6.2 Line integrals (Page 7/20)

Calculus volume 3 Page 7 / 20

\int_{a_{1}}^{a_{n}} f (x) d x = \int_{a_{1}}^{a_{2}} f (x) d x + \int_{a_{1}}^{a_{3}} f (x) d x + \dots + \int_{a_{n - 1}}^{a_{n}} f (x) d x,

which is analogous to property iv.

Using properties to compute a vector line integral

Find the value of integral $\int_{C} F \cdot T d s,$ where C is the rectangle (oriented counterclockwise) in a plane with vertices $(0, 0), (2, 0), (2, 1), and (0, 1),$ and where $F = ⟨ x - 2 y, y - x ⟩$ ( [link] ).

A vector field in two dimensions. The arrows following roughly a 90-degree angle to the origin in quadrants 1 and 3 point to the origin. As the arrows deviate from this angle, they point away from the angle ad become smaller. Above, they point up and to the left, and below, they point down and to the right. A rectangle is drawn in quadrant 1 from 0 to 2 on the x axis and from 0 to 1 on the y axis. C_1 is the base, C_2 is the right leg, C_3 is the top, and C_4 is the left leg. — Rectangle and vector field for [link] .

Note that curve C is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let $C_{1}$ represent the side from $(0, 0)$ to $(2, 0),$ let $C_{2}$ represent the side from $(2, 0)$ to $(2, 1),$ let $C_{3}$ represent the side from $(2, 1)$ to $(0, 1),$ and let $C_{4}$ represent the side from $(0, 1)$ to $(0, 0)$ ( [link] ). Then,

\int_{C} F \cdot T d r = \int_{C_{1}} F \cdot T d r + \int_{C_{2}} F \cdot T d r + \int_{C_{3}} F \cdot T d r + \int_{C_{4}} F \cdot T d r .

We want to compute each of the four integrals on the right-hand side using [link] . Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse C counterclockwise):

\begin{matrix} C_{1} : ⟨ t, 0 ⟩, 0 \leq t \leq 2 \\ C_{2} : ⟨ 2, t ⟩, 0 \leq t \leq 1 \\ C_{3} : ⟨ 2 - t, 1 ⟩, 0 \leq t \leq 2 \\ C_{4} : ⟨ 0, 1 - t ⟩, 0 \leq t \leq 1. \end{matrix}

Therefore,

\begin{matrix} \int_{C_{1}} F \cdot T d r & = \int_{0}^{2} F (r (t)) \cdot r^{'} (t) d t \\ = \int_{0}^{2} ⟨ t - 2 (0), 0 - t ⟩ \cdot ⟨ 1, 0 ⟩ d t = \int_{0}^{1} t d t \\ = {[\frac{t^{2}}{2}]}_{0}^{2} = 2. \end{matrix}

Notice that the value of this integral is positive, which should not be surprising. As we move along curve C ₁ from left to right, our movement flows in the general direction of the vector field itself. At any point along C ₁ , the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along C ₁ , and the line integral will have positive value.

The calculations for the three other line integrals are done similarly:

\begin{matrix} \int_{C_{2}} F \cdot d r & = \int_{0}^{1} ⟨ 2 - 2 t, t - 2 ⟩ \cdot ⟨ 0, 1 ⟩ d t \\ = \int_{0}^{1} (t - 2) d t \\ = {[\frac{t^{2}}{2} - 2 t]}_{0}^{1} = - \frac{3}{2}, \end{matrix}

\begin{matrix} \int_{C_{3}} F \cdot T d s & = \int_{0}^{2} ⟨ (2 - t) - 2, 1 - (2 - t) ⟩ \cdot ⟨ −1, 0 ⟩ d t \\ = \int_{0}^{2} t d t = 2, \end{matrix}

and

\begin{matrix} \int_{C_{4}} F \cdot d r & = \int_{0}^{1} ⟨ −2 (1 - t), 1 - t ⟩ \cdot ⟨ 0, −1 ⟩ d t \\ = \int_{0}^{1} (t - 1) d t \\ = {[\frac{t^{2}}{2} - t]}_{0}^{1} = - \frac{1}{2} . \end{matrix}

Thus, we have $\int_{C} F \cdot d r = 2 .$

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Calculate line integral $\int_{C} F \cdot d r,$ where F is vector field $⟨ y^{2}, 2 x y + 1 ⟩$ and C is a triangle with vertices $(0, 0),$ $(4, 0),$ and $(0, 5),$ oriented counterclockwise.

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Applications of line integrals

Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.

Suppose that a piece of wire is modeled by curve C in space. The mass per unit length (the linear density) of the wire is a continuous function $ρ (x, y, z) .$ We can calculate the total mass of the wire using the scalar line integral $\int_{C} ρ (x, y, z) d s .$ The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by $ρ (x *, y *, z *) Δ s$ for some point $(x *, y *, z *)$ in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral $\int_{C} ρ (x, y, z) d s .$

Calculating the mass of a wire

Calculate the mass of a spring in the shape of a curve parameterized by $⟨ t, 2 cos t, 2 sin t ⟩,$ $0 \leq t \leq \frac{π}{2},$ with a density function given by $ρ (x, y, z) = e^{x} + y z$ kg/m ( [link] ).

A three dimensional diagram. An increasing, then slightly decreasing concave down curve is drawn from (0,2,0) to (pi/2, 0, 2). The arrow on the curve is pointing to the latter endpoint. — The wire from [link] .

To calculate the mass of the spring, we must find the value of the scalar line integral $\int_{C} (e^{x} + y z) d s,$ where C is the given helix. To calculate this integral, we write it in terms of t using [link] :

\begin{matrix} \int_{C} e^{x} + y z d s & = \int_{0}^{π / 2} ((e^{t} + 4 cos t sin t) \sqrt{1 + {(−2 cos t)}^{2} + {(2 sin t)}^{2}}) d t \\ = \int_{0}^{π / 2} ((e^{t} + 4 cos t sin t) \sqrt{5}) d t \\ = \sqrt{5} {[e^{t} + 2 {sin}^{2} t]}_{t = 0}^{t = π / 2} \\ = \sqrt{5 (e^{π / 2} + 1)} . \end{matrix}

Therefore, the mass is $\sqrt{5} (e^{π / 2} + 1)$ kg.

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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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