3.36pt

Line IntegralsMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Spring 2019

MotivationSuppose we wish to find the mass of a piece of wire bent intothe shape of a curve C.

If the density of the wire at (x , y , z) is given by ρ(x , y , z), then ina section of the wire of length ∆s the mass is approximately

∆m = ρ(x , y , z) ∆s.

The mass of the wire is

m ≈n∑

i=1

ρ(xi , yi , zi)∆si .

The exact mass of the wire is

m = lim‖P‖→0

n∑i=1

ρ(xi , yi , zi)∆si

provided the limit exists and is the same for every choice ofevaluation points.

Line Integral

We may adapt this approach to find the integral of otherfunctions defined along curve C.

DefinitionThe line integral of f (x , y , z) with respect to arc length alongthe oriented curve C in three-dimensional space is∫

Cf (x , y , z) ds = lim

‖P‖→0

n∑i=1

f (xi , yi , zi)∆si

provided the limit exists and is the same for every choice ofevaluation points.

Line Integral

We may adapt this approach to find the integral of otherfunctions defined along curve C.

DefinitionThe line integral of f (x , y , z) with respect to arc length alongthe oriented curve C in three-dimensional space is∫

Cf (x , y , z) ds = lim

‖P‖→0

n∑i=1

f (xi , yi , zi)∆si

provided the limit exists and is the same for every choice ofevaluation points.

Evaluation Theorem (1 of 2)

Theorem (Evaluation Theorem)Suppose that f (x , y , z) is continuous in a region D containingthe curve C and that C is described parametrically by(x(t), y(t), z(t)) for a ≤ t ≤ b, where x(t), y(t), and z(t) havecontinuous first derivatives. Then∫

Cf (x , y , z) ds

=

∫ b

af (x(t), y(t), z(t))

√[x ′(t)]2 + [y ′(t)]2 + [z ′(t)]2 dt .

Evaluation Theorem (2 of 2)

Theorem (Evaluation Theorem)Suppose that f (x , y) is continuous in a region D containing thecurve C and that C is described parametrically by (x(t), y(t))for a ≤ t ≤ b, where x(t) and y(t) have continuous firstderivatives. Then∫

Cf (x , y) ds =

∫ b

af (x(t), y(t))

√[x ′(t)]2 + [y ′(t)]2 dt .

Example

Evaluate the line integral ∫C

2x ds

where C is the portion of the parabola y = x2 with endpoints(0,0) and (1,1).

Remark: we must describe the parabola parametrically.

Solution

x(t) = ty(t) = t2

for 0 ≤ t ≤ 1.∫C

2x ds =

∫ 1

02t√

(1)2 + (2t)2 dt

=

∫ 1

02t√

1 + 4t2 dt

=14

∫ 5

1u1/2 du

=16

(5√

5− 1)

Example

A wire takes the shape of a semicircle x2 + y2 = 1 with y ≥ 0.The linear density of the wire at any point is proportional todistance of the point from the line y = 1. Find the mass of thewire.

Solution

ρ(x , y) = k(1− y) (density of wire)x(t) = cos ty(t) = sin t

for 0 ≤ t ≤ π.

Therefore the mass of the wire is

m =

∫Cρ(x , y) ds

=

∫ π

0k(1− sin t)

√(− sin t)2 + (cos t)2 dt

= k∫ π

0(1− sin t) dt

= k(t + cos t)|π0= k(π − 2)

Solution

ρ(x , y) = k(1− y) (density of wire)x(t) = cos ty(t) = sin t

for 0 ≤ t ≤ π.Therefore the mass of the wire is

m =

∫Cρ(x , y) ds

=

∫ π

0k(1− sin t)

√(− sin t)2 + (cos t)2 dt

= k∫ π

0(1− sin t) dt

= k(t + cos t)|π0= k(π − 2)

Arc Length

TheoremFor any piecewise-smooth curve C,∫

C1 ds

gives the arc length of the curve C.

Example

Find the arc length of the helix (sin t , t , cos t) for 0 ≤ t ≤ π.

s =

∫C

1 ds

=

∫ π

0

√(cos t)2 + (1)2 + (− sin t)2 dt

=

∫ π

0

√2 dt

= π√

2

Example

Find the arc length of the helix (sin t , t , cos t) for 0 ≤ t ≤ π.

s =

∫C

1 ds

=

∫ π

0

√(cos t)2 + (1)2 + (− sin t)2 dt

=

∫ π

0

√2 dt

= π√

2

Smoothness

DefinitionWe say that curve C is smooth if C is described parametricallyby (x(t), y(t), z(t)) for a ≤ t ≤ b, where x(t), y(t), and z(t)have continuous first derivatives and

[x ′(t)]2 + [y ′(t)]2 + [z ′(t)]2 6= 0

for a ≤ t ≤ b.

DefinitionSuppose that curve C is

C = C1 ∪ C2 ∪ · · · ∪ Cn

where each of the C1, C2, . . . , Cn is smooth and the terminalpoint of Ci is the initial point of Ci+1 for i = 1,2, . . . ,n − 1. Inthis case we say that C is piecewise-smooth.

Smoothness

DefinitionWe say that curve C is smooth if C is described parametricallyby (x(t), y(t), z(t)) for a ≤ t ≤ b, where x(t), y(t), and z(t)have continuous first derivatives and

[x ′(t)]2 + [y ′(t)]2 + [z ′(t)]2 6= 0

for a ≤ t ≤ b.

DefinitionSuppose that curve C is

C = C1 ∪ C2 ∪ · · · ∪ Cn

where each of the C1, C2, . . . , Cn is smooth and the terminalpoint of Ci is the initial point of Ci+1 for i = 1,2, . . . ,n − 1. Inthis case we say that C is piecewise-smooth.

Result

TheoremSuppose that f (x , y , z) is a continuous function in some regionD containing the oriented curve C. Then if C ispiecewise-smooth with C = C1 ∪ C2 ∪ · · · ∪ Cn, where C1, C2,. . . , Cn are smooth and the terminal point of Ci is the initialpoint of Ci+1 for i = 1,2, . . . ,n − 1, we have∫

−Cf (x , y , z) ds =

∫C

f (x , y , z) ds

and ∫C

f (x , y , z) ds =

∫C1

f (x , y , z) ds +

∫C2

f (x , y , z) ds

+ · · ·+∫

Cn

f (x , y , z) ds.

Example (1 of 3)

Evaluate the line integral ∫C

(x + y) ds

where C is the right-angled path from (1,0) to (1,1) to (0,1).

Example (2 of 3)

. H1,0L

H1,1LH0,1L

C1

C2

-1.0 -0.5 0.5 1.0 x

-1.0

-0.5

0.5

1.0y

C1 :

{x = 1y = t

for 0 ≤ t ≤ 1

C2 :

{x = 1− ty = 1

for 0 ≤ t ≤ 1

Example (3 of 3)

∫C

(x + y) ds =

∫C1

(x + y) ds +

∫C2

(x + y) ds

=

∫ 1

0(1 + t) dt +

∫ 1

0(1− t + 1) dt

=32

+

∫ 1

0(2− t) dt

=32

+32

= 3

Additional Line Integrals (1 of 3)

DefinitionThe line integral of f (x , y , z) with respect to x along theoriented curve C in three-dimensional space is∫

Cf (x , y , z) dx = lim

‖P‖→0

n∑i=1

f (xi , yi , zi)∆xi

provided the limit exists and is the same for every choice ofevaluation points.

Additional Line Integrals (2 of 3)

DefinitionThe line integral of f (x , y , z) with respect to y along theoriented curve C in three-dimensional space is∫

Cf (x , y , z) dy = lim

‖P‖→0

n∑i=1

f (xi , yi , zi)∆yi

provided the limit exists and is the same for every choice ofevaluation points.

Additional Line Integrals (3 of 3)

DefinitionThe line integral of f (x , y , z) with respect to z along theoriented curve C in three-dimensional space is∫

Cf (x , y , z) dz = lim

‖P‖→0

n∑i=1

f (xi , yi , zi)∆zi

provided the limit exists and is the same for every choice ofevaluation points.

Evaluation Theorem

Theorem (Evaluation Theorem)Suppose that f (x , y , z) is continuous in a region D containingthe curve C and that C is described parametrically by(x(t), y(t), z(t)) for a ≤ t ≤ b, where x(t), y(t), and z(t) havecontinuous first derivatives. Then∫

Cf (x , y , z) dx =

∫ b

af (x(t), y(t), z(t))x ′(t) dt ,∫

Cf (x , y , z) dy =

∫ b

af (x(t), y(t), z(t))y ′(t) dt ,∫

Cf (x , y , z) dz =

∫ b

af (x(t), y(t), z(t))z ′(t) dt .

Examples

Compute the following line integrals over the path C which isthe portion of the parabola x = y2 from (1,1) to (1,−1).

I∫

C(x + y) dx

I∫

C(x + y) dy

I∫

C(x + y) ds

Illustration of Curve C

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0y

Solution

The curve C is parameterized for −1 ≤ t ≤ 1 as

x = t2 y = −t

and then∫C

(x + y) dx =

∫ 1

−1(t2 − t)2t dt = −4

3∫C

(x + y) dy =

∫ 1

−1(t2 − t)(−1) dt = −2

3∫C

(x + y) ds =

∫ 1

−1(t2 − t)

√(2t)2 + (−1)2 dt ≈ 1.21267

Result

TheoremSuppose that f (x , y , z) is a continuous function in some regionD containing the oriented curve C. Then, the following hold.

1. If C is piecewise-smooth, then∫−C

f (x , y , z) dx = −∫

Cf (x , y , z) dx .

2. If C = C1 ∪C2 ∪ · · · ∪Cn, where C1, C2, . . . , Cn are smoothand the terminal point of Ci is the initial point of Ci+1 fori = 1,2, . . . ,n − 1, then∫

Cf (x , y , z) dx =

n∑i=1

∫Ci

f (x , y , z) dx

Remark: similar results hold for integrals with respect to y andz.

Example

Calculate the line integral∫C

2x dy + 4y dx

where C consists of the line segment from (0,0) to (1,0)followed by the quarter circle to (0,1) followed by the linesegment to (0,0).

Parameterization of C

. H1,0LH0,0L

H0,1L

C1

C2C3

-1.0 -0.5 0.5 1.0 x

-1.0

-0.5

0.5

1.0y

C1 :

{x = ty = 0

for 0 ≤ t ≤ 1

C2 :

{x = cos ty = sin t

for 0 ≤ t ≤ π2

C3 :

{x = 0y = 1− t

for 0 ≤ t ≤ 1

Solution

∫C

2x dy + 4y dx =

∫C1

2x dy + 4y dx +

∫C2

2x dy + 4y dx

+

∫C3

2x dy + 4y dx

=

∫C1

2x dy + 4y dx︸ ︷︷ ︸=0

+

∫C2

2x dy + 4y dx

+

∫C3

2x dy + 4y dx︸ ︷︷ ︸=0

=

∫C2

2x dy + 4y dx

=

∫ π/2

0(2 cos2 t − 4 sin2 t) dt

= −π2

WorkSuppose F(x , y , z) = 〈f1(x , y , z), f2(x , y , z), f3(x , y , z)〉represents a vector field function representing the force presenton an object at location (x , y , z).

The work done in moving the object along curve Cparametrized by (x(t), y(t), z(t)) where a ≤ t ≤ b is

W =

∫ b

af1(x , y , z)x ′(t) dt +

∫ b

af2(x , y , z)y ′(t) dt

+

∫ b

af3(x , y , z)z ′(t) dt

=

∫C

f1(x , y , z) dx +

∫C

f2(x , y , z) dy +

∫C

f3(x , y , z) dz

=

∫C〈f1(x , y , z), f2(x , y , z), f3(x , y , z)〉 · 〈dx ,dy ,dz〉

=

∫C

F(x , y , z) · dr

Example

Compute the work done by the force fieldF(x , y , z) = 〈xy ,3z,1〉 along the quarter ellipse parametrizedby

x = 2 cos ty = 3 sin tz = 1

from (2,0,1) to (0,3,1).

Illustration of Vector Field and Curve

. 0 1 2 3x

0 1 2 3y

0.5

1.0

1.5

z

Calculation of Work

W =

∫C

F(x , y , z) · dr

=

∫ π/2

0〈6 cos t sin t ,3,1〉 · 〈−2 sin t ,3 cos t ,0〉dt

=

∫ π/2

09 cos t − 12 cos t sin2 t dt

=

∫ π/2

09 cos t dt − 12

∫ π/2

0cos t sin2 t dt

= 9− 12∫ 1

0u2 du

= 5

Remarks

If the orientation of the curve is generally in the same directionas the vector field, the force adds energy to the object and thusdoes positive work.

If the orientation of the curve is generally in the directionopposite the vector field, the force opposes the motion of theobject and thus does negative work.

Homework

I Read Section 14.2.I Exercises: 1–41 odd.