### 1. Meaning and technique

complete a duty of a solitary variable f(x) end an expression , x takes every the values in this interval starting at a and ending in ~ b. Double-integrating a role of a two variable f(x,y) end a domain D, x and y take every the worths in this domain beginning at a and ending at b for x; and beginning atc and also ending in ~ d for y. with line integrals we integrate the role f(x,y), a duty of 2 variables, and also the values of x and also y will certainly be the points, (x,y), that lie top top a curve C. notice that this is different from the twin integrals where the integration is made over a region R or D. In heat integrals, we incorporate over a curve made from the clues of the the duty itself. Let�s consider the curve C that the point out come from, and assume assume the the curve is smooth. The curve is given by the parametric equations: x = g(t), y = h(t) a ≤ t ≤ b v the parameterization of the curve as a vector function, the curve is provided by: (t) = g(t) + h(t) a ≤ t ≤ b Let's recall the the curve is called smooth
if (t) is consistent and (t) ≠ for every t. The heat integral of f(x,y) along C is denoted by: The differential facet is ds. This is the reality that us are relocating along the curve, C, rather dx for the x-axis, or dy for the y-axis. The over formula is called the line integral that f v respect come arc length. Let's recall that the arc length of a curve is given by the parametric equations: l = ∫ab ds width ds = √<(dx/dt)2 + (dy/dt)2> dt Therefore, to compute a line integral we transform everything end to the parametric equations. The heat integral is then: instance 1

advice ∫C 3x2 ds where C is the heat segment native (-1, 1) to (1,2). we have currently seen, in the equation of a line in 3D an are section, the the parameterization formula the the line segment beginning at the point(xo, yo) and ending in ~ the allude (x1,y1) is : (t) = (1 - t) 〈xo, yo〉 + t 〈x1,x2〉 that is: x = xo (1 - t) + x1 t y = yo (1 - t) + y1 t So, the parameterization formula that the line segment starting at (-1, -1) and also ending at (1,2) is : (t) = (1 - t) 〈-1, 1〉 + t 〈1,2〉. the is: x = (1 - t) (- 1) + t(1) = 2t - 1 y = (1 - t) (1) + t(2) = t + 1 at (-1, 1): x = - 1, y = 1 → t = 0
, and also in ~ (1,2): x = 1 , y = 2 → t = 1 for this reason 0 ≤ t ≤ 1 us have: dx/dt = 2 , and also dy/dt = 1. Hence: ds = √<22 + 12> dt = √5 dt. ds = √5 dt thus ∫C 3x2 ds = ∫01 3(2t - 1)2 √5 dt = 3 √5 ∫01 (4t2 - 4t + 1) dt = 3√5 <(4/3)t3 - 2t2 + t> 01 = 3√5 <(4/3) - 2 + 1> = 3√5 <1/3> = √5 ∫C 3x2 = √5

instance 2

advice ∫Cxy2 ds where C is the right fifty percent of the one x2+ y2 = 16, rotated in the counter clockwise direction. we parameterize the curve i beg your pardon is the circle, climate the integrand, and also then the differial element: x = 4 cos t, y = 4 sin t - π/2 ≤ t ≤ + π/2. f(x,y) = xy2 = 4 cos t (4 sint)2 = 64 cos t sin2t dx/dt = - 4 sin t , dy/dt = 4 cos t ds = 4 dt Therefore ∫C xy2 ds = ∫- π/2 + π/2 64 cos t sin2t (4 dt) = 256 ∫- π/2 + π/2 cos t sin2t dt = 256 <(1/3) sin3 t>- π/2 + π/2 = (256/3) = 2 (256/3) sin3 (+ π/2) = 2 (256/3) = 512/3.∫C xy2 ds = 512/3

### 2. Integrals end piecewise smooth curves

Now, we space going to incorporate line integrals over piecewise smooth curves. A piecewise smooth curve C is any kind of curve that have the right to be written as the union the a finite number of smooth curves, C1, C2, C3, , ... Cn, whereby the end allude of Ci is the beginning point the Ci+1. To advice the line integrals over piecewise smooth curves , we evaluate the line integral end each the the pieces and also then add them up: ∫C f(x,y) ds = ∫C1 f(x,y) ds ∫C2 f(x,y) ds ∫C3 f(x,y) ds ∫C4 f(x,y) ds ∫C5 f(x,y) ds + ...

instance 3

advice ∫C 2x ds wherein C is the curve displayed below. C1: x = t , y = 3 - 4 ≤ t ≤ 0 C2: x = 3 cos t , y = 3 sin t π/2 ≤ t ≤ 0 C3: x = 3 , y = t - 3 ≤ t ≤ 0 C1: dx/dt = 1 dy/dt = 0 ds = dt ∫C1 2x ds = ∫- 40 2t dt = (t2)|- 40 = 16 ∫C1 2x ds = 16 C2: dx/dt = - 3 sint dy/dt = 3 coos t ds = 3 dt ∫C2 2x ds = ∫π/20 2 cos t (3 dt) = 6 (sin t)|π/20 = 6 (sin t)|π/20 = - 6 ∫C2 2x ds = - 6 C3: dx/dt = 0 dy/dt = 1 ds = dt ∫C3 2x ds = ∫0-3 2 (3) dt = 6 (t)|0-3 = 6(- 3) = - 18 ∫C3 2x ds = - 18 Therefore: ∫C 2x ds = ∫C1 2x ds + ∫C2 2x ds +∫C3 2x ds . = 16 - 6 - 18 = - 8 ∫C 2x ds = - 8

example 4

In the instance 1, we have have found: ∫C 3x2 ds whereby C is the heat segment native (-1, 1) come (1,2). below we move the direction that the curve to view whether le liine integral change: us evaluate then ∫-C 3x2 ds where -C is the heat segment from (1,2) (- 1, 1). The parameterization formula that the line segment starting at (1, 2) and ending at (- 1, 1) is : (t) = (1 - t) 〈1, 2〉 + t 〈- 1,1〉. the is: x = (1 - t) (1) + t(- 1) = 1 - 2t y = (1 - t) (2) + t(1) = 2 - t in ~ (1,2): x = 1 , y = 2 → t = 0
, and at (-1, 1): x = - 1, y = 1 → t = 1 so 0 ≤ t ≤ 1 us have: dx/dt = - 2 , and also dy/dt = - 1. Hence: ds = √<22 + 12> dt = √5 dt. ds = √5 dt thus ∫-C 3x2 ds = ∫01 3(1 - 2t)2 √5 dt = √5 ∫-C 3x2 = √5 , as in the instance 1. So, because that this sort of heat integrals, the is because that the integrals with respect to the arc size ds , as soon as we switch the direction the the curve, the heat integral (with respect to arc length) will certainly not change. But it does not host for all the line integrals. Because that a heat integrals with respect to arc length, us have: ∫C f(x,y) ds = ∫-C f(x,y) ds

### 3. Heat integrals end a three-dimensional curve

The heat integrals end a three-dimensional curve can be prolonged from the line integrals over atwo-dimensional curve.

You are watching: Line integral with respect to arc length

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Let�s intend that the three-dimensional curve C is provided by the parameterization: x = x(t), y = y(t), z = z(t) a ≤ t ≤ b
. The line integral is then provided by:

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