The document discusses double integrals in polar coordinates. It defines a domain D bounded by functions r1(θ) and r2(θ) between angles A and B. It partitions the domain into tiles of widths Δr and Δθ and approximates the area and volume integrals using sums over the tiles. The volume integral over the domain D of a function z=f(r,θ) is written as a double integral in polar coordinates from r1 to r2 and θ from A to B.

1. Double Integrals Over Polar Equations

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

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

4. Given a domain D defined by r1(θ) and r2(θ) where r1
< r2 and A < θ < B as shown in the figure.
…
Double Integrals Over Polar Equations
Y
Z
A
B
X
Partition the xy-plane
with Δr and Δθ.
D
r2(θ)
r1(θ)

5. Given a domain D defined by r1(θ) and r2(θ) where r1
< r2 and A < θ < B as shown in the figure.
…
Δr
Double Integrals Over Polar Equations
Y
Z
A
B
X
Partition the xy-plane
with Δr and Δθ.
D
r2(θ)
r1(θ)

6. Given a domain D defined by r1(θ) and r2(θ) where r1
< r2 and A < θ < B as shown in the figure.
…
Δθ
Δr
Double Integrals Over Polar Equations
Y
Z
A
B
X
Partition the xy-plane
with Δr and Δθ.
D
r2(θ)
r1(θ)

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

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

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

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

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

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

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

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

15. 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.

16. 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.

17. 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.

18. 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.

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)
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.

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)
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.

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.
θ = 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.

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.
(√18, 315o
, 0)
θ = 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.

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)
θ = 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.

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

26. 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

27. 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
θ = 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

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
θ = 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. 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, θ). Here are some examples of
cylindrical graphs.

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, θ). Here are some examples of
cylindrical graphs.
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤ θ ≤ 2π}
→ 0 ≤ z ≤ 1

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, θ). Here are some examples of
cylindrical graphs.
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤ θ ≤ 2π}
→ 0 ≤ z ≤ 2 π
x
y
z = sin(r)

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, θ). Here are some examples of
cylindrical graphs.
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤ θ ≤ 2π}
→ 0 ≤ z ≤ 2 π
x
y
z = sin(r)
b. z = θ sin(r),
D = {0 ≤ r ≤ π, 0 ≤ θ ≤ 2π}
→ 0 ≤ z ≤ 2 π

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, θ). Here are some examples of
cylindrical graphs.
Example C. a. z = sin(r),
D = {0 ≤ r ≤ π, 0 ≤ θ ≤ 2π}
→ 0 ≤ z ≤ 2 π
x
y
z = sin(r)
x y
z = θsin(r)
b. z = θ sin(r),
D = {0 ≤ r ≤ π, 0 ≤ θ ≤ 2π}
→ 0 ≤ z ≤ 2 π

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

35. 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 ≤ θ ≤ 2π}.
I II

36. ∫ ∫ f(r, θ) rdrdθ
r=r1(θ)
r2(θ)
θ=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 ≤ θ ≤ 2π}. Using the
formula
I II
V =
Vol(II) =
, their volumes are

37. ∫ ∫ f(r, θ) rdrdθ
r=r1(θ)
r2(θ)
θ=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 ≤ θ ≤ 2π}. Using the
formula
I II
V =
∫ ∫ sin(r) r drdθ,
r=0
π
θ=0
2π
Vol(II) =
, their volumes are

38. ∫ ∫ f(r, θ) rdrdθ
r=r1(θ)
r2(θ)
θ=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 ≤ θ ≤ 2π}. Using the
formula
I II
V =
∫ ∫ sin(r) r drdθ,
r=0
π
θ=0
2π
Vol(II) =∫ ∫ θ sin(r) r drdθ. (HW: Finish the problems.)
r=0
π
θ=0
2π
, their volumes are

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

40. Example D. Let z = f(r, θ) = cos(θ) over the domain
D = {(r, θ)| r = sin(θ) where 0 ≤ θ ≤ π/2}.
Find the volume of the solid defined by z over D.
Double Integrals Over Polar Equations
x
1
D :
r=sin(θ)
x

41. Example D. Let z = f(r, θ) = cos(θ) over the domain
D = {(r, θ)| r = sin(θ) where 0 ≤ θ ≤ π/2}.
Find the volume of the solid defined by z over D.
= ∫ ∫ cos(θ)r drdθ
r=0θ=0
Convert the integral to
iterated integral, we get
∫∫cos(θ)dA
D
r=sin(θ)π/2
Double Integrals Over Polar Equations
x
1
D :
r=sin(θ)

42. Example D. Let z = f(r, θ) = cos(θ) over the domain
D = {(r, θ)| r = sin(θ) where 0 ≤ θ ≤ π/2}.
Find the volume of the solid defined by z over D.
= ∫ ∫ cos(θ)r drdθ
r=0θ=0
Convert the integral to
iterated integral, we get
∫∫cos(θ)dA
D
r=sin(θ)π/2
= ∫ cos(θ)r2
/2 | dθ
r=0θ=0
π/2 sin(θ)
Double Integrals Over Polar Equations
x
1
x
D :
r=sin(θ)

43. Example D. Let z = f(r, θ) = cos(θ) over the domain
D = {(r, θ)| r = sin(θ) where 0 ≤ θ ≤ π/2}.
Find the volume of the solid defined by z over D.
= ∫ ∫ cos(θ)r drdθ
r=0θ=0
Convert the integral to
iterated integral, we get
∫∫cos(θ)dA
D
r=sin(θ)π/2
= ∫ cos(θ)r2
/2 | dθ
r=0θ=0
π/2 sin(θ)
= ½ ∫ cos(θ)sin2
(θ)dθ
θ=0
π/2
Change variable
Set u = sin(θ):
= sin3
(θ)/6 | =1/6
θ=0
π/2
Double Integrals Over Polar Equations
x
1
x
D :
r=sin(θ)

44. 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

45. 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

46. 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

47. 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

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. Hence the integral is
r=2
Its defined by the polar
equation r = 2 & r=0
∫ ∫ r3
* rdrdθ
r= 00
2π r= 2
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
* rdrdθ
r= 00
2π r= 2
∫ r5
/5 | dθ
r= 00
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
* rdrdθ
r= 00
2π r= 2
∫ r5
/5 | dθ
r= 00
2π
r= 2
=
∫ 32/5 dθ
0
2π
=
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
∫ ∫ (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
* rdrdθ
r= 00
2π r= 2
∫ r5
/5 | dθ
r= 00
2π
r= 2
=
∫ 32/5 dθ
0
2π
=
= 64π/5
Double Integrals Over Polar Equations

52. For more integration examples of changing from
dxdy form to the polar rdrdθr –form may be found at
the following link:
Double Integrals Over Polar Equations
http://ltcconline.net/greenl/courses/202/multipleIntegration/d
oublePolarIntegration.htm