Surfaces, surface area and surface integrals Our main objective here - - PowerPoint PPT Presentation

Surfaces, surface area and surface integrals

Our main objective here is the construction of surface integrals. Surface integrals are essential for stating in completely general terms the basic laws of electromagnetism. Also, the theorems of Gauss-Ostrogradskii and Stokes on vector calculus rely in an essential way on surface integrals.

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Parametric representation of surfaces

As an example, let’s consider the surface of the north hemisphere

• f radius r > 0 in R3 centred at the origin.

The points of the surface are defined by: x2 + y2 + z2 = r2, z ≥ 0.

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Parametric representation of surfaces (cont.)

Alternatively, the same surface can be traced out by the point (x, y, f(x, y)), when x and y vary across the domain D =

• (x, y) ∈ R2 | x2 + y2 ≤ r2

⊂ R2

xy,

while f : D → R is defined as f(x, y) =

• r2 − x2 − y2,

∀(x, y) ∈ D.

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Parametric representation of surfaces (cont.)

The previous example can be generalized in the following way. Suppose that f : D → R is a given continuous function defined on D ⊂ R2

xy in the x − y plane.

Then, the graph of the function f defines a surface S given by S =

• (x, y, f(x, y)) ∈ R3 | (x, y) ∈ D
• .

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Parametric representation of surfaces (cont.)

The graphs of functions f : D → R represent an important class

• f surfaces.

Unfortunately, not every surface in R3 can be represented by the graph of a function. Neither of the above surfaces can be represented by a graph of a single valued function f : D → R for some D ∈ R2

xy.

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Parametric surfaces

Definition A parametric function for a surface S is a given function or mapping Φ : D → R3 defined on some given region D in a u − v-plane R2

uv according to

Φ(u, v) = (x(u, v), y(u, v), z(u, v)) = = x(u, v)i + y(u, v)j + z(u, v)k, ∀(u, v) ∈ D. The surface S of the parametric function is the set of points in R3 traced by Φ(u, v) as (u, v) traverses the region D. S :=

• Φ(u, v) ∈ R3 | (u, v) ∈ D
• .

The region D is defined is called the parametric domain, the variable (u, v) is called the parametric variable, and the function Φ : D → R3 is called a parametric representation of the surface.

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C1 surfaces

Suppose that the parametric function Φ is a C1-function, that is ∂x(u, v) ∂u , ∂y(u, v) ∂u , ∂z(u, v) ∂u , ∂x(u, v) ∂v , ∂y(u, v) ∂v , ∂z(u, v) ∂v exist and are continuous functions of (u, v) in D. In this case, the parametric representation Φ : D → R3 is called a C1-parametric representation and the related surface S is called a C1-surface.

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Special case revised

Suppose we are given the continuous function f : D → R, where D ⊂ R2

xy is some region in the x − y plane.

We can take the u − v plane R2

uv (from the definition) to be iden-

tical to the x − y plane R2

xy and define

Φ(u, v) = ui + vj + f(u, v)k = (u, v, f(u, v)) , ∀(u, v) ∈ D. In this case, the scalar components of Φ(u, v) are given by x(u, v) = u, y(u, v) = v, z(u, v) = f(u, v). Then it is clear that the corresponding surface S coincides with the graph of the function f : D → R.

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Example: Northern hemisphere

The length of OA is the radius r of the sphere. From the right-angle triangle OAE we find OE = r cos(φ), OB = AE = r sin(φ). From the right-angle triangle OBC we find OC = OB cos(θ) = r sin(φ) cos(θ), OD = OB sin(θ) = r sin(φ) sin(θ).

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Example: Northern hemisphere (cont.)

OC, OD and OE give the x, y and z coordinates of the point A in terms of the angles θ and φ so that x(θ, φ) = r sin(φ) cos(θ), y(θ, φ) = r sin(φ) sin(θ), z(θ, φ) = r cos(φ). If θ varies through the range 0 ≤ θ ≤ 2π and φ varies through the range 0 ≤ φ ≤ π/2 then the point A having the above coordinates traverses the surface S of the northern hemisphere. In this case, the domain D is defined as D = {(θ, φ) | 0 ≤ θ ≤ 2π & 0 ≤ φ ≤ π/2} = [0, 2π] × [0, π/2], while the parametric function Φ : D → R3 is defined as Φ(θ, φ) = (x(θ, φ), y(θ, φ), z(θ, φ)) = = r sin(φ) cos(θ)i + r sin(φ) sin(θ)j + r cos(φ)k.

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Example: Whole sphere

If θ varies through the range 0 ≤ θ ≤ 2π and φ varies through the range 0 ≤ φ ≤ π then the point A traverses the entire sphere. In this case, the domain D is given by D = {(θ, φ) | 0 ≤ θ ≤ 2π & 0 ≤ φ ≤ π} = [0, 2π] × [0, π], while the parametric function Φ : D → R3 is still defined as Φ(θ, φ) = (x(θ, φ), y(θ, φ), z(θ, φ)) = = r sin(φ) cos(θ)i + r sin(φ) sin(θ)j + r cos(φ)k. Note that, as opposed to the case of hemisphere, the surface S of the sphere cannot be represented by the graph of a function.

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Example: Helicoid

For a “radius” r in the range 0 ≤ r ≤ 1 and an “angle” θ in the range 0 ≤ θ ≤ 2π, define Φ(r, θ) = (x(r, θ), y(r, θ), z(r, θ)) = r cos(θ)i + r sin(θ)j + θk. The parametric mapping Φ : D → R3, with D = [0, 1] × [0, 2π], represents the surface S of a helicoid.

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Tangents to a surface

Suppose that we are given a C1-parametric function Φ : D → R3 with D = [a, b] × [c, d], and fix some (u0, v0) ∈ D. Then the mapping γv0 : [a, b] → R3 defined by γv0(u) := Φ(u, v0) for u ∈ [a, b] is the parametric representation of a curve Γv0.

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Tangents to a surface (cont.)

The derivative of the parametric function γv0(u) at u = u0 is given by dγv0 du (u0) = ∂Φ ∂u (u0, v0) = ∂x ∂u(u0, v0), ∂y ∂u(u0, v0), ∂z ∂u(u0, v0)

• .

This vector is tangent to the curve Γv0 at the point Φ(u0, v0). Similarly, the mapping γu0 : [c, d] → R3 with γu0(v) := Φ(u0, v) and v ∈ [c, d] is the parametric representation of a curve Γu0. In this case, the vector dγu0 dv (v0) = ∂Φ ∂v (u0, v0) = ∂x ∂v (u0, v0), ∂y ∂v (u0, v0), ∂z ∂v (u0, v0)

• is tangent to the curve Γu0 at the point Φ(u0, v0).

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Normal vector

Define the vector cross product of the vector ∂Φ

∂u (u0, v0) and the

vector ∂Φ

∂v (u0, v0), that is

N(u0, v0) := ∂Φ ∂u (u0, v0) × ∂Φ ∂v (u0, v0). More specifically, N(u0, v0) =

• i

j k

∂x ∂u(u0, v0) ∂y ∂u(u0, v0) ∂z ∂u(u0, v0) ∂x ∂v (u0, v0) ∂y ∂v(u0, v0) ∂z ∂v(u0, v0)

• =

= i ∂y ∂u(u0, v0)∂z ∂v (u0, v0) − ∂y ∂v (u0, v0) ∂z ∂u(u0, v0)

• − j

∂x ∂u(u0, v0)∂z ∂v (u0, v0) − ∂x ∂v (u0, v0) ∂z ∂u(u0, v0)

• +

+ k ∂x ∂u(u0, v0)∂y ∂v (u0, v0) − ∂x ∂v (u0, v0)∂y ∂u(u0, v0)

• .

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Normal vector (cont.)

Another way to define the normal vector is N(u0, v0) = i∂(y, z) ∂(u, v)(u0, v0) + j∂(z, x) ∂(u, v)(u0, v0) + k∂(x, y) ∂(u, v)(u0, v0), where ∂(y, z) ∂(u, v)(u0, v0) =

• ∂y

∂u(u0, v0) ∂y ∂v(u0, v0) ∂z ∂u(u0, v0) ∂z ∂v(u0, v0)

• ,

∂(z, x) ∂(u, v)(u0, v0) =

• ∂z

∂u(u0, v0) ∂z ∂v(u0, v0) ∂x ∂u(u0, v0) ∂x ∂v (u0, v0)

• ,

∂(x, y) ∂(u, v)(u0, v0) =

• ∂x

∂u(u0, v0) ∂x ∂v (u0, v0) ∂y ∂u(u0, v0) ∂y ∂v(u0, v0)

• .

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Smooth surfaces

Definition A surface is called smooth at the point Φ(u0, v0) when N(u0, v0) = 0. The surface is called smooth when it is smooth at Φ(u, v), ∀(u, v) ∈ D, that is N(u, v) = 0, ∀(u, v) ∈ D. For a smooth surface S, one can define the unit vector n(u, v) := N(u, v) N(u, v), ∀(u, v) ∈ D, where

N(u, v) := ∂(x, y) ∂(u, v)(u, v) 2 + ∂(y, z) ∂(u, v)(u, v) 2 + ∂(z, x) ∂(u, v)(u, v) 2 .

As the tangent vectors span the plane that is tangent to the sur- face at Φ(u, v), the unit vector n(u, v) is normal to the plane, & therefore also normal to the surface S at Φ(u, v).

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Area of a surface

Fix some u0, v0, and small ∆u > 0, ∆v > 0, such that a ≤ u0 < u0 + ∆u ≤ b, c ≤ v0 < v0 + ∆v ≤ d. Then we can define a small rectangle ∆D according to ∆D := [u0, u0 + ∆u] × [v0, v0 + ∆v], which is mapped by Φ onto the small “piece of surface” ∆S := {Φ(u, v) | (u, v) ∈ ∆D} .

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Area of a surface (cont.)

∆S is approximately a flat parallelogram with edges AB and AC. The edges AB and AC are approximately given by the vectors v1 and v2, where v1 := Φ(u0, v0 + ∆v) − Φ(u0, v0) ≈ ∂Φ ∂v (u0, v0)∆v, v2 := Φ(u0 + ∆u, v0) − Φ(u0, v0) ≈ ∂Φ ∂u (u0, v0)∆u.

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Area of a surface (cont.)

The area of the parallelogram ∆S is given by area {∆S} = v1 × v2. On the other hand, v1 × v2 ≈ ∂Φ ∂u (u0, v0) × ∂Φ ∂v (u0, v0)

• ∆u∆v = N(u0, v0)∆u∆v.

Thus we have area {∆S} ≈ N(u0, v0)∆u∆v.

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Area of a surface (cont.)

As ∆u and ∆v shrink to the infinitesimals du and dv, the area ∆D shrinks to the infinitesimal rectangle dD, and the piece of surface ∆S shrinks to the infinitesimal parallelogram dS, and so area {dS} = N(u0, v0) dudv. Consequently, the total area of the surface S is the “sum” or in- tegral the elemental areas, namely area {S} =

• D

N(u, v) dudv = =

• D
• ∂Φ

∂u (u, v) × ∂Φ ∂v (u, v)

• dudv =

=

• D

∂(x, y) ∂(u, v)(u, v) 2 + ∂(y, z) ∂(u, v)(u, v) 2 + ∂(z, x) ∂(u, v)(u, v) 2 dudv. To get the area of S we, in fact, need to compute a 2-D integral.

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Area of a surface (cont.)

The area of a surface is intrinsic, and should not depend on the particular parametric representation we have chosen for the sur- face. This fact is guaranteed by the following theorem. Theorem Suppose that Φ : D → R3 and ˜ Φ : ˜ D → R3 are alternative C1-parametric representations of a surface S, so that S := {Φ(u, v) | (u, v) ∈ D} ≡

• ˜

Φ(˜ u, ˜ v) | (˜ u, ˜ v) ∈ ˜ D

• .

Then,

• D
• ∂Φ

∂u (u, v) × ∂Φ ∂v (u, v)

• dudv =
• ˜

D

• ∂ ˜

Φ ∂˜ u (˜ u, ˜ v) × ∂ ˜ Φ ∂˜ v (˜ u, ˜ v)

• d˜

ud˜ v.

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Area of a surface: special case

Suppose a surface S is the graph of a function f : D → R. Then, this surface has the parametric representation Φ : D → R3 in which Φ(u, v) = ui + vj + f(u, v)k, ∀(u, v) ∈ D. Next, we calculate ∂Φ ∂u (u, v) = 1i + 0j + ∂f ∂u(u, v)k, ∂Φ ∂v (u, v) = 0i + 1j + ∂f ∂v (u, v)k.

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Area of a surface: special case (cont.)

Consequently, we obtain ∂Φ ∂u (u, v) × ∂Φ ∂v (u, v) = −∂f ∂u(u, v)i − ∂f ∂v (u, v)j + k, which leads to

• ∂Φ

∂u (u, v) × ∂Φ ∂v (u, v)

• =
• 1 +

∂f ∂u(u, v) 2 + ∂f ∂v (u, v) 2 . Therefore, we have area {S} =

• D
• 1 +

∂f ∂u(u, v) 2 + ∂f ∂v (u, v) 2 dudv.

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Surface integral of a scalar field

Suppose that S has the parametric representation Φ : D → R3, where D ⊂ R2

uv is given by D = [a, b] × [c, d].

Suppose also that f : R3 → R is a given continuous scalar field. Recall that at a fixed (yet arbitrary) point (u0, v0) ∈ D we had area{∆S} ≈ N(u0, v0)∆u∆v, with N(u0, v0) := ∂Φ ∂u (u0, v0) × ∂Φ ∂v (u0, v0). Then, we can compute f(Φ(u0, v0)) area{∆S} ≈ f(Φ(u0, v0)) N(u0, v0)∆u∆v. What is the significance of this quantity?

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Surface integral of a scalar field (cont.)

As the piece of surface ∆S shrinks to the infinitesimal parallelo- gram dS, we get f(Φ(u0, v0)) area{dS} = f(Φ(u0, v0)) N(u0, v0)dudv, and hence one can define the surface integral of the scalar field f

• ver the surface S as
• S

fdS =

• D

f(Φ(u, v)) N(u, v)dudv. Thus, for example, if f(x, y, z) is the charge density at (x, y, z) on the surface, then the surface integral gives the total charge on S.

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Invariance under re-parameterization

Theorem Suppose that f : R3 → R is a continuous scalar field and that Φ : D → R3 and ˜ Φ : ˜ D → R3 are alternative C1-parametric representations of a surface S, so that S := {Φ(u, v) | (u, v) ∈ D} ≡

• ˜

Φ(˜ u, ˜ v) | (˜ u, ˜ v) ∈ ˜ D

• .

Then,

• D

f(Φ(u, v))

• ∂Φ

∂u (u, v) × ∂Φ ∂v (u, v)

• dudv =

=

• ˜

D

f( ˜ Φ(˜ u, ˜ v))

• ∂ ˜

Φ ∂˜ u (˜ u, ˜ v) × ∂ ˜ Φ ∂˜ v (˜ u, ˜ v)

• d˜

ud˜ v.

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Notations

Some standard notations for the integral are

• S

f(x, y, z)dA,

• S

f(x, y, z)dσ,

• S

f(x, y, z)dS, as well as

• S

f dA,

• S

f dσ,

• S

f dS. Note that the notations remind us that we are integrating over S with respect to an underlying space variable in R3 generically de- noted by (x, y, z). Also observe that substituting f(x, y, z) = 1 recovers the formula for the area of S.

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Examples

Example 1: Suppose that S is the helicoid and the scalar field is f(x, y, z) :=

• x2 + y2 + 1,

∀(x, y, z) ∈ R3. Determine the surface integral

• S

f dA. Example 2 (Mass of a conical shell): Consider a thin shell in the shape of a right circular cone S of height h. Assuming that the mass density distribution ρ (mass per unit surface area) is known, compute the total mass M according to M :=

• S

ρ dA.

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Surface integral of vector fields

Suppose S has the parametric representation Φ : D → R3, with D := [a, b] × [c, d] for concreteness. Also, let F : R3 → R3 be a given continuous vector field. Φ maps ∆D onto the piece of surface ∆S given by ∆S = {Φ(u, v) | (u, v) ∈ ∆D} , with area(∆S) = N(u0, v0)∆u∆v. and N(u0, v0) = ∂Φ ∂u (u0, v0) × ∂Φ ∂v (u0, v0).

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Surface integral of vector fields (cont.)

We now multiply area(∆S) by F(Φ(u0, v0)) · n(u0, v0), where n(u0, v0) is the unit normal to S at Φ(u0, v0). F(Φ(u0, v0)) · n(u0, v0) area(∆S) ≈ F(Φ(u0, v0)) · n(u0, v0) N(u0, v0)∆u∆v ≈ F(Φ(u0, v0)) · N(u0, v0)∆u∆v. Thus, F(Φ(u0, v0)) · n(u0, v0) area(∆S) ≈ F(Φ(u0, v0)) · N(u0, v0)∆u∆v. Let F(x, y, z) (= J(x, y, z)) represent the current density at the point (x, y, z), then the above is the total current passing through ∆S. So, what is the total current passing through the whole surface S?

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Surface integral of vector fields (cont.)

When ∆u → du and ∆v → dv, we have F(Φ(u0, v0)) · n(u0, v0) area(dS) ≈ F(Φ(u0, v0)) · N(u0, v0) dudv. Now, we “sum” over the whole D to obtain

• D

F(Φ(u, v)) · N(u, v) dudv, that is nothing else but the the surface integral of the vector field F over the surface S. In our physical interpretation, the integral would give us the total current passing through the surface S.

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Invariance under re-parameterization

Theorem Suppose that F : R3 → R3 is a continuous vector field and that Φ : D → R3 and ˜ Φ : ˜ D → R3 are alternative C1-parametric representations of a surface S, so that S := {Φ(u, v) | (u, v) ∈ D} ≡

• ˜

Φ(˜ u, ˜ v) | (˜ u, ˜ v) ∈ ˜ D

• .

Then,

• D

F(Φ(u, v))· ∂Φ ∂u (u, v) × ∂Φ ∂v (u, v)

• dudv =

=

• ˜

D

F( ˜ Φ(˜ u, ˜ v)) ·

• ∂ ˜

Φ ∂˜ u (˜ u, ˜ v) × ∂ ˜ Φ ∂˜ v (˜ u, ˜ v)

• d˜

ud˜ v.

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Notations

Some standard notations for the integral are

• S

F(x, y, z) · dA,

• S

F(x, y, z) · dσ,

• S

F(x, y, z) · dS, as well as

• S

F · dA,

• S

F · dσ,

• S

F · dS. The notations are quite explicit, and remind us that we are integ- rating over S with respect to an underlying space variable in R3 generically denoted by (x, y, z).

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Remark

Suppose that S has the special form of the graph of a function f : D → R. Then, we can parametrize S according to: Φ(u, v) = ui + vj + f(u, v)k, in which case ∂Φ ∂u (u, v) × ∂Φ ∂v (u, v) = −∂f ∂u(u, v)i − ∂f ∂v (u, v)j + k. Consequently, in this case we have:

• S

F · dA =

• D
• F3(u, v, f(u, v)) − F1(u, v, f(u, v))∂f

∂u(u, v) − F2(u, v, f(u, v))∂f ∂v (u, v)

• dudv.

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Flux

Definition The surface integral

• S

F · dA is called the flux of the vector field F through the surface S, and has the interpretation of the “aggregate flow” of F through the surface S. When F is identified with the current density J, the surface in- tegral

• S J · dA gives the total current passing through S.

Clearly, this current should not depend in any way on the parti- cular parametric representation we use for the surface S.

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Examples

Example 1: Let the surface S be represented by Φ(u, v) = u2i + uj + vk, ∀(u, v) ∈ D := [0, 2] × [0, 3]. Determine the total current passing through S, when the current density is given by J(x, y, z) = 3z2i + 6j + 6xzk, ∀(x, y, z) ∈ R3. Example 1: For a single point charge Q located at the origin, find its corresponding electric flux, i.e.

• S E · dA, through the

northern hemisphere of radius r. Recall that E(x, y, z) = Q 4πǫ0[x2 + y2 + z2]3/2 (xi + yj + zk).

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Amp` ere’s circuital law

It is known that a current i passing through a conductor causes a magnetic field vector B(x, y, z) at all points (x, y, z) in the space surrounding the conductor. Is there any quantitative relationship between i and B? Suppose that i is a time-constant current flowing through a long and very thin metallic conductor. Let Γ be a fixed simple curve that “loops” just once around the conductor (with the direction of Γ defined by the right hand rule). It has been determined by experiment that

• Γ

B · dr = µ0i, with µ0 being the magnetic permeability of free space. This physical law is known as Amp` ere’s circuital law.

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Amp` ere’s circuital law (cont.)

Despite its generality the circuital law has the disadvantage that the cause of the magnetic field is assumed to be current though a conductor. From the point of view of electromagnetism, it is much more use- ful to have a circuital law in which the cause of the magnetic field is a current density arising from the movement of charge through space.

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Amp` ere’s circuital law (cont.)

Let J : R3 → R3 be a time-constant current density vector field, and let S be an arbitrary (fixed) surface with boundary curve Γ. Note that S is not in any sense a physical surface or barrier that impedes or disturbs the movement of charge described by J. It has been determined by experiment that

• Γ

B · dr = µ0

• S

J · dA. This physical law is also known as Amp` ere’s circuital law. Notice that the above relation between J and B holds for every possible choice of S with boundary Γ. Notice also that, for a given S, the surface integral on the right is nothing but the total current flowing through S.

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Faraday’s law

Let B be a time varying magnetic field, that is at each point (x, y, z) ∈ R3, the magnetic field vector is B(t, x, y, z) for each instant t. In essence Faraday’s law of electromagnetic induction states that a time varying magnetic field B causes a time varying electric field E. Naturally we would like a quantitative or mathematical relation between the time varying fields B and E.

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Faraday’s law (cont.)

Let’s fix some surface S in R3 with boundary curve Γ. Then, the (time-dependent) magnetic flux can be defined as Φmag(t) :=

• S

B · dA, where B = B(t, x, y, z).

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 42/43

Faraday’s law (cont.)

Then, Faraday’s law of electromagnetic induction states that the electric field E caused by the time varying magnetic field B always satisfies the relation

• Γ

E · dr = −∂Φmag(t) ∂t , ∀t. Note that this relation holds regardless of how one chooses the surface S with boundary Γ. Usually, Faraday’s law of electromagnetic induction is written in the following “combined” form

• Γ

E · dr = − ∂ ∂t

• S

B · dA = −

• S

∂B ∂t · dA, which displays the relation between the given time varying mag- netic field B and the resulting time varying electric field E.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 43/43