 0 ≤  ≤ π/2}.
Find the volume of the solid defined by z over D.
Double Integrals Over Polar Equations
x

41. Example D. Let z = f(r, ) = cos() over the domain
D = {(r, )| 0 < r < sin(), 0 ≤  ≤ π/2}.
Find the volume of the solid defined by z over D.
Double Integrals Over Polar Equations
1
D
r=sin()
The domain D = {0 < r < sin(), 0 ≤  ≤ π/2}
is the right half of a circle as shown.
x

42. Example D. Let z = f(r, ) = cos() over the domain
D = {(r, )| 0 < r < sin(), 0 ≤  ≤ π/2}.
Find the volume of the solid defined by z over D.
Double Integrals Over Polar Equations
1
D
r=sin()
The domain D = {0 < r < sin(), 0 ≤  ≤ π/2}
is the right half of a circle as shown.
The surface z = cos() makes a 90o twist and
the solid with base D below z are shown here. x
y
x
z = f(r, ) = cos()
D
(r, 0, 1)

43. Example D. Let z = f(r, ) = cos() over the domain
D = {(r, )| 0 < r < sin(), 0 ≤  ≤ π/2}.
Find the volume of the solid defined by z over D.
Double Integrals Over Polar Equations
1
D
r=sin()
The domain D = {0 < r < sin(), 0 ≤  ≤ π/2}
is the right half of a circle as shown.
The surface z = cos() makes a 90o twist and
the solid with base D below z are shown here. x
y
x
x
z = f(r, ) = cos()
D
D
y
(r, 0, 1)
(r, 0, 1)

44. = ∫ ∫ cos()r drd
r=0
=0
Convert the integral to iterated integral, we get
∫∫cos()dA
D
r=sin()
π/2
= ∫ cos()r2/2 | d
r=0
=0
π/2 sin()
= ½ ∫ cos()sin2()d
=0
π/2
Change variable,
set u = sin().
= sin3()/6 | =1/6
=0
π/2
Double Integrals Over Polar Equations
y
x
x
z = f(r, ) = cos()
D
D
y
(r, 0, 1)
(r, 0, 1)
1
D
r=sin()
x

45. Example E. Evaluate
by converting it into polar integral
Double Integrals Over Polar Equations
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2

46. Example E. Evaluate
by converting it into polar integral
The domain D is:
r=2
Double Integrals Over Polar Equations
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2

47. Example E. Evaluate
by converting it into polar integral
The domain D is:
r=2
Its defined by the polar
equation r = 2 & r=0
Double Integrals Over Polar Equations
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2

48. Example E. Evaluate
by converting it into polar integral
The domain D is: (x2 + y2)3/2 = (r2)3/2 = r3 in polar
form.
r=2
Its defined by the polar
equation r = 2 & r=0
Double Integrals Over Polar Equations
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2

49. Example E. Evaluate
by converting it into polar integral
The domain D is: (x2 + y2)3/2 = (r2)3/2 = r3 in polar
form. Hence the integral is
r=2
Its defined by the polar
equation r = 2 & r=0
∫ ∫ r3 * rdrd
r= 0
0
2π r= 2
Double Integrals Over Polar Equations
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2

50. Example E. Evaluate
by converting it into polar integral
The domain D is: (x2 + y2)3/2 = (r2)3/2 = r3 in polar
form. Hence the integral is
r=2
Its defined by the polar
equation r = 2 & r=0
∫ ∫ r3 * rdrd
r= 0
0
2π r= 2
∫ r5/5 | d
r= 0
0
2π
r= 2
=
Double Integrals Over Polar Equations
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2

51. Example E. Evaluate
by converting it into polar integral
The domain D is: (x2 + y2)3/2 = (r2)3/2 = r3 in polar
form. Hence the integral is
r=2
Its defined by the polar
equation r = 2 & r=0
∫ ∫ r3 * rdrd
r= 0
0
2π r= 2
∫ r5/5 | d
r= 0
0
2π
r= 2
=
∫ 32/5 d
0
2π
=
Double Integrals Over Polar Equations
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2

52. Example E. Evaluate
by converting it into polar integral
The domain D is: (x2 + y2)3/2 = (r2)3/2 = r3 in polar
form. Hence the integral is
∫ ∫ (x2 + y2)3/2 dydx
y= -4 – x2
-2
2 y= 4 – x2
r=2
Its defined by the polar
equation r = 2 & r=0
∫ ∫ r3 * rdrd
r= 0
0
2π r= 2
∫ r5/5 | d
r= 0
0
2π
r= 2
=
∫ 32/5 d
0
2π
=
= 64π/5
Double Integrals Over Polar Equations

53. For more integration examples of changing from
dxdy form to the polar rdrdr –form may be found at
the following link:
Double Integrals Over Polar Equations
http://ltcconline.net/greenl/courses/202/multipleIntegration/
doublePolarIntegration.htm