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Multivariable Calculus 1 / 130 Section 16.2 Line Integrals Types of Integrals

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Multivariable Calculus 1 / 130 Section 16.2 Line Integrals Types of Integrals Section 16.2 Line Integrals Section 16.2 Line Integrals Goals: Compute line integrals of multi variable functions. Compute line integrals of vector functions. Interpret the physical implications of a line integral. Multivariable Calculus 1 / 130 Section 16.2 Line Integrals Types of Integrals We have integrated a function over The real number line The plane R b f (x)dx RR Three space a D f (x; y)dA RRR R f (x; y; z)dV Multivariable Calculus 2 / 130 Section 16.2 Line Integrals Line Integrals In this chapter we will integrate a function over a curve (in either two or three dimensions, though more are possible). A two-variable function f (x; y) over a plane curve r(t). Multivariable Calculus 3 / 130 Section 16.2 Line Integrals Parameterizations and the Line Integral The naive approach to integrating a function f (x; y) over a curve r(t) = x(t)i + y(t)j would be to plug in x(t) and y(t). Now we are integrating a function of t. But this is dependent on our choice of parameterizations. x(2t)i + y(2t)j defines the same curve but moves twice as fast. Integrating this composition would give half the ∆t and half the area: Multivariable Calculus 4 / 130 Section 16.2 Line Integrals Integrating Independent of Parameterization Instead, our variable of integration should measure something geometrically inherent to the curve: Multivariable Calculus 5 / 130 Section 16.2 Line Integrals Arc Length A more geometrically relevant measure is to integrate with respect to distance traveled. The rectangles approximate the (signed) area between C and the graph z = f (x; y). Multivariable Calculus 6 / 130 Section 16.2 Line Integrals Integrating with Respect to Arc Length X ∗ ∗ p 2 2 Area ≈ f (xi ; yi ) ∆x + ∆y s 2 2 X ∆x ∆y = f (x(t∗); y(t∗)) + ∆t i i ∆t ∆t s Z dx 2 dy 2 −−−−!∆t!0 f (x(t); y(t)) + dt dt dt Z Alternately: f (r(t))jr0(t)jdt Multivariable Calculus 7 / 130 Section 16.2 Line Integrals Integrating with Respect to x or y We can also integrate with respect to change in just x or just y. Z X X ∆x ∆t!0 f (x ∗; y ∗)∆x = f (x(t∗); y(t∗)) ∆t −−−−! f (x(t); y(t))x 0(t)dt i i i i ∆t Multivariable Calculus 8 / 130 Section 16.2 Line Integrals Comparison to Integrating a Vector Function Remark A plane curve C is defined by a vector function r(t), but we should not confuse integrating r(t) itself with integrating a function f (x; y) over C. These two types of integrals are almost completely unrelated. Z b We visualize r(t)dt as the total displacement achieved by a traveling with velocity r(t) (a vector). Z We visualize f (x; y)ds as the area between the graph of f and the C curve C (a number). Multivariable Calculus 9 / 130 Section 16.2 Line Integrals Applications R We defined C fds as an area. It can also be useful for integrating any function that is a rate with respect to distance: Example Over varied terrain, if p(x; y) gives the price per mile to build railroad R tracks at point (x; y), then C p(x; y)ds gives the total cost to construct a railroad following C. Example Over varied terrain, if f (x; y) gives the fuel consumption per mile traveled R at the point (x; y), then C f (x; y)ds gives the total fuel consumption to travel along C. Multivariable Calculus 10 / 130 Section 16.2 Line Integrals Example 1 Let C be the line segment from (0; 0) to (3; 4). Let f (x; y) = x2 + cos(πy). Compute the line integral Z f (x; y)ds: C Multivariable Calculus 11 / 130 Section 16.2 Line Integrals Parameterizations to Know In Chapter 13 we learned parameterizations for the following plane curves: A line segment from A to B A circle of radius a The graph of an explicit function y = f (x) Multivariable Calculus 12 / 130 Section 16.2 Line Integrals Example 2 R What does integrating C 1ds compute? Formula Z 1ds = arc length × height C = arc length Multivariable Calculus 13 / 130 Section 16.2 Line Integrals Example 3 2 2 3=2 Calculate the arc length of r(t) = (t − t)i + 3 (2t) j on the interval 0 ≤ t ≤ 4. Multivariable Calculus 14 / 130 Section 16.2 Line Integrals Summary Questions Why do we convert to a different differential when setting up a line integral? What does ds mean? What is its differential in terms of dt? How do we compute arc length? Multivariable Calculus 15 / 130 Section 16.1 Vector Fields Section 16.1 Vector Fields Goals: Recognize real world phenomena that are modeled by vector fields. Determine the geometric behavior of a vector field from its equation. Compute line integrals of a vector field over a curve. Multivariable Calculus 16 / 130 Section 16.1 Vector Fields The Definition of a Vector Field Definition 2 A vector field in R is a function that assigns a two-dimensional vector 2 F(x; y) to each point in R . 3 A vector field in R is a function that assigns a three-dimensional vector 3 F(x; y; z) to each point in R . Multivariable Calculus 17 / 130 Section 16.1 Vector Fields Drawing a Vector Field We draw a vector field by attaching the vectors F(x; y) to the points (x; y) by the tail. For obvious reasons, we only draw these vectors from a finite set of points. Multivariable Calculus 18 / 130 Section 16.1 Vector Fields Notation for a Vector Field A vector field is defined by component functions P and Q (and R in three dimensions) F(x; y) = P(x; y)i + Q(x; y)j or F(x; y) = hP(x; y); Q(x; y)i : F(x; y; z) = (0:2x + 0:04y)i + (0:03z − 0:1)j + 0:2 sin(xz)k Multivariable Calculus 19 / 130 Section 16.1 Vector Fields Examples of Vector Fields Wind speed at each point on the ground. The force exerted by gravity (or magnetism or charge) at each point in space. The gradient of a differentiable function. Multivariable Calculus 20 / 130 Section 16.1 Vector Fields Example 1 y 1 Sketch the two dimensional vector field F(x; y) = 2 i − 2 j. Multivariable Calculus 21 / 130 Section 16.1 Vector Fields Example 2 How can we visualize the vector field xi + yj F(x; y) = ? px2 + y 2 Multivariable Calculus 22 / 130 Section 16.1 Vector Fields Example 3 How can we visualize the vector field F(x; y) = −yi + xj? Multivariable Calculus 23 / 130 Section 16.1 Vector Fields Exercise Consider the vector field −xi − yj F = : (x2 + y 2)3=2 1 What direction do the vectors of this field point? 2 How do the lengths depend on the location of (x; y)? 3 Can you sketch the field? Multivariable Calculus 24 / 130 Section 16.2 Line Integrals Dot Products and Work The dot product measures the angle of two vectors, as well as their magnitude. F · s = jFjjsj cos θ This models the work done by a force F on a displacement s, since only Fproj contributes to work. W = Fproj · s = F · s Multivariable Calculus 25 / 130 Section 16.2 Line Integrals Line Integrals of Vector Fields The formula W = F · s assumes that F is constant, and the displacement s is along a straight line. A vector field introduces the possibility that F is different at different points. To compute the work done by a vector field, we use an integral. Definition The line integral of the vector field F(x; y) over the vector function r(t) is defined: Z Z F · dr = F(r(t)) · r0(t)dt C This is sometimes called a work integral. We'll see later that line integral of a vector field has applications beyond physics. Multivariable Calculus 26 / 130 Section 16.2 Line Integrals Alternate Notation If F(x; y) = P(x; y)i + Q(x; y)j we can rewrite this integral without a vector operation: Z Z F(x; y) · dr = hP; Qi · x0(t); y 0(t) dt C C Z = Px0(t)dt + Qy 0(t)dt C Z = Pdx + Qdy C Notation The line integral of a vector field can also be written: Z Z P(x; y)dx + Q(x; y)dy: C C Multivariable Calculus 27 / 130 Section 16.2 Line Integrals Example 4 Suppose an object travels once anticlockwise around the unit circle and is y 1 acted on by a force field F(x; y) = 2 i − 2 j. Does F do positive or negative work on the object? Calculate the total work done. Multivariable Calculus 28 / 130 Section 16.2 Line Integrals Independence of Parameterization Theorem If C1 and C2 are two parameterizations of the same curve then Z Z F(x; y)dr = F(x; y)dr C1 C2 To prove this, let C1 be given by r1(t). Then there is some function u(t) such that C2 is given by r2(t) = r1(u(t)). The proof is a u-substitution. This makes physical sense, because work does not care about speed, only displacement. Thus traveling more quickly or slowly along a curve will not change the total work done. Multivariable Calculus 29 / 130 Section 16.2 Line Integrals Summary of Line Integrals Geometric Notation Calculation Interpretation Z b Z p f (x; y) ds Area below the graph f (x(t); y(t)) (x 0(t))2 + (y 0(t))2dt C a Z Z b f (x; y) dx Area when projected f (x(t); y(t))x0(t)dt onto xz plane C a Z Z b F(x; y) · dr Work done by F F(x(t); y(t)) · x0(t); y 0(t) dt C a Z b or P(x(t); y(t))x0(t) + Q(x(t); y(t))y 0(t) dt a Z b Not a line integral, Z b r(t) dt computes change in x(t)i + y(t)j dt a position from velocity a Multivariable Calculus 30 / 130 Section 16.2 Line Integrals Summary Questions How do we represent a vector field, graphically? What does a line integral of a vector field measure? How can we see whether a vector field is doing positive or negative work on a path? What does dr mean? What is its differential in terms of dt? Multivariable Calculus 31 / 130 Section 16.3 The Fundamental Theorem for Line Integrals Section 16.3 The Fundamental Theorem for Line Integrals Goals: Use the fundamental theorem to evaluate line integrals of conservative vector fields.
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