6.2 Line integrals (Page 6/20)

Calculus volume 3 Page 6 / 20

Reversing orientation

Find the value of integral $\int_{C} F \cdot d r,$ where $C$ is the semicircle parameterized by $r (t) = ⟨ cos t + π, sin t ⟩, 0 \leq t \leq π$ and $F = ⟨ − y, x ⟩ .$

Notice that this is the same problem as [link] , except the orientation of the curve has been traversed. In this example, the parameterization starts at $r (0) = ⟨ π, 0 ⟩$ and ends at $r (π) = ⟨ 0, 0 ⟩ .$ By [link] ,

\begin{matrix} \int_{C} F \cdot d r & = \int_{0}^{π} ⟨ − sin t, cos t + π ⟩ \cdot ⟨ − sin t + π, cos t ⟩ d t \\ = \int_{0}^{π} ⟨ − sin t, − cos t ⟩ \cdot ⟨ sin t, cos t ⟩ d t \\ = \int_{0}^{π} (− {sin}^{2} t - {cos}^{2} t) d t \\ = \int_{0}^{π} −1 d t \\ = − π . \end{matrix}

Notice that this is the negative of the answer in [link] . It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.

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Let C be an oriented curve and let − C denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:

\int_{C} F \cdot d r = − \int_{C} F \cdot d r .

That is, reversing the orientation of a curve changes the sign of a line integral.

Let $F = x i + y j$ be a vector field and let C be the curve with parameterization $⟨ t, t^{2} ⟩$ for $0 \leq t \leq 2 .$ Which is greater: $\int_{C} F \cdot T d s$ or $\int_{− C} F \cdot T d s ?$

$\int_{C} F \cdot T d s$

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Another standard notation for integral $\int_{C} F \cdot d r$ is $\int_{C} P d x + Q d y + R d z .$ In this notation, P , Q , and R are functions, and we think of d r as vector $⟨ d x, d y, d z ⟩ .$ To justify this convention, recall that $d r = T d s = r^{'} (t) d t = ⟨ \frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t} ⟩ d t .$ Therefore,

F \cdot d r = ⟨ P, Q, R ⟩ \cdot ⟨ d x, d y, d z ⟩ = P d x + Q d y + R d z .

If $d r = ⟨ d x, d y, d z ⟩,$ then $\frac{d r}{d t} = ⟨ \frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t} ⟩,$ which implies that $\frac{d r}{d t} = ⟨ \frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t} ⟩ d t .$ Therefore

\begin{matrix} \int_{C} F \cdot d r & = \int_{C} P d x + Q d y + R d z \\ = (P (r (t)) \frac{d x}{d t} + Q (r (t)) \frac{d y}{d t} + R (r (t)) \frac{d z}{d t}) d t . \end{matrix}

Finding the value of an integral of the form $\int_{C} P d x + Q d y + R d z$

Find the value of integral $\int_{C} z d x + x d y + y d z,$ where C is the curve parameterized by $r (t) = ⟨ t^{2}, \sqrt{t}, t ⟩, 1 \leq t \leq 4 .$

As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t . In this case, [link] allows us to make this change:

\begin{matrix} \int_{C} z d x + x d y + y d z & = \int_{1}^{4} (t (2 t) + t^{2} (\frac{1}{2 \sqrt{t}}) + \sqrt{t}) d t \\ = \int_{1}^{4} (2 t^{2} + \frac{t^{3 / 2}}{2} + \sqrt{t}) d t \\ = {[\frac{2 t^{3}}{3} + \frac{t^{5 / 2}}{5} + \frac{2 t^{3 / 2}}{3}]}_{t = 1}^{t = 4} \\ = \frac{793}{15} . \end{matrix}

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Find the value of $\int_{C} 4 x d x + z d y + 4 y^{2} d z,$ where $C$ is the curve parameterized by $r (t) = ⟨ 4 cos (2 t), 2 sin (2 t), 3 ⟩, 0 \leq t \leq \frac{π}{4} .$

$−26$

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We have learned how to integrate smooth oriented curves. Now, suppose that C is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve . To be precise, curve C is piecewise smooth if C can be written as a union of n smooth curves $C_{1}, C_{2},…, C_{n}$ such that the endpoint of $C_{i}$ is the starting point of $C_{i + 1}$ ( [link] ). When curves $C_{i}$ satisfy the condition that the endpoint of $C_{i}$ is the starting point of $C_{i + 1},$ we write their union as $C_{1} + C_{2} + \dots + C_{n} .$

Three curves: C_1, C_2, and C_3. One of the endpoints of C_2 is also an endpoint of C_1, and the other endpoint of C_2 is also an end point of C_3. C_1’s and C_3’s other endpoints are not connect to any other curve. C_1 and C_3 appear to be nearly straight lines while C_2 is an increasing concave down curve. There are three arrowheads on each curve segment all pointing in the same direction: C_1 to C_2, C_2 to C_3, and C_3 to its other endpoint. — The union of $C_{1}, C_{2}, C_{3}$ is a piecewise smooth curve.

The next theorem summarizes several key properties of vector line integrals.

Properties of vector line integrals

Let F and G be continuous vector fields with domains that include the oriented smooth curve C . Then

$\int_{C} (F + G) \cdot d r = \int_{C} F \cdot d r + \int_{C} G \cdot d r$
$\int_{C} k F \cdot d r = k \int_{C} F \cdot d r,$ where k is a constant
$\int_{C} F \cdot d r = \int_{− C} F \cdot d r$
Suppose instead that C is a piecewise smooth curve in the domains of F and G , where $C = C_{1} + C_{2} + \dots + C_{n}$ and $C_{1}, C_{2},…, C_{n}$ are smooth curves such that the endpoint of $C_{i}$ is the starting point of $C_{i + 1} .$ Then

$\int_{C} F \cdot d s = \int_{C_{1}} F \cdot d s + \int_{C_{2}} F \cdot d s + \dots + \int_{C_{n}} F \cdot d s .$

Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along C , then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation $\int_{a}^{b} f (x) d x = − \int_{a}^{b} f (x) d x .$ Finally, if $[a_{1}, a_{2}], [a_{2}, a_{3}],…, [a_{n - 1}, a_{n}]$ are intervals, then

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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6.2 Line integrals (Page 6/20)

Reversing orientation

Finding the value of an integral of the form ∫ C P d x + Q d y + R d z

Properties of vector line integrals

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Finding the value of an integral of the form $\int_{C} P d x + Q d y + R d z$