18.1 the Idea of a Line Integral

Line Integrals

18.1 The Idea of a Line Integral

As the integral dx integrates over the line y = 0, the line integral ds is a single integral that integrates over an oriented curve or path c instead than over a line, where ds is the element of arc length. The integral represents, graphically, the area under the curve y = f (x)above the line y = 0. The line integral represents, graphically, the area of a curtain formed by the projection of path c in the x-y plane on the surface z = f (x, y).

A line integral can be expressed as ∫c f (x, y)dx; ∫c f (x, y)dy; ∫c f (x, y)ds in or ∫c f (x, y, z)dx;

∫c f (x, y, z)dy; ∫c f (x, y, z)dz; ∫c f (x, y, z)ds in ; with c a piecewise smooth oriented function.

The line integral of the Vector Field along an oriented curve c is given by

, where the direction of is in the direction of the orientation.

The lineintegral measures the extent to which the path c is going with the field or against the field.

eg 1 Find the line integral of the vector field = < – 2, 3 > along the path given by the linear segment from (1, 0) to (3, 0).

Since = < – 2, 3 >, , and , = .

eg 2 Find the line integral of the vector field = < x, y > along the path given by the linear segments from (0,0) to (1,1) and from (1,1) to (1, – 1).

Let c1 be the linear segment from (0,0) to (1, 1). Since = < x, y >, = < x, y > and d = < dx, dy >, =x, y

Let c2 be the linear segment from (1,1) to (1, 1). =

So + .

Work: The work done on an object by a constant force while it moves on a straight line is given by W = where is the displacement vector in the direction of motion.

If the force is not constant or the path is not a straight line, the element of work becomes

. The work along the curve will be the line integral .

eg3 A mass lying flat on a table is attached to a spring that is compressed 10 cm beyond its rest position. Find the work done by the force exerted by the spring to move the mass back to its equilibrium position.

If we consider the equilibrium position to be zero, and the spring is compressed in the negative direction, by Hook’s law where k is a positive constant. The line integral becomes

= 50k. Since the force acts in the direction of motion, the work is positive.

eg 4 Find the work done by the Earth’s gravity to lift a 4000Nweight communications satellite to an altitude of 24,500km. Consider the radius of the earth to be r = 6500km.

Since the force done by the earth gravity is F = in the direction of the center of the earth,

F = 4000 = , k = 1.69 X 1011. The work will be given by W = = 20,548,387.1N-km = −20,548,387,100J. Since the force acts in the direction opposite to the direction of motion, the work is negative.

Homework 18.1

18.2 Computing Line Integrals Over Parameterized Curves

We can describe integrals along a curve c parametrically by = < x(t), y(t) > in or by = < x(t), y(t), z(t) > in . Let c be a smooth curve, in t : [a, b].

eg 5 Find over c : y = x2 in x : [0, 1].

This integral can be seen as

Since c : = < t, t2 > , t : [0, 1], = < = < 1, 2t >, and = < 0, t + t2 > ,

0, t + t2< 1, 2t > dt =

eg 6 dx + xdy over the line segment c : x = 2 + t, y = 3 t in t : [0, 1].

We write this integral as , xdx, dy >, so

, (2 + t) > < 1, 1 > dt = (2 + t)dt = -

Vector form of Line Integrals

Let c be a smooth curve, (x, y)ds = f (x(t), y(t)) dt in t : [a, b] where ds is the infinitesimal element of arc length, in , and f (x, y, z)ds= dt in t : [a, b] where = in .

Recall: Since s(t) = du,by the Fundamental Theorem of Calculus = .

So f (x, y, z)ds = , in t: [a, b].

eg7 Consider wherec : x2 + y2 = 1, y > 0 in with a ccw orientation.

Since this is the top of the unit semicircle, = < cos(t),sin(t) > in t : [0, ; = < sin(t), cos(t) > ;

so + cos2(t)sin(t))dt = 2

If C is the curve with pieces c1andc2, .

eg 8 Consider 12 xds where C =

Let c1 be represented by = t, t2 t : [0, 1]; = < 1, 2t > ;

Let c2 be represented by = < 1, t > t : [1, 2]; = < 0, 1 > ;

12xds = 12xds + 12xds =

A change in orientation will change the sign of the line integral if it is along a coordinate axis, but will not change the sign along arc length.

f (x, y, z)dx = f (x, y, z)dx; f (x, y, z)ds = (x, y, z)ds.

eg 9 Consider (x + 1)dx along both c1 : y = with CW orientation and c2 : y = with CCWorientation.

For c1, = < cos(t), sin(t)t : [0, ; = sin(t)

(x + 1) dt =

For c2, = < cos(t), sin(t) > t : [0, ;

(x + 1) dt =

eg 10 Now consider the same integrals along the arc length of the curve

ds along both c1 : y = with cw orientation and c2 : y = with ccw orientation.

For c1, = < cos(t), sin(t) >t : [0, ; = sin(t), cos(t) > ; = 1

For c2,

Average Value Along a Path C

The average value of f (x, y) on c will be fave = (x, y)ds.

eg 11 Find the average value of f (x, y) = x2y3 on c : y = x3 , x [0, 2]

Let = < t, t3 > ; = , t [0, 2]

So fave= . This integral can be evaluated with a CAS.

Line Integrals of Vector Fields

Line integrals of vector fields will be given by

eg12 Find the line integral if along the semicircle or radius 1 oriented counterclockwise from the point (x, y) = (1, 0) to (x, y) = (− 1, 0).

Since the path is = < cos(t), sin(t) >,

Since the field is perfectly aligned with the path, will be the highest value a line integral will obtain between those two points for any path C.

eg13 Find the line integral if along the line segments from (x, y) = (1, 0) to (0, 1), and from (0, 1) to (− 1, 0).

Let C1 be the line segment from (1, 0) to (0, 1). If we parameterize the curve we obtain C1 = < 1 – t, t>,

Let C2 be the line segment from (0, 1) to (−1, 0). If we parameterize the curve we obtain

C2 = < – t, 1 – t >, , or

The.

eg 14 Find the line integral if = < y, x > along the semicircle or radius 1 oriented clockwise from

x = − 1 to x = 1.

Since the path is = < − cos(t), sin(t) > 0 =

Work: The work done on an object in a constant force field while it moves on a straight line is given by W = where is the unit tangent vector in the direction of motion and is the length of the line segment.

The work done by a variable force field on an object that traverses the curve C is given by

W = dt = dt

, ,

eg 15 Find the work needed to move an object in a force field from P(5, 10) to

Q (20, 30) with distance measured in meters.

W = = < 4, 7 > < 15, 20 > = 200N-m = 200J

If the force is not constant or the path is not a straight line, the element of work becomes

where the element of arc length of the curve is . The work will be the line integral

eg 16 Find the work done to move an object in the force field along C : y = ,

x [0, 1] in a CCW direction with the force in N and the distance in m.

Since = < cos(t), sin(t) > ; d = < − sin(t), cos(t) > dt; t [0,

eg 17 Find the work done to move an object in the force field

along c: < t4, − t2, t3 >, t [0, 1] with distance in ft.

Circulation: If is a velocity vector field, the circulation of a fluid due to the velocity field along the curve is given by where is the unit tangent vector in the direction of motion and is the length of the line segment.

Flux of a Velocity Field in

Flux: The flux (rate of flow) of a fluid across a line segment of a constant velocity field is given by

flux = where is the density of the fluid is length of the line segment and is the unit normal vector to the line segment.

eg 18 Find the flux of a fluid through a line segment of 20m in a constant velocity field of 50m/min at an angle of /3 with the normal.

If the velocity field is not constant or the path is not a straight line, the element of flux becomes

where as is the element of arc length of the curve. The flux will be the line integral

= = .

Let = < p, q > be a velocity vector field in . The flux across the curve c along the direction of the normal will be given by .

Another form we can express this integral is as.

Proof: Let be the unit tangent vector. The unit normal vector will be

(not such that = 0 with to the right of ( into the paper).

Since and

to the right of

eg 19 Find the rate of flow of a fluid in the x-y plane, with distances measured in m, of a fluid with a velocity field = < x3 , 1 > m/sec across the parabola x = y2 y : [−1, 1].

Since

= 2/7m2/sec. This means that the net flow across the parabola is 2/7m2/sec.

If we use the other form =2/7m2/sec.

Homework 18.2

18.3 Gradient Fields and Path-Independent Fields

If can be represented as the gradient of the potential U, then the line integral. The Fundamental theorem of Line Integrals states that if C is a smooth curve given by a vector function, and if U (x,y) is a differentiable function whose gradient vector is continuous in C,

This means that the value of the integral depends on the end points and not on the path. If a force field can be represented as a gradient of a potential, we say that the field is independent of the path.

Proof:

Let describe the curve C,

eg 20 Evaluate

with C1 : y = x, 0 ≤ x ≤ 1 and C2 : x = 2y2 – y, 0 ≤ y ≤ 1.

. We see that this line integral is independent of the paths C1 and C2.

Any vector field can be represented as a gradient of a potential U (that is ), if .

Proof:

If .

So . If we take the partials.

Since the second mixed partial are equal,.

For the previous example, since so the line integral is independent of the path.

Finding the Potential Function

eg 21 Evaluate for

C : y = x in x : [0, 1] (a) without finding the potential function, (b) by finding the potential function.

a. Since

b. Since there is a potential function.

Since and by equating components,

The potential.

eg 22 Evaluate between the points (1, 2) and (3, 4).

Since, there is a potential function.

Since

, so

The potential and

Conservative Vector Fields and Closed Curves

If C is a smooth-closed curve in the x-y plane .

If is a force field, is the work done by the force field .

If is a velocity field, will give the flow across the closed curve, and is the circulation to the fluid around C, and

If is a conservative vector field, the line integral along a closed curve C,

, since its initial and final points are the same.

If is a conservative force field, the work along a closed curve is zero.

If is a conservative velocity field, the circulation along a closed curve is zero. A fluid like this is called irrotational (no vortices in the flow).

NOTE: A line integral independent of the path can be zero with different initial and final points. (Open curve)

eg 23 Evalute .

Since

The potential and

Conservation of Energy

. The work done by the force field along a path C on an object is the change in its kinetic energy.

Since is conservative, with potential P (x, y, z),

the opposite of the change in potential energy

Since . Since the two integrals are the same,

. “Conservation of energy”.

Homework 18.3

Line Integrals Independent of the Path Worksheet

1)Evaluate the line integrals (a) by direct integration along any path, (b) by finding the potential.

a) Ans:

b) Ans:

2) Integrate by any method.

a) (any path)Ans:

b) Ans:

c) Ans:

3) Find the work done by the forceNon a particle that moves from the point to the point with distance measured in meters. Ans:

4) Find the circulation of the velocity field of a fluid from the point (1, 0) to the point (-1, 0), without finding the potential function,

a) along the x-axis.Ans:

b) along the semicircle .Ans:

5) Find the flux of the velocity field of a fluid with density from the point (1, 0) to the point (-1, 0), without finding the potential function,

a) along the x-axis.Ans:

b) along the semicircle .Ans:

18.4 Green’s Theorem

Green’s Theorem is a consequence of the First Fundamental Theorem of Calculus (Leibniz’s Rule) applied to double integrals. It give the relationship between a line integral of a simply connected closed curve C, with positive orientation (the region is at the left as the curve is traversed) and a double integral over a plane region R.

Green’s Theorem:

Let C be a positively oriented, piecewise smooth simply connected closed curve in and let R be the region bounded by C. If P and Q have continuous partial derivatives on an open region with no holes containing R, then the line integral of around the closed curve C is given by

, where is called the two dimensional or scalar Curl of the vector field

eg 24 Given R, the square find .

by Leibniz’s Rule.

Also,

From this example, we see that .

eg 25 Given R the square , find

by Leibniz’s Rule.

Also

From this example, we see that .

If we add the last two results we get Green’s Theorem.

eg 26 Express as a double integral over R where c is a simply connected closed curve.

eg 27 Evaluate where C is the triangle with vertices (0, 0), (1, 0), (0, 1).

To evaluate without using Green’s Theorem,

Let c1 : < t, 0 >, 0 < t < 1 the line segment from (0, 0) to (1, 0).

Let c2 : < 1 – t, t >, 0 < t < 1 the line segment from (1, 0) to (0, 1).

Let c3 : < 0, 1 – t >, 0 < t < 1 the line segment from (1, 0) to (0, 0).

If we apply Green’s Theorem, .

A consequence of Green’s Theorem is:

or if we combine both formulas,

eg 28 Find the area of the ellipse by a) double integration; b) using Green’s Theorem.

a) .

Green’s Theorem in Vector Form

Let

Since , Green’s Theorem in vector form becomes

eg 29 Evaluate over the unit circle

a) by using Green’s Theorem, b) by direct integration.

b) Since

Curl Test for Vector Fields in 2-Space

Let be a vector field with continuous partial derivatives such that the domain of has the property that every closed curve that encircles it, encircles a region the lies entirely in the domain ( has no holes), and the curl of is zero , then is path independent, so is a gradient of some potential.

By Green’s Theorem , if the curl of is zero

, then , so the vector field (x, y) is path independent (conservative).

eg 30 Is the vector field (x, y) = < y,x > conservative over the unit circle?

Since, by the curl test for vector fields in 2-space is zero, so ((x, y) = < y,xis a conservative vector Field. By Green’s Theorem, .

eg 31 Is independent of the path for over the unit circle?

By Direct integration so the line integral is not independent of the path.

Since , by the curl test for vector fields in 2-space is zero. This does not imply that the integral is independent of the path even though by Green’s Theorem,

The discrepancy of the two results is that has a hole at (0, 0) so Green’s Theorem can not be used, or since the curl test does not apply.

A vector field can be path independent even if the domain constrain of Green’s Theorem does not apply.

eg 32 Consider

Even though the curl test does not apply over any path containing the point (0, 0). Since

↓ integrate in xdiff wrt y↓ comparec(y) = k, so

The potential so the vector field is conservative.

To find a line integral for this vector field, we can use the potential function (since the integral is independent of the path), but we should not use Green’s Theorem for any closed path that includes the origin (the curl test will not apply).