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:
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 :


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. 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 3advice ∫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 4In 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 :

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.
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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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