MATH Harrell. An integral workout. Lecture 21. Copyright 2013 by Evans M. Harrell II.

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1 MATH Harrell An integral workout Lecture 21 Copyright 2013 by Evans M. Harrell II.

2 This week s learning plan and announcements We ll integrate over curves Then we ll integrate over... curved surfaces

3 What about that test? 75 th %-ile: 37 median: th %-ile: 31 If you were in the bottom 25%, it is a good idea to consult with me about your grade. Office period Thu 11:30-12:30 or by appt.

4 Two of you have asked...

5 Now, just what were we doing before the break?

6 What can this man do in 3D?.or 17D? It s time for change!

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9 Jacobi in 3D x=x(u,v,w), y=y(u,v,w), z=z(u,v,w), Example: x = ρ sin(φ) cos(θ) y = ρ sin(φ) sin(θ) z = ρ cos(φ) The volume of a little 3 D box is again a determinant, up to a sign.

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11 Jacobi in 3D The Jacobian when you change variables from {x 1, x n } to {u 1, u n } is always the absolute value of the determinant of the matrix { x i / u j } Because this is the volume of an n-dimensional parallelopiped with sides r/ u j. This r is the position vector as a function of the u s!

12 Jacobi in 3D Example. Find the integral of (x+y+z) 1/2 on Ω bounded by 0 x+y+z 16, 1 2x+y 4, 1 y-3z 5 Hint: Do you really have to work out x,y,z in terms of u,v,w? How are (x,y,z)/ (u,v,w) and (u,v,w)/ (x,y,z) related?

13 Jacobi in 3D Example. Find the integral of (x+y+z) 1/2 on Ω bounded by 0 x+y+z 16, 1 2x+y 4, 1 y-3z 5 Hint: Do you really have to work out x,y,z in terms of u,v,w? Nope. The two Jacobians are reciprocals. However, remember that one of them is a function of the old variables and the other is a function of the new variables.

14 4 3 = 512/5

15 fn of x,y,z fn of u,v,w 4 3 = 512/5

16 And now for something completely different

17 Integrating on curves One kind of integral is straightforward:

18 Integrating on curves One kind of integral is straightforward:

19 Integrating on curves One kind of integral is straightforward:

20 Line integrals and arc length Example: mass of a wire. Suppose that the mass density of a wire is the distance from the origin and the wire is in the form of a spiral. What would the total mass be?

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22 A different kind of integral in 2 or 3D: The line integral Work done by a force on a moving object:

23 Line integrals

24 One definition

25 where F = P i + Q j, and dr = dx i + dy j Line integrals

26 Line integrals Important! In a line integral we do not hold dx or dy fixed while letting the other one vary.!

27 Examples In all cases, let the curve be the unit circle, traversed counterclockwise. 1. F(x) = x i + y j 2. F(x) = y i + x j 3. F(x) = y i - x j

28 Examples In all cases, let the curve be the unit circle, traversed counterclockwise. 1. F(x) = x i + y j

29 Examples In all cases, let the curve be the unit circle, traversed counterclockwise. 1. F(x) = x i + y j If x = cos t, y = sin t, then dr = (-sin t i + cos t j) dt, then F dr = 0

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31 Examples In all cases, let the curve be the unit circle, traversed counterclockwise. 1. F(r) = x i + y j. F dr = 0 at each point For a practical consequence of this fact, look at

32 Examples In all cases, let the curve be the unit circle, traversed counterclockwise. 1. F(x) = x i + y j 2. F(x) = y i + x j If x = cos t, y = sin t, then dr = (-sin t i + cos t j) dt, then F dr =( - sin 2 t + cos 2 t) dt = cos(2 t) dt. The integral of this for t from 0 to π/2 is π/4. If we integrate around the whole circle, so the beginning and end points are the same, we get 0.

33 Examples So, it s kind of like in kindergarten calculus, when the integral from a to a of any function is 0. Or, is it?

34 Examples In all cases, let the curve be the unit circle, traversed counterclockwise. 1. F(x) = x i + y j 2. F(x) = y i + x j 3. F(x) = y i - x j. This time F dr =( - sin 2 t - cos 2 t) dt = - dt. The integral around the whole circle is -2π, even though the beginning and end points are the same. Under what conditions is it true that an integral from a to b=a gives us 0 - as in one dimension?

35 Under what conditions is it true that an integral from a to b=a gives us 0 - as in one dimension? Related questions: 1. Is there such a thing as an antiderivative? 2. Does the value of the integral depend on the path you take?

36 The fundamental theorem for line integrals, at least some of them

37 Path-independence Can we ever reason that if the curve C goes from a to b, then the integral is just of the form f(b) - f(a), as in one dimension?

38 The fundamental theorem Assuming C is?, f is?, and h = f on a set that is? :

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41 Path-independence The technicalities of the theorem that path-independence is equivalent to the fact that integrals over all loops are zero. Paths stay within an open, simply connected domain. Curve and vector function F are sufficiently nice to change variables. Say, continuously differentiable.

42 The fundamental theorem Assuming C is, f is, and h = f on a set that is :

43 The fundamental theorem Assuming C is a piecewise smooth curve, f is continuously differentiable, and h = f on a set that is open and simply connected:

44 Examples 1. F(r) = x i + y j. F(r) = (x 2 +y 2 )/2 at each point 2. F(x) = y i + x j. F(r) = xy 3. F(x) = y i - x j. F(r) is not a gradient.

45 , F(r) = y i + x j.

46 The fundamental theorem Why is this true? Strategy: reduce this question to a one-dimensional integral: f(r(t)) is a scalar-valued function of one variable. What s its derivative?

47 The fundamental theorem Why is this true? Strategy: reduce this question to a one-dimensional integral: f(r(t)) is a scalar-valued function of one variable. What s its derivative? f(r(t)) r (t). By the fundamental theorem of alculus, the integral of this function from t 1 to t 2 is f(r(t 2 )) - f(r(t 1 )) = f(b) - f(a). Quoth a rat demon, strand em!

48 The easy way to do line integrals, if h = f 1. h(r) = x i + y j. h(r) = (x 2 +y 2 )/2 at each point 2. h(x) = y i + x j. h(r) = xy Remind me - how do you find f if h = f?

49 A typical example h(r) = (2 x y 3-3 x 2 ) i + (3 x 2 y y) j Integral would be (2 x y 3-3 x 2 )dx + (3 x 2 y y)dy 1. Check that h(r) is a gradient. 2. Fix y, integrate P w.r.t. x. 3. Fix x, integrate Q w.r.t. y. 4. Compare and make consistent.

50 A typical example h(r) = (2 x y 3-3 x 2 ) i + (3 x 2 y y) j P y = 6 x y 2 = Q x, so we know h = f for some f. To find f, integrate P in x, treating y as fixed. We get x 2 y 3 - x 3 + φ, but we don t really know φ is constant as regards y. It can be any function φ(y) and we still have φ/ x = 0.

51 A typical example h(r) = (2 x y 3-3 x 2 ) i + (3 x 2 y y) j Now that we know f(x,y) = x 2 y 3 - x 3 + φ, let s figure out φ by integrating Q in the variable y: The integral of Q in y, treating x as fixed is x 2 y 3 + y 2 + ψ, but ψ won t necessarily be constant as regards x. It can be any function ψ(x) and we still have ψ/ y = 0. Compare: f(x,y) = x 2 y 3 - x 3 + φ(x) = x 2 y 3 + y 2 + ψ(x) So we can take φ(y) = y 2 + C 1, ψ(x) = x 3 + C 2, Conclusion: f(x,y) = x 2 y 3 - x 3 + y 2 + C (combining the two arbitrary constants C 1,2 into one).

52 The fundamental theorem Assuming C is a piecewise smooth curve, f is continuously differentiable, and h = f on a set that is open and simply connected:

53 Meet George Green Portrait from Mactutor archive.

54 Application: The planimeter Picture by Paul E. Kunkel, with his kind permission.

55 What does Green tell us when 1. P = 0, Q = x? 2. P = -y, Q = 0? 3. P = -y/2, Q = x/2?

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57 Green is a two-way street

58 Green is a two-way street

59 The End

60 Clicker quiz If we change variables from (x,y) to (u,v) such that x = u+v and y = πv 2π u, what is da =? A da = du dv B da = 2 π du dv C da = 3 π du dv D da = (1/π) du dv E none of the above.

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