6.3 Vector Line integrals: Calc 1 review: Suppose f : R → R, Fundamental Theorem of Calculus: ∫_{a}^{b} f′(t)dt = f(b)−f(a). ∫_{a}^{b}(velocity) dt = distance traveled. ∫_{a}^{b}(rate of change)dt = total change. Given f′(t), can find f(b) − f(a). Given velocity, can find distance traveled. Given rate of change, can find total change. Calc III: Suppose f : R^{2} → R, Given ∂f = (\frac{∂f}{∂x}, \frac{∂f}{∂y}), find f(p) − f(q).

6.1 vector field integral review: Let F (x, y) = (x, y), let x(t) = Notation 1: Work definition ∫_{C} F · ds = ∫_{C_{1}} F · ds + ∫_{C_{2}} F · ds = ∫_{0}^{1} F (x_{1}(t)) · x′_{1}(t)dt + ∫_{0}^{1} F (x_{2}(t)) · x′_{2}(t)dt = ∫_{0}^{1} F (t, 0) · (1, 0)dt + ∫_{0}^{1} F (1, t) · (0, 1)dt = ∫_{0}^{1}(t, 0) · (1, 0)dt + ∫_{0}^{1}(1, t) · (0, 1)dt = ∫_{0}^{1} tdt + ∫_{0}^{1} tdt = 2(\frac{1}{2}t^{2})|_{0}^{1} = 1 Notation 2: differential form ∫_{C} F · ds = ∫_{C_{1}} F · ds + ∫_{C_{2}} F · ds = ∫_{C_{1}} F · (dx, dy) + ∫_{C_{2}} F · (dx, dy) = ∫_{C_{1}}(x, y) · (dx, dy) + ∫_{C_{2}}(x, y) · (dx, dy) = ∫_{C_{1}}(xdx + ydy) + ∫_{C_{2}}(xdx + ydy) = ∫_{0}^{1} tdt + ∫_{0}^{1}(1(0) + tdt) = 2(\frac{1}{2}t^{2})|_{0}^{1} = 1 Note: Both of the above methods are algebraically equivalent, so it doesn’t matter which notation you use.

Notation 3: Tangent vector (circulation) ∫_{C} F · ds = ∫_{x} F (x, y) · T (x, y)ds where T is the unit tangent vector to the path x. x_{1}(t) = (t, 0), x′_{1}(t) = (1, 0), ||x′_{1}(t)|| = 1.

Thus T_{1}(x, y) = (1, 0)

x_{2}(t) = (1, t), x′_{2}(t) = (0, 1), ||x′_{2}(t)|| = 1.

Thus T_{2}(x, y) = (0, 1)

∫_{C} F · ds = ∫_{x_{1}} F · T_{1}ds + ∫_{x_{2}} F · T_{2}ds = ∫_{x_{1}} F (x, 0) · (1, 0)ds + ∫_{x_{2}} F (1, y) · (0, 1)ds = ∫_{x_{1}}(x, 0) · (1, 0)ds + ∫_{x_{2}}(1, y) · (0, 1)ds = ∫_{x_{1}} xds + ∫_{x_{2}} yds [Note: these are scalar line integrals] = ∫_{0}^{1} t||x′_{1}(t)||dt + ∫_{0}^{1} t||x′_{2}(t)||dt = ∫_{0}^{1} t(1)dt + ∫_{0}^{1} t(1)dt = 2(\frac{1}{2}t^{2})|_{0}^{1} = 1 Note: The algebra for notation 3 is slightly messier than for notations 1 and 2, thus this notation is normally used only when it is possible to determine circulation geometrically.

Can use any parametrization for the path x. For example: x_{1} : [0, 1] → R^{2}, x_{1}(t) = (t^{2}, 0) x_{2} : [0, 1] → R^{2}, x_{2}(t) = (t^{3}, 0) Notation 1: Work definition ∫_{C} F · ds = ∫_{C_{1}} F · ds + ∫_{C_{2}} F · ds = ∫_{0}^{1} F (x_{1}(t)) · x′_{1}(t)dt + ∫_{0}^{1} F (x_{2}(t)) · x′_{2}(t)dt = ∫_{0}^{1} F (t^{2}, 0) · (2t, 0)dt + ∫_{0}^{1} F (1, t^{3}) · (0, 3t^{2})dt = ∫_{0}^{1}(t^{2}, 0) · (2t, 0)dt + ∫_{0}^{1}(1, t^{3}) · (0, 3t^{2})dt = ∫_{0}^{1} 2t^{3}dt + ∫_{0}^{1} 3t^{5}dt = \frac{1}{2}t^{4}|_{0}^{1} + \frac{1}{2}t^{6}|_{0}^{1} = 1 Notation 2: differential form ∫_{C} F · ds = ∫_{C_{1}} F · ds + ∫_{C_{2}} F · ds = ∫_{C_{1}} F · (dx, dy) + ∫_{C_{2}} F · (dx, dy) = ∫_{C_{1}}(x, y) · (dx, dy) + ∫_{C_{2}}(x, y) · (dx, dy) = ∫_{C_{1}}[xdx + ydy] + ∫_{C_{2}}[xdx + ydy] = ∫_{0}^{1} t^{2}(2t)dt + ∫_{0}^{1} 1(0) + t^{3}3t^{2}dt = \frac{1}{2}t^{4}|_{0}^{1} + \frac{1}{2}t^{6}|_{0}^{1} = 1 Note: Both of the above methods are algebraically equivalent, so it doesn’t matter which notation you use.

Notation 3: Tangent vector (circulation) ∫_{C} F · ds = ∫_{x} F (x, y) · T (x, y)ds where T is the unit tangent vector to the path x. x_{1}(t) = (t^{2}, 0), x′_{1}(t) = (2t, 0), ||x′_{1}(t)|| = 2t.

Thus T_{1}(x, y) = (1, 0)

x_{2}(t) = (1, t^{3}), x′_{2}(t) = (0, 3t^{2}), ||x′_{2}(t)|| = 3t^{2}.

Thus T_{2}(x, y) = (0, 1)

∫_{C} F · ds = ∫_{x_{1}} F · T_{1}ds + ∫_{x_{2}} F · T_{2}ds = ∫_{x} F (x, 0) · (1, 0)ds + ∫_{x} F (1, y) · (0, 1)ds = ∫_{x}(x, 0) · (1, 0)ds + ∫_{x}(1, y) · (0, 1)ds = ∫_{x} xds + ∫_{x} yds [Note: these are scalar line integrals] = ∫_{0}^{1} t^{2}||x′_{1}(t)||dt + ∫_{0}^{1} t^{3}||x′_{2}(t)||dt = ∫_{0}^{1} t^{2}(2t)dt + ∫_{0}^{1} t^{3}3t^{2}dt = \frac{1}{2}t^{4}|_{0}^{1} + \frac{1}{2}t^{6}|_{0}^{1} = 1 Note: The algebra for notation 3 is slightly messier than for notations 1 and 2, thus this notation is normally used only when it is possible to determine circulation geometrically.

Method 2: 6.2 Green’s Theorem Curve is not closed, so can’t use Green’s Theorem.

Method 3: 6.3 Claim F has path independent line integrals Claim F is a gradient field submethod 1: F (x, y) = (x, y) = (\frac{∂f}{∂x}, \frac{∂f}{∂y}) = ∇f = ∇(\frac{1}{2}x^{2} + \frac{1}{2}y^{2}).

I.e, f(x, y) = \frac{1}{2}x^{2} + \frac{1}{2}y^{2}.

∫_{C} F · ds = f(1, 1) − f(0, 0) = \frac{1}{2} + \frac{1}{2} − 0 = 1.

submethod 2: Claim (∇ × F)(x) = 0 for all x ∈ R^{n} Since n = 2, (∇ × F)(x) = 0 for all x ∈ R^{n} iff \frac{∂N}{∂x} = \frac{∂M}{∂y}, where F = (M, N).

\frac{∂y}{∂x} = 0 = \frac{∂x}{∂y} Thus can choose any path starting at (0, 0) and ending at (1, 1) Ex: Let x : [0, 1] → R^{2}, x(t) = (t, t). ∫_{C} F · ds = ∫_{0}^{1} F (x(t)) · x′(t)dt = ∫_{0}^{1} F (t, t) · (1, 1)dt = ∫_{0}^{1}(t, t) · (1, 1)dt = ∫_{0}^{1} 2tdt = t^{2}|_{0}^{1} = 1 Closed curve example Suppose x : [0, 2π] → R^{n}, x(t) = (cos(t), sin(t)) Notation 1: Work definition ∫_{C} F · ds = ∫_{0}^{2π} F (x(t)) · x′(t)dt = ∫_{0}^{2π} F (cos(t), sin(t)) · (−sin(t), cos(t))dt = ∫_{0}^{2π}(cos(t), sin(t)) · (−sin(t), cos(t))dt = ∫_{0}^{2π}[−cos(t)sin(t) + sin(t)cos(t)]dt = ∫_{0}^{2π} 0 dt = 0 Notation 2: differential form ∫_{C} F · ds = ∫_{C} F · (dx, dy) = ∫_{C}(x, y) · (dx, dy) = ∫_{C}[xdx + ydy] = ∫_{0}^{2π}[cos(t)(−sin(t))dt + sin(t)cos(t)dt] = ∫_{0}^{2π} 0 dt = 0

Notation 3: Tangent vector (circulation) Circulation = 0. Thus ∫_{C} F · ds = 0 Method 2: 6.2 Green’s Theorem

Let F = (M, N) = (x, y)

∫_{C} F · ds = ∫_{C} Mdx + Ndy = ∫ ∫_{D}(∇ × f) · k dA = ∫ ∫_{D}[\frac{∂N}{∂x} − \frac{∂M}{∂y}]dA = ∫ ∫_{D}[\frac{∂y}{∂x} − \frac{∂x}{∂y}]dA = ∫ ∫_{D} 0dA Use chapter 5 methods to evaluate double integral.

Note this method is algebraically different than using the definition of vector line integrals. Thus if one method is algebraically difficult, try the other method. Method 3: 6.3 Claim F has path independent line integrals Claim F is a gradient field submethod 1: F (x, y) = (x, y) = (\frac{∂f}{∂x}, \frac{∂f}{∂y}) = ∇f = ∇(\frac{1}{2}x^{2} + \frac{1}{2}y^{2}).

I.e, f(x, y) = \frac{1}{2}x^{2} + \frac{1}{2}y^{2}.

∫_{C} F · ds = 0 since C is a closed curve. submethod 2: Claim (∇ × F)(x) = 0 for all x ∈ R^{n} Since n = 2, (∇ × F)(x) = 0 for all x ∈ R^{n} iff \frac{∂N}{∂x} = \frac{∂M}{∂y}, where F = (M, N).

\frac{∂y}{∂x} = 0 = \frac{∂x}{∂y} ∫_{C} F · ds = 0 since C is a closed curve.