MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora - - PowerPoint PPT Presentation

Line integral and Green’s Theorem Path independence of line integral

MA102: Multivariable Calculus

Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example:

Evaluate

• V 2xdV where V is the region bounded by

the planes x = 0, y = 0, z = 0 and 2x + 3y + z = 6. Note that V is Type-I: 0 ≤ z ≤ 6 − 2x − 3y and (x, y) ∈ D, where D is special domain given by 0 ≤ x ≤ 3 and 0 ≤ y ≤ −2 3x + 2. Thus

• V

2xdV =

• D

( 6−2x−3y 2x dz)dA = 3 − 2

3 x+2

(6 − 2x − 3y)2xdydx = 9.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example:

Find the volume of the region bounded by the surfaces z = x2 + 3y2 and z = 8 − x2 − y2. The volume is V =

• Ω dzdydx, where Ω is bounded

above by the surface z = 8 − x2 − y2 and below by the surface z = x2 + 3y2. Therefore, the limits of z are from z = x2 + 3y2 to z = 8 − x2 − y2. The projection of Ω on xy-plane is the solution of 8 − x2 − y2 = x2 + 3y2 = ⇒ x2 + 2y2 = 4. Therefore the limits of x and y are to be determined by R : x2 + 2y 2 = 4.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example (cont.):

V =

• R

8−x2−y2

z=x2+3y2 dzdA

= 2

−2

√

(4−x2)/2 −√ (4−x2)/2

(8 − 2x2 − 4y2)dydx = 2

−2

• (8 − x2)y − 4

3y3 √

(4−x2)/2 y=−√ (4−x2)/2

= 4 √ 2 3 2

−2

(4 − x2)3/2dx = 8π √ 2.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example: Change of order of integration

Consider the evaluation of integral

• R

sin x x dA over the

triangle formed by y = 0, x = 1 and y = x. If we write R = {(x, y) : 0 ≤ y ≤ 1, y ≤ x ≤ 1}, then

• R

sin x x dA = 1 1

x=y

sin x x dx

• dy.

The innermost integral is difficult to evaluate. If we change the order of integration, by taking R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x}, then

• R

sin x x dA = 1 x

y=0

sin x x dydx = 1 sin xdx = 1−cos 1.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example: Change of order of integration

Find the volume of the region bounded by x + z = 1, y + 2z = 2 in the first quadrant.

Figure : Region bounded by x + z = 1 and y + 2z = 2, x, y, z ≥ 0

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example: Change of order of integration

Draw line parallel to z-axis and note that the upper surfaces are: 2z + y = 2 over triangle bounded by x = 0, y = 1y = 2x and z = 1 − x over the triangle bounded by y = 0, x = 1, y = 2x. Therefore, V = 2

y=0

y/2

x=0

• 2−y

2

z=0

dz dx dy+ 1

x=0

2x

y=0

1−x

z=0

dz dy dx = 2 3 On the other hand, by first drawing the line parallel to x-axis, we get V = 1

z=0

2−2z

y=0

1−z

x=0

dx dy dz = 2 3

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example (cont.):

Taking the line parallel to y-axis we get V = 1

x=0

1−x

z=0

2−2z

y=0

dy dz dx = 2 3 Example: Evaluate I = 4

z=0

1

y=0

2

x=2y

2 cos(x2) √z dx dy dz. The integral is difficult to evaluate in the given order of

• integration. We change the order of integration and

evaluate the integral: I = 4

z=0

2

x=0

x/2

y=0

2 cos(x2) √z dy dx dz. = 4

z=0

2

x=0

x cos(x2) √z dx dz = 2 sin 4.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Change of variable

Let T : R2 → R2 be C1 given by T(u, v) = (x(u, v), y(u, v)). Then the Jacobian matrix J(u, v) of T is given by J(u, v) :=

• xu

xv yu yv

• .

Define the Jacobian of T by ∂(x, y) ∂(u, v) := xuyv − xvyu = det J(u, v). Polar coordinates: Define T(r, θ) := (r cos θ, r sin θ). Then ∂(x, y) ∂(r, θ) =

• cos θ

−r sin θ sin θ r cos θ

• = r.
• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Change of variable

Let T : R2 → R2 be C1 given by T(u, v) = (x(u, v), y(u, v)). Let T −1 : R2 → R2 be the inverse given by T −1(x, y) = (u(x, y), v(x, y)). Let J(u, v) and J′(x, y) be the Jacobian matrices of T and T −1. Applying Chain rule, we have 1 = xu ux + xv vx ; 0 = xu uy + xv vy = yu ux + yv vx ; 1 = yu uy + yv vy. Thus, J′(x, y)·J(u, v) =

• 1

1

• ⇒ det(J′(x, y)) =

1 det(J(u, v)).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Change of variable for double integrals

Suppose T is injective and J(u, v) is nonsingular. Let D ⊂ R2 and G := T(D). Suppose that f is integrable on G. Then dA = dxdy =

• ∂(x, y)

∂(u, v)

• dudv

and

• G

f(x, y)dxdy =

• D

f(x(u, v), y(u, v))

• ∂(x, y)

∂(u, v)

• dudv.

Polar coordinates:

• G

f(x, y)dxdy =

• D

f(r cos θ, r sin θ)rdrdθ.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Evaluate the integral I = 1 1−x √x + y(y − 2x)2dA. The given domain is the triangle bounded by x = 0, y = 0 and x + y = 1. We take the transformation u = x + y and v = y − 2x. Under this transformation, the given triangle will be transformed into triangle bounded by v = u, v = −2u and u = 1. The inverse of this transformation is x = u−v

3

and y = 2u+v

3 . Hence the Jacobian

J =

• 1/3

−1/3 2/3 1/3

• = 1/3.

Hence I = 1 3 1 u

v=−2u

√ uv 2dv du

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Using the transformation u = 2x + 3y and v = x − 3y, find the value of the integral I =

• R

e2x+3y cos(x − 3y) dx dy where R is the region bounded by the parallelogram with vertices (0, 0), (1, 1/3), (4/3, 1/9), (1/3, −2/9). Under the given transformations, R will be transformed into the rectangle with vertices (0, 0), (3, 0), (3, 1) and (0, 1). Also, |J| = 1

• 9. Thus,

I = 1 9 1

v=0

3

u=0

eu cos v du dv = 1 9(e3 − 1) sin 1.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Evaluate

• G

√ x2 + z2dV where G is the region bounded by the paraboloid y = x2 + z2 and y = 4. We have

• G

f(x, y, z)dV =

• D

4

x2+z2

• x2 + z2dy
• dxdz,

where D = {(x, z) : x2 + z2 ≤ 4}. Setting x = r cos θ and z = r sin θ for (r, θ) ∈ [0, 2] × [0, 2π],

• G

f(x, y, z)dV = 2π 2 r(4 − r 2)rdrdθ = 128π 5 .

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Change of variable for multiple integrals

Let D ⊂ Rn be open and bounded. Let T : D → Rn be such that T is C1, injective and the Jacobian J(U) is nonsingular for U ∈ D. Let G := T(D) and f : G → R be integrable over G. Then dx1 · · · dxn =

• ∂(x1, · · · , xn)

∂(u1, · · · , un)

• du1 · · · dun

and

• G

f(X)dx1 · · · dxn =

• D

f(X(U))

• ∂(x1, · · · , xn)

∂(u1, · · · , un)

• du1 · · · dun

=

• D

f(X(U))

• dX

dU

• dU.
• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Cylindrical coordinates

Consider T(r, θ, z) = (r cos θ, r sin θ, z). Then ∂(x, y, z) ∂(r, θ, z) =

• cos θ

−r sin θ sin θ r cos θ 1

• = r.

Thus dV = rdrdθdz and

• G

f(x, y, z)dV =

• D

f(r cos θ, r sin θ, z)rdrdθdz.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Evaluate

• G
• x2 + y2dV, where G is the region

bounded by x2 + y 2 = 1, z = 4 and z = 1 − x2 − y2. Consider cylindrical coordinates D := {(r, θ, z) : (r, θ) ∈ [0, 1] × [0, 2π], 1 − r 2 ≤ z ≤ 4}. Then

• G

f(x, y, z)dV = 1 2π 4

1−r 2 dz

• r rdrdθ = 12π

5 .

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Spherical coordinates

Consider T(ρ, φ, θ) = (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ). Then ∂(x, y, z) ∂(ρ, φ, θ) =

• sin φ cos θ

ρ cos φ cos θ −ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ cos φ −ρ sin φ

• =

ρ2 sin φ. Thus dV = ρ2 sin φdρdφdθ and

• G

f(x, y, z)dV =

• D

f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φdρdφdθ.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Evaluate

• G e(x2+y2+z2)3/2dV, where

G := {(x, y, z) : x2 + y2 + z2 ≤ 1}. Using spherical coordinates we have

• D

f(x, y, z)dV = 2π π 1 eρ3ρ2 sin φdρdφdθ = 4 3π(e − 1).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Line Integral: Partition of curves

Let Γ be a curve in Rn paramatrized by r : [a, b] → Rn. Then a partition P := (a = t0 < . . . < tm = b) of [a, b] induces a partition of Γ into m subarcs with arclengths ∆s1, . . . , ∆sm. Define µ(P) := max

1≤j≤m ∆sj.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Riemann sum w.r.t. arclength

Let f : Γ → R. Then for any Pj in the j-th subarc, consider the Riemann sum of f w.r.t. to the partition P S(P, f) :=

m

• j=1

f(Pj)∆sj.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Line integral

Definition: Suppose that Γ is a piecewise smooth curve in Rn paramatrized by r : [a, b] → Rn and f : Γ → R. Then the line integral of f along Γ is defined by

• Γ

f ds := lim

µ(P)→0 S(P, f) =

lim

µ(P)→0 m

• j=1

f(Pj)∆sj if the limit exists (independent of the partitions P and the chosen points Pj).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Line integral

Definition: Suppose that Γ is a piecewise smooth curve in Rn paramatrized by r : [a, b] → Rn and f : Γ → R. Then the line integral of f along Γ is defined by

• Γ

f ds := lim

µ(P)→0 S(P, f) =

lim

µ(P)→0 m

• j=1

f(Pj)∆sj if the limit exists (independent of the partitions P and the chosen points Pj). Fact: If f is continuous and r(t) is piecewise smooth then we have

• Γ

f ds = b

a

f(r(t))r ′(t)dt.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Line integral

For the plane curve Γ : r(t) = (x(t), y(t)), t ∈ [a, b] we have

• Γ

f(x, y)ds = b

a

f(x(t), y(t))

• x′(t)2 + y′(t)2 dt.
• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Line integral

For the plane curve Γ : r(t) = (x(t), y(t)), t ∈ [a, b] we have

• Γ

f(x, y)ds = b

a

f(x(t), y(t))

• x′(t)2 + y′(t)2 dt.

Example: Evaluate

• Γ(2 + x2y)ds, where Γ is the upper

half of the circle x2 + y2 = 1. Considering x(t) = cos t, y(t) = sin t, 0 ≤ t ≤ π, we have

• Γ

(2 + x2y)ds = π (2 + cos2 t sin t)dt = 2π + 2/3.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Properties of line integrals

Fact: Let Γ be parametrized by a piecewise smooth curve r : [a, b] → Rn and f, g : Γ → R be continuous. Then the following hold:

Γ(f + αg)ds =

• Γ fds + α
• Γ gds for α ∈ R.
• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Properties of line integrals

Fact: Let Γ be parametrized by a piecewise smooth curve r : [a, b] → Rn and f, g : Γ → R be continuous. Then the following hold:

Γ(f + αg)ds =

• Γ fds + α
• Γ gds for α ∈ R.
• Let Γ = Γ1 + · · · + Γm, where Γi is parametrized by

smooth curve ri : [ai, bi] → Rn. Then

• Γ

fds =

• Γ1

fds + · · · +

• Γm

fds.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Evaluate

• Γ 2xds, where Γ consists of the arc C1 of the

parabola y = x2 from (0, 0) to (1, 1) followed by the line segment C2 from (1, 1) to (1, 2). Then

• Γ

2xds =

• C1

2xds +

• C2

2xds = 1 6(5 √ 5 + 11).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Application

Suppose a thin wire in the shape of a curve Γ parametrized by a smooth path r : [a, b] → R2 has density ρ(x, y). Then the total mass of the wire is given by m =

• Γ

ρ(x, y)ds The center of mass (x, y) is given by x = 1 m

• Γ

xρ(x, y)ds and y = 1 m

• Γ

yρ(x, y)ds

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Application: Example

The wire W has the shape Γ = Γ1 + Γ2, where Γ1 is parametrized by γ1(t) := (cos t, sin t), t ∈ [0, π] and Γ2 is parametrized by γ2(t) := (t, 0), t ∈ [−1, 1]. Let the density function be given by ρ(x, y) =

• x2 + y2.

We have γ′

i(t) = 1, ρ(γ1(t)) = 1 and ρ(γ2(t)) = |t|.

m =

• Γ

ρ(x, y)ds = π dt −

−1

t dt + 1 t dt = π + 1. The center of mass (x, y) is given by x = 0 and y = 2 1 + π.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector line integral

Definition: Let Γ be a curve in Rn parametrized by a piecewise smooth path r : [a, b] → Rn and let F be a continuous function on an open set containing Γ to Rn. Then the line integral of F over Γ is defined by

• Γ

F • dr := b

a

F(r(t)) • r ′(t)dt = b

a

F(r(t)), r ′(t)dt.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector line integral

Definition: Let Γ be a curve in Rn parametrized by a piecewise smooth path r : [a, b] → Rn and let F be a continuous function on an open set containing Γ to Rn. Then the line integral of F over Γ is defined by

• Γ

F • dr := b

a

F(r(t)) • r ′(t)dt = b

a

F(r(t)), r ′(t)dt. Note that [a, b] − → R, t − → F(r(t)) • r ′(t) is piecewise continuous and hence Riemann integrable.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector line integrals and scalar line integrals

Suppose that Γ is (piecewise) smooth parametrized by r. Then r ′(t) = 0. Define the tangent vector T(r(t)) :=

r ′(t) r ′(t) to Γ at r(t).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector line integrals and scalar line integrals

Suppose that Γ is (piecewise) smooth parametrized by r. Then r ′(t) = 0. Define the tangent vector T(r(t)) :=

r ′(t) r ′(t) to Γ at r(t).

Then F • T is the tangential component of F and

• Γ

F • dr = b

a

F(r(t)) • r ′(t)dt = b

a

F(r(t)) • T(r(t))r ′(t)dt =

• Γ

F • Tds =

• Γ

F, Tds.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector line integrals and scalar line integrals

Suppose that Γ is (piecewise) smooth parametrized by r. Then r ′(t) = 0. Define the tangent vector T(r(t)) :=

r ′(t) r ′(t) to Γ at r(t).

Then F • T is the tangential component of F and

• Γ

F • dr = b

a

F(r(t)) • r ′(t)dt = b

a

F(r(t)) • T(r(t))r ′(t)dt =

• Γ

F • Tds =

• Γ

F, Tds. Vector line integral of F = line integral of F • T.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Notations for vector line integrals

• When Γ is closed, that is, r(a) = r(b), the line integral
• Γ

F • dr is denoted by

• Γ

F • dr.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Notations for vector line integrals

• When Γ is closed, that is, r(a) = r(b), the line integral
• Γ

F • dr is denoted by

• Γ

F • dr.

• When n = 2 and F = (P, Q) the line integral is written as
• Γ

F • dr =

• Γ

(P(x, y)dx + Q(x, y)dy).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Notations for vector line integrals

• When Γ is closed, that is, r(a) = r(b), the line integral
• Γ

F • dr is denoted by

• Γ

F • dr.

• When n = 2 and F = (P, Q) the line integral is written as
• Γ

F • dr =

• Γ

(P(x, y)dx + Q(x, y)dy).

• For n = 3 and F = (P, Q, R) the line integral is written as
• Γ

F • dr =

• Γ

(P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Examples

• Evaluate
• Γ F • dr, where F(x, y, z) := (xy, yz, zx) and

r(t) := (t, t2, t2), t ∈ [0, 1]. We have

• Γ

F•dr = 1 F(r(t))•r ′(t)dt = 1 (t3+2t5+2t4)dt = 59 60.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Examples

• Evaluate
• Γ F • dr, where F(x, y, z) := (xy, yz, zx) and

r(t) := (t, t2, t2), t ∈ [0, 1]. We have

• Γ

F•dr = 1 F(r(t))•r ′(t)dt = 1 (t3+2t5+2t4)dt = 59 60.

• Evaluate
• Γ(yx2dx + sin(πy)dy), where Γ is the line

segment from (0, 2) to (1, 4). We have r(t) = (t, 2 + 2t), t ∈ [0, 1]. Thus

• Γ

(yx2dx + sin(πy)dy) = = 1 2 sin(π(2 + 2t))dt + 1 (2 + 2t)t2dt = 7 6

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Oriented path

• A parametrization r : [a, b] → Rn determines an
• rientation or a direction of the curve Γ = r([a, b]).

Indeed, as t varies from a to b, r(t) traverses the path from r(a) to r(b).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Oriented path

• A parametrization r : [a, b] → Rn determines an
• rientation or a direction of the curve Γ = r([a, b]).

Indeed, as t varies from a to b, r(t) traverses the path from r(a) to r(b).

• Let Γ be an oriented path. Denote the reverse
• rientation of Γ by −Γ. If r : [a, b] → Rn is a

parametrization of the oriented path Γ then ρ : [a, b] → Rn given by ρ(t) := r(a + b − t) is a parametrization of −Γ.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Oriented path

• A parametrization r : [a, b] → Rn determines an
• rientation or a direction of the curve Γ = r([a, b]).

Indeed, as t varies from a to b, r(t) traverses the path from r(a) to r(b).

• Let Γ be an oriented path. Denote the reverse
• rientation of Γ by −Γ. If r : [a, b] → Rn is a

parametrization of the oriented path Γ then ρ : [a, b] → Rn given by ρ(t) := r(a + b − t) is a parametrization of −Γ.

• Let Γ be an oriented path. Then
• −Γ

F • dr = −

• Γ

F • dr.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Work done

Definition: The work done by a force F on a particle traversing an oriented path Γ is the line integral Work =

• Γ

F • dr.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Work done

Definition: The work done by a force F on a particle traversing an oriented path Γ is the line integral Work =

• Γ

F • dr. Remark: The total work done by F on a particle traversing the path Γ and then reversing back to the initial point is W =

• Γ

F • dr +

• −Γ

F • dr =

• Γ

F • dr −

• Γ

F • dr = 0.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Green’s Theorem

Let D ⊂ R2 be a simply connected (no holes) region with positively oriented (counter clockwise direction) boundary ∂D. Let F = (P, Q) be C1 on D. Then

• D

(∂Q ∂x − ∂P ∂y )dA =

• ∂D

(P(x, y)dx + Q(x, y)dy) =

• ∂D

F • dr. Let C be a circle of radius a centered at the origin. Find

• C F • dr for F = (−y, x) using Green’s theorem.
• C

F • dr =

• D

2dA = 2

• D

dA = 2πa2.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Applications of Green’s Theorem

• Evaluation of area

Area(D) =

• D

dA = 1 2

• ∂D

(xdy − ydx). Taking F = (0, x) and G = (y, 0), we have

• D

dA =

• D

(∂Q ∂x − ∂P ∂y )dA =

• ∂D

xdy −

• D

dA =

• D

(∂Q ∂x − ∂P ∂y )dA =

• ∂D

ydx

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector fields

A vector field in Rn is a function F : Rn → Rn that assigns to each X ∈ Rn a vector F(X). A vector field in Rn with domain U ⊂ Rn is called a vector field on U.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector fields

A vector field in Rn is a function F : Rn → Rn that assigns to each X ∈ Rn a vector F(X). A vector field in Rn with domain U ⊂ Rn is called a vector field on U. Geometrically, a vector field F on U is interpreted as attaching a vector to each point of U. Thus, there is a subtle difference between a vector field in Rn and a function from Rn to Rn.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Vector fields

A vector field in Rn is a function F : Rn → Rn that assigns to each X ∈ Rn a vector F(X). A vector field in Rn with domain U ⊂ Rn is called a vector field on U. Geometrically, a vector field F on U is interpreted as attaching a vector to each point of U. Thus, there is a subtle difference between a vector field in Rn and a function from Rn to Rn. When a function F : Rn → Rn is viewed as a vector field, for each X the vector F(X) is identified with the vector that starts at the point X with the magnitude and direction

• f F(X).

Thus every vector field on U ⊂ Rn is uniquely determined by a function from U → Rn.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Examples of vector fields

• The gravitational force field describes the force of

attraction of the earth on a mass m and is given by F(X) = −mMG r 3 X, where X = (x, y, z), r := X. The vector field F points to the centre of the earth.

• The vector field F : R2 → R2 given by

F(x, y) := (−y, x) is a rotational vector field in R2 which rotates a vector in the anti-clockwise direction by an angle π/2.

• Let r : [0, 1] → Rn be C1 and Γ := r([0, 1]). Then

F : Γ ⊂ Rn → Rn given by F(r(t)) = r ′(t) is a tangent vector field on Γ.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Figure : Examples of vector fields

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Figure : Vector field representing Hurricane Katrina

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Integral Curves of a Vector Field

Let F be a vector field on domain D. An oriented smooth curve C : r(t) for t ∈ [a, b] in the domain D is called an integral curve of F, if, at every point P on C,

• F(P) = 0,
• F(r(t)) = r ′(t) for t ∈ [a, b].

Obviously, F is a tangent (velocity) vector field on the integral curve. Thus, integral curves provide a geometric picture of a vector field. In a force field, integral curves are called lines of force. In a fluid flow, integral curves of the velocity field are called streamlines or flow-lines.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Finding Integral Curves of a Vector Field

For a vector field F(x, y) = M(x, y) ˆ ı + N(x, y) ˆ  , the integral curve at X = (x, y) is the solution of the differential equation dy dx = N(x, y) M(x, y) at the point X. For a vector field F(x, y, z) = M(x, y, z)ˆ ı + N(x, y, z) ˆ  + P(x, y, z), an integral curve may be viewed as the solution of a system of ODEs x′(t) = M(x, y, z) , y′(t) = N(x, y, z) , z′(t) = P(x, y, z) .

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Examples of Vector Fields and their Integral Curves

Consider the vector field F(x, y) = −y ˆ ı + x ˆ  for (x, y) ∈ R2. At any point P = (x, y) with y = 0, we solve the differential equation dy dx = −x y Its solution is x2 + y2 = a2. These circles x2 + y2 = a2 are the integral curves of F.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

More example

Consider the vector field G(x, y) = y ˆ ı + x ˆ  for (x, y) ∈ R2. At any point P = (x, y) with y = 0, we solve the differential equation dy dx = x y Its solution is y2 − x2 = a. These hyperbolas y2 − x2 = a are the integral curves of G.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Path Independence

Let F : D ⊆ Rn → Rn be a continuous vector field on D. We say that the vector field F has independence of path

• n D if for every pair of piecewise smooth, oriented

curves C1 and C2 in D with a common initial point and a common final point, we have

• C1

F • dr =

• C2

F • dr. Fact: Let F : D ⊆ Rn → Rn be a continuous vector field on

• D. The vector field F has independence of path on D if

and only if the vector line integral

• C

F • dr = 0 for every piecewise smooth, oriented, closed curve C in D.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Important Example

Let F(x, y) = −y ˆ ı + x ˆ  x2 + y2 for (x, y) ∈ D∗ = R2 \ {(0, 0)}. Let C : r(t) = (cos t, sin t) for t ∈ [0, 2π]. Then r ′(t) = (− sin t, cos t) for t ∈ [0, 2π].

• C

F • dr = 2π

t=0

F(r(t)) • r ′(t) dt = 2π

t=0

(sin2 t + cos2 t) dt = 2π

t=0

dt = 2π = 0 So, F is NOT independent of path in D∗.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Gradient vector fields

If f : Rn → R is a C1 scalar field then ∇f : Rn → Rn is a vector field in Rn.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Gradient vector fields

If f : Rn → R is a C1 scalar field then ∇f : Rn → Rn is a vector field in Rn.

• A vector field F in Rn is said to be a gradient vector

field or a conservative vector field if there is a scalar field f : Rn → R such that F = ∇f. In such a case, f is called a scalar potential of the vector field F.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Path independence and gradient vector field

Let F be a continuous vector field on an open set U ⊂ Rn. Consider the following statements:

• 1. F is a gradient vector field on U.

2.

• Γ F • dr is path independent in U.

3.

• Γ F • dr = 0 for any closed path in U.
• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Path independence and gradient vector field

Let F be a continuous vector field on an open set U ⊂ Rn. Consider the following statements:

• 1. F is a gradient vector field on U.

2.

• Γ F • dr is path independent in U.

3.

• Γ F • dr = 0 for any closed path in U.

We also know that (2) ⇔ (3). The implication (3) ⇒ (1) holds under a suitable assumption on U. This is called the first fundamental theorem for line integrals.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Path independence implies gradient field

Definition: A subset U ⊂ Rn is said to be path connected if for any two points X and Y in U there is a path γ : [a, b] → Rn such that γ(a) = X, γ(b) = Y and γ([a, b]) ⊂ U.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Path independence implies gradient field

Definition: A subset U ⊂ Rn is said to be path connected if for any two points X and Y in U there is a path γ : [a, b] → Rn such that γ(a) = X, γ(b) = Y and γ([a, b]) ⊂ U. Theorem (1st Fundamental Thm for line integral): Let U ⊂ Rn be open and path connected and F be a continuous vector field on U. Suppose

• Γ F • dr is independent of Γ for any PC1

path Γ in U. Then there exists a C1 function f : U → R such that F = ∇f.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Path independence implies gradient field

Definition: A subset U ⊂ Rn is said to be path connected if for any two points X and Y in U there is a path γ : [a, b] → Rn such that γ(a) = X, γ(b) = Y and γ([a, b]) ⊂ U. Theorem (1st Fundamental Thm for line integral): Let U ⊂ Rn be open and path connected and F be a continuous vector field on U. Suppose

• Γ F • dr is independent of Γ for any PC1

path Γ in U. Then there exists a C1 function f : U → R such that F = ∇f. Further, for X0 ∈ U, define f : U → R by f(X) := X

X0

F • dr where the integral is taken over any PC1 path joining X0 to X. Then f is well defined, f is C1 and F = ∇f.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

2nd Fundamental Theorem for line integrals

If f : [a, b] → R is C1 then by FTI b

a f ′(x)dx = f(b) − f(a).

Theorem: Let U ⊂ Rn be open and f : U → R be C1. Let r : [a, b] → Rn be PC1 such that r([a, b]) ⊂ U. Then

• Γ

∇f • dr = f(r(b)) − f(r(a)).

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

2nd Fundamental Theorem for line integrals

If f : [a, b] → R is C1 then by FTI b

a f ′(x)dx = f(b) − f(a).

Theorem: Let U ⊂ Rn be open and f : U → R be C1. Let r : [a, b] → Rn be PC1 such that r([a, b]) ⊂ U. Then

• Γ

∇f • dr = f(r(b)) − f(r(a)). Proof: We have

• Γ

∇f • dr = b

a

∇f(r(t)) • r ′(t)dt = b

a

d dt f(r(t))dt = f(r(b)) − f(r(a)). Thus, gradient field implies path independence.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Gradient vector fields and path independence

In summary, we have the following necessary and sufficient condition for a continuous vector field to be gradient vector field. Let U ⊂ Rn be open and path connected and F be a continuous vector field on U. Then, F is a gradient vector field if and only if F has the path independence property in U. We now find a necessary and sufficient condition for continuously differentiable vector field to be gradient vector field.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Necessary condition

Let F be a vector field on U with a scalar potential f, that is, F = ∇f. Suppose F = (F1, . . . , Fn). Fact: If a C1 vector field F = (F1, . . . , Fn) on U is conservative then for all i and j ∂Fi ∂xj = ∂Fj ∂xi .

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Necessary condition

Let F be a vector field on U with a scalar potential f, that is, F = ∇f. Suppose F = (F1, . . . , Fn). Fact: If a C1 vector field F = (F1, . . . , Fn) on U is conservative then for all i and j ∂Fi ∂xj = ∂Fj ∂xi . Proof: We have Fi = ∂if ⇒ ∂jFi = ∂j∂if = ∂i∂jf = ∂iFj.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Consider F(x, y) := (3 + 2xy, x2 − 3y2) =: (P, Q). Then Qx = 2x = Py so the necessary condition is satisfied. We wish to find f such that F = ∇f. If f exists then fx(x, y) = 3 + 2xy ⇒ f(x, y) = 3x + x2y + h(y). Thus fy(x, y) = x2 + h′(y) = x2 − 3y2 ⇒ h′(y) = −3y2. Hence h(y) = −y3 + c for some constant c. Consequently, f(x, y) = 3x + x2y − y3 + c and F = ∇f.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Consider F(x, y) := (

−y x2+y2, x x2+y2) = (P, Q) for

(x, y) = (0, 0). Then we have Qx = Py so the necessary condition is satisfied.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Consider F(x, y) := (

−y x2+y2, x x2+y2) = (P, Q) for

(x, y) = (0, 0). Then we have Qx = Py so the necessary condition is satisfied. For the path Γ : r(t) = (cos t, sin t), t ∈ [0, 2π], we have

• Γ

F • dr = 2π dt = 2π. This shows that F is not conservative.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Example

Consider F(x, y) := (

−y x2+y2, x x2+y2) = (P, Q) for

(x, y) = (0, 0). Then we have Qx = Py so the necessary condition is satisfied. For the path Γ : r(t) = (cos t, sin t), t ∈ [0, 2π], we have

• Γ

F • dr = 2π dt = 2π. This shows that F is not conservative. Remark: The necessary condition ∂iFj = ∂jFi is also sufficient for conservativeness of F when the domain of F is simply connected. This is a consequence of Green’s theorem.

• R. Barman & S. Bora

MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral

Necessary and sufficient condition for C1 vector field

Let F(x, y) := (P, Q) be C1 defined in a simply connected domain U. Then F is a gradient field (conservative field) if and only if Qx = Py. Proof: We have alrady proved that Qx = Py is a necessary

• condition. To prove that this condition is sufficient, we

apply Green’s theorem. Let C be a closed path (positively

• riented) in U. Let D be the region bounded by C. Since

U is simply connected, so D ⊆ U. By Green’s theorem,

• C

F • dr =

• D

(∂Q ∂x − ∂P ∂y )dA = 0. Thus, F is conservative.

• R. Barman & S. Bora

MA-102 (2017)