16.2: Line Integrals 16.3: The Fundamental Theorem for Line Integrals

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16.2: Line Integrals 16.3: The Fundamental Theorem for Line Integrals 16.5: Curl and Divergence

16.2 Line Integrals

Line Integrals A Line Integral is similar to a single integral except that instead of integrating over an interval [a, b], we integrate over a curve C. Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f (x, y)  0, C f (x, y) ds represents the area of one side of the “fence” or “curtain”, whose base is C and whose height above the point (x, y) is f (x, y):

Line Integrals or “Curve Integrals” Line integrals were invented in the early 19th century to solve problems involving fluid flow, forces, electricity, and magnetism. We start with a plane curve C given by: the parametric equations: x = x(t) y = y(t) a  t  b or the vector equation: r(t) = x(t) i + y(t) j, and we assume that C is a smooth curve. [Meaning r is continuous and r(t)  0.]

Example: Consider the function f (x,y)= x+y and the parabola y=x2 in the x-y plane, for 0≤ x ≤2. Imagine that we extend the parabola up to the surface f, to form a curved wall or curtain: The base of each rectangle is the arc length along the curve: ds The height is  f  above the arc length: f (x,y) If we add up the areas of these rectangles as we move along the curve C, we get the area: Parabola is the curve C

Example (cont.’) Next: Rewrite the function f(x,y) after parametrizing x and y: For example in this case, we can choose parametrization: x(t)=t and y(t)=t2 Therefore r = <t, t2> and f = t +t2 using the fact that (recall from chapter 13! ) ds = | r’ | dt then ds = And the integral becomes:

Line Integrals in 2D: The Line integral of f along C is given by: NOTE: The value found for the line integral will not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b.

Practice 1: Evaluate C (2 + x2y) ds, where C is the upper half of the unit circle x2 + y2 = 1. Solution: In order to use Formula 3, we first need parametric equations to represent C. Recall that the unit circle can be parametrized by means of the equations x = cos t y = sin t and the upper half of the circle is described by the parameter interval 0  t  .

Practice 1 – Solution cont’d Therefore Formula 3 gives

Application: If f (x, y) = 2 + x2y represents the density of a semicircular wire, then the integral in Example 1 would represent the mass of the wire! The center of mass of the wire with density function  is located at the point , where:

Line Integrals in 3D Space: We now suppose that C is a smooth space curve given by the parametric equations: x = x(t) y = y(t) z = z(t) a  t  b or by a vector equation: r(t) = x(t) i + y(t) j + z(t) k. If f is a function of three variables that is continuous on some region containing C, then we define the line integral of f along C (with respect to arc length) in a manner similar to that for plane curves:

Line Integrals (in general) Notice that in both 2D or 3D the compact form of writing a line integral is: For the special case f (x, y, z) = 1, we get: Which is simply the length of the curve C!

Example: Evaluate C y sin z ds, where C is the circular helix given by the equations: x = cos t, y = sin t, z = t, 0  t  2.

Example – Solution Formula 9 gives:

Line Integrals of Vector Fields Suppose that F = P i + Q j + R k is a continuous force field. The work done by this force in moving a particle along a smooth curve C is: Equation 12 says that work is the line integral with respect to arc length of the tangential component of the force. This integral is often abbreviated as C F  dr and occurs in other areas of physics as well.

Example Find the work done by the force field F(x, y) = x2 i – xy j in moving a particle along the quarter-circle r(t) = cos t i + sin t j, 0  t  /2. Solution: Since x = cos t and y = sin t, we have F(r(t)) = cos2t i – cos t sin t j and r(t) = –sin t i + cos t j

Example – Solution cont’d Therefore the work done is

The Fundamental Theorem for Line Integrals 16.3 The Fundamental Theorem for Line Integrals

The Fundamental Theorem for Line Integrals Recall that Part 2 of the Fundamental Theorem of Calculus can be written as: where F  is continuous on [a, b]. This is also called: the Net Change Theorem.

The Fundamental Theorem for Line Integrals If we think of the gradient vector f of a function f of two or three variables as a sort of derivative of f, then the following theorem can be regarded as a version of the Fundamental Theorem for line integrals:

16.5 Curl and Divergence

Vector Field Consider a vector function (vector field) in 3D: F = P i + Q j + R k Meaning that there exists a potential function f, such that f =F Recall that  is the operator: F is called a “conservative” vector field if it can be written as: F = f

Curl of a F Here is a better way to remember this formula! Given that: F = P i + Q j + R k and the partial derivatives of P, Q, and R all exist, then the curl of F is the vector field defined by: Here is a better way to remember this formula! “curl F” can also be written: XF (cross product), so we can write: curl F is a vector!

Example 1 If F(x, y, z) = xz i + xyz j – y2 k, find curl F. Solution:

Property of Curl: curl(f) =  x f = 0 F is called a “conservative” vector field if it can be written as: F = f Theorem: If f is function of three variables that has continuous second order partial derivatives, then: curl(f) =  x f = 0 This is a great way to find out if a field F is conservative or not! If XF = 0 then there exists a function f such that F= f, so F is conservative.

Divergence   div F is a scalar! Property of Divergence:

Example 2 If F(x, y, z) = xz i + xyz j + y2 k, find div F. Solution: By the definition of divergence (Equation 9 or 10) we have div F =   F = z + xz

Summary of operators:

Maxwell’s equations: Definition of terms: