1. Surface Area

2. Given two 3D vectors
u and v,
|u x v| = area of the
parallelogram
defined by u and v.
Surface Area

3. Given two 3D vectors
u and v,
|u x v| = area of the
parallelogram
defined by u and v.
Surface Area
u x v
v
u
length of u x v
= area of the parallelogram

4. Given two 3D vectors
u and v,
|u x v| = area of the
parallelogram
defined by u and v.
Surface Area
1
1
x
y
z
M
L
L
u x v
v
u
length of u x v
= area of the parallelogram
Theorem: Given a 3D
parallelogram over the
domain [0,1] x [0, 1].

5. Given two 3D vectors
u and v,
|u x v| = area of the
parallelogram
defined by u and v.
Surface Area
1
1
x
y
z
M
L
L
u x v
v
u
length of u x v
= area of the parallelogram
Theorem: Given a 3D
parallelogram over the
domain [0,1] x [0, 1].
Let M and L be the slopes in
the x-direction and the
y-direction respectively,

6. Given two 3D vectors
u and v,
|u x v| = area of the
parallelogram
defined by u and v.
Surface Area
1
1
x
y
z
M
L
L
u x v
v
u
length of u x v
= area of the parallelogram
Theorem: Given a 3D
parallelogram over the
domain [0,1] x [0, 1].
Let M and L be the slopes in
the x-direction and the
y-direction respectively, then
the area of the parallelogram
is √M2
+ L2
+ 1.

7. Surface Area
1
1
x
y
z
M
L
L
Proof: We may assume the
parallelogram is as shown in
the figure.

8. Surface Area
1
1
x
y
z
M
L
L
Proof: We may assume the
parallelogram is as shown in
the figure. It is defined by the
two vectors:
<1, 0, M> and <0, 1, L>.

9. Surface Area
1
1
x
y
z
M
L
L
Proof: We may assume the
parallelogram is as shown in
the figure. It is defined by the
two vectors:
<1, 0, M> and <0, 1, L>.
Hence the area of the
parallelelogram is
|<0, 1, L> x <1, 0, M>|

10. Surface Area
1
1
x
y
z
M
L
L
Proof: We may assume the
parallelogram is as shown in
the figure. It is defined by the
two vectors:
<1, 0, M> and <0, 1, L>.
Hence the area of the
parallelelogram is
|<0, 1, L> x <1, 0, M>| = | <M, L, -1> | = √M2
+ L2
+ 1.

11. Surface Area
1
1
x
y
z
M
L
L
Proof: We may assume the
parallelogram is as shown in
the figure. It is defined by the
two vectors:
<1, 0, M> and <0, 1, L>.
Hence the area of the
parallelelogram is
|<0, 1, L> x <1, 0, M>| = | <M, L, -1> | = √M2
+ L2
+ 1.
In general, if the domain is
[0,Δx] x [0, Δy] instead,
then the area is
√M2
+ L2
+ 1 ΔxΔy
Δx
Δy
x
y
z
MΔx
LΔy

12. Surface Area
y
x
z = f(x, y)
D
Given a smooth surface defined by z = f(x, y) over a
rectangular domain D = [a,b] x [c, d], partition D into
sub–rectangles.

13. Surface Area
y
x
z = f(x, y)
D
Given a smooth surface defined by z = f(x, y) over a
rectangular domain D = [a,b] x [c, d], partition D into
sub–rectangles.

14. Surface Area
x
z = f(x, y)
D
Given a smooth surface defined by z = f(x, y) over a
rectangular domain D = [a,b] x [c, d], partition D into
sub–rectangles. Let (xi, yi) be a point at the corner of a
sub–rectangle and ΔSi be the corresponding
parallelogram–surface area in the tangent plane at the
point (xi, yi, f(xi,yi)).
(xi, yi)
(xi, yi, z i)

15. Surface Area
x
z = f(x, y)
D
Tangent plane over (xi, yi)
Given a smooth surface defined by z = f(x, y) over a
rectangular domain D = [a,b] x [c, d], partition D into
sub–rectangles. Let (xi, yi) be a point at the corner of a
sub–rectangle and ΔSi be the corresponding
parallelogram–surface area in the tangent plane at the
point (xi, yi, f(xi,yi)).
(xi, yi)
(xi, yi, z i)

16. Surface Area
Given a smooth surface defined by z = f(x, y) over a
rectangular domain D = [a,b] x [c, d], partition D into
sub–rectangles. Let (xi, yi) be a point at the corner of a
sub–rectangle and ΔSi be the corresponding
parallelogram–surface area in the tangent plane at the
point (xi, yi, f(xi,yi)).
y
x
z = f(x, y)
D
Tangent plane over (xi, yi)
(xi, yi)
(xi, yi, z i)
ΔSi

17. Let Δx and Δy be the sides
be the sub–rectangle, then
ΔSi = √M2
+ L2
+ 1 ΔxΔy
where
M = fx(xi, yi) =
L = fy(xi, yi) =
dz dx|/
dz dy|/
(xi, yi)
(xi, yi)
Surface Area
Given a smooth surface defined by z = f(x, y) over a
rectangular domain D = [a,b] x [c, d], partition D into
sub–rectangles. Let (xi, yi) be a point at the corner of a
sub–rectangle and ΔSi be the corresponding
parallelogram–surface area in the tangent plane at the
point (xi, yi, f(xi,yi)).
y
x
z = f(x, y)
D
Δy
Tangent plane over (xi, yi)
(xi, yi)
Δx
(xi, yi, z i)
ΔSi

18. Let Δx and Δy be the sides
be the sub–rectangle, then
ΔSi = √M2
+ L2
+ 1 ΔxΔy
where
M = fx(xi, yi) =
L = fy(xi, yi) =
dz dx|/
dz dy|/
(xi, yi)
(xi, yi)
Surface Area
Given a smooth surface defined by z = f(x, y) over a
rectangular domain D = [a,b] x [c, d], partition D into
sub–rectangles. Let (xi, yi) be a point at the corner of a
sub–rectangle and ΔSi be the corresponding
parallelogram–surface area in the tangent plane at the
point (xi, yi, f(xi,yi)).
y
x
z = f(x, y)
D
Δy
Tangent plane over (xi, yi)
(xi, yi)
Δx
(xi, yi, z i)
ΔSi

19. Hence the surface S area over D is the sum of all
the areas i.e. S ≈ ΣΔSi = Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy2 2
Surface Area

20. Hence the surface S area over D is the sum of all
the areas i.e. S ≈ ΣΔSi = Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy
Let Δx, Δy  0 we get the real surface area,
2 2
Surface Area

21. Hence the surface S area over D is the sum of all
the areas i.e. S ≈ ΣΔSi = Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy
Let Δx, Δy  0 we get the real surface area, i.e.
S = lim
2 2
Δx, Δy  0
Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy
2 2
Surface Area

22. Hence the surface S area over D is the sum of all
the areas i.e. S ≈ ΣΔSi = Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy
Let Δx, Δy  0 we get the real surface area, i.e.
S = lim
2 2
Δx, Δy  0
Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy or
2 2
S =
2 2
∫∫ D
Surface Area
√fx + fy + 1 dA where dA = dxdy

23. Hence the surface S area over D is the sum of all
the areas i.e. S ≈ ΣΔSi = Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy
Let Δx, Δy  0 we get the real surface area, i.e.
S = lim
2 2
Δx, Δy  0
Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy or
2 2
x=a
b
S =
y=c
d
2 2
to compute via iterated integrals:
√fx + fy + 1 dxdy or
2 2
∫∫ x=a
b
y=c
d
√fx + fy + 1 dydx
2 2
∫∫
Surface Area
√fx + fy + 1 dA where dA = dxdy∫∫ D

24. Hence the surface S area over D is the sum of all
the areas i.e. S ≈ ΣΔSi = Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy
Let Δx, Δy  0 we get the real surface area, i.e.
S = lim
2 2
Δx, Δy  0
Σ√fx(xi, yi)+fy(xi, yi)+1 ΔxΔy or
2 2
x=a
b
S =
y=c
d
√fx + fy + 1 dA where dA = dxdy
2 2
to compute via iterated integrals:
√fx + fy + 1 dxdy or
2 2
∫∫ x=a
b
y=c
d
√fx + fy + 1 dydx
2 2
∫∫
x=f(y)y=c
Over a general domain D:
√fx + fy + 1 dxdy
2 2
∫∫√fx + fy + 1 dydx
2 2
∫∫
x=g(y)d
y=f(x)
y=g(x)
x=a
b
(Type I) (Type II)
or
Surface Area
∫∫ D

25. The integrals for surface area are difficult to
calculate in general.
Surface Area

26. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area

27. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]

28. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
1
4

29. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
z2
= 4 – x2
, its easier to use implicit differentiation for
fx and fy.
1
4

30. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
z2
= 4 – x2
, its easier to use implicit differentiation for
fx and fy. Treat z as a function of x, both sides:
1
4
dx
d

31. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
z2
= 4 – x2
, its easier to use implicit differentiation for
fx and fy. Treat z as a function of x, both sides:
dx
d z2
= dx
d (4 – x2
)
1
4
dx
d

32. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
z2
= 4 – x2
, its easier to use implicit differentiation for
fx and fy. Treat z as a function of x, both sides:
dx
d z2
= dx
d (4 – x2
)
2z dx
dz = -2x
1
4
dx
d

33. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
z2
= 4 – x2
, its easier to use implicit differentiation for
fx and fy. Treat z as a function of x, both sides:
dx
d
dx
dz
z2
= dx
d (4 – x2
)
2z dx
dz = -2x
= -x
z = 1
4
dx
d

34. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
z2
= 4 – x2
, its easier to use implicit differentiation for
fx and fy. Treat z as a function of x, both sides:
dx
d
dx
dz
z2
= dx
d (4 – x2
)
2z dx
dz = -2x
= -x
z = -x
√4 – x2
= fx
1
4
dx
d

35. The integrals for surface area are difficult to
calculate in general. Two examples are given here,
one utilizes the inverse-trig functions, the other uses
polar integration.
Surface Area
Example: Find the surface area of the cylinder
x2
+ z2
= 4 over the domain [0, 1] x [0, 4]
z2
= 4 – x2
, its easier to use implicit differentiation for
fx and fy. Treat z as a function of x, both sides:
dx
d
dx
dz
z2
= dx
d (4 – x2
)
2z dx
dz = -2x
= -x
z = -x
√4 – x2
By the same method dy
dz = 0
= fx
= fy
1
4
dx
d

36. Surface Area
Hence fx + fy + 1
2 2
=

37. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1

38. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2

39. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2

40. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1

41. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1
= 2 π/6 dy = 4π/3
y=0
4
∫

42. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1
= 2 π/6 dy = 4π/3
y=0
4
∫
Example: Find the surface area of the cone
z = √ x2
+ y2
over D = unit circle.

43. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1
= 2 π/6 dy = 4π/3
y=0
4
∫
Example: Find the surface area of the cone
z = √ x2
+ y2
over D = unit circle.

44. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1
= 2 π/6 dy = 4π/3
y=0
4
∫
Example: Find the surface area of the cone
z = √ x2
+ y2
over D = unit circle.
fx= x/√ x2
+ y2
fy= y/√ x2
+ y2

45. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1
= 2 π/6 dy = 4π/3
y=0
4
∫
Example: Find the surface area of the cone
z = √ x2
+ y2
over D = unit circle.
fx= x/√ x2
+ y2
fy= y/√ x2
+ y2
fx + fy + 1 =
22

46. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1
= 2 π/6 dy = 4π/3
y=0
4
∫
Example: Find the surface area of the cone
z = √ x2
+ y2
over D = unit circle.
fx= x/√ x2
+ y2
fy= y/√ x2
+ y2
fx + fy + 1 =
22 x2
x2
+ y2 +
y2
x2
+ y2 + 1

47. Surface Area
Hence fx + fy + 1
2 2
= ( -x
√4 – x2
)2
+ 1 =
x2
4 – x2 + 1 =
4
4 – x2
So the integral for the surface is:
y=0
4
x=0
1
dxdy = 2∫∫
2
√4 – x2
y=0
4
x=0
sin-1
(x/2) |dy∫
x=1
= 2 π/6 dy = 4π/3
y=0
4
∫
Example: Find the surface area of the cone
z = √ x2
+ y2
over D = unit circle.
fx= x/√ x2
+ y2
fy= y/√ x2
+ y2
fx + fy + 1 =
22 x2
x2
+ y2 +
y2
x2
+ y2 + 1 = 2

48. Surface Area
So the integral for the surface is:
√2 dA∫∫ D

49. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D)∫∫ D

50. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D) = π√2∫∫ D

51. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D) = π√2∫∫ D
Or by polar integration:

52. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D) = π√2∫∫ D
Or by polar integration:
the polar equation for the unit
disc is r = 1 and 0 < θ < 2π.
So the polar integral is:

53. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D) = π√2∫∫ D
Or by polar integration:
the polar equation for the unit
disc is r = 1 and 0 < θ < 2π.
So the polar integral is:
√2 dA =∫∫D
√2 r dr dθ∫∫ r=0
1
θ=0
2π

54. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D) = π√2∫∫ D
Or by polar integration:
the polar equation for the unit
disc is r = 1 and 0 < θ < 2π.
So the polar integral is:
√2 dA =∫∫D
√2 r dr dθ = ½∫∫ r=0
1
θ=0
2π
∫
θ=0
2π
√2 r2
| dθ
r=0
1

55. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D) = π√2∫∫ D
Or by polar integration:
the polar equation for the unit
disc is r = 1 and 0 < θ < 2π.
So the polar integral is:
√2 dA =∫∫D
√2 r dr dθ = ½∫∫ r=0
1
θ=0
2π
∫
θ=0
2π
√2 r2
| dθ
r=0
1
∫
θ=0
2π
= ½ √2 dθ

56. Surface Area
So the integral for the surface is:
√2 dA = √2 * (area of D) = π√2∫∫ D
Or by polar integration:
the polar equation for the unit
disc is r = 1 and 0 < θ < 2π.
So the polar integral is:
√2 dA =∫∫D
√2 r dr dθ = ½∫∫ r=0
1
θ=0
2π
∫
θ=0
2π
√2 r2
| dθ
r=0
1
∫
θ=0
2π
= ½ √2 dθ
= π√2