Find the value of the line integral. F · dr C (Hint: If F is conservative, the integration may be easier on an alternative path.) F(x, y) = yi − xj?

(a) r1(t) = ti + tj, 0 ≤ t ≤ 1

(b) r2(t) = ti + t2j, 0 ≤ t ≤ 1

(c) r3(t) = ti + t3j, 0 ≤ t ≤ 1

1 Answer
Jul 27, 2018

Answer:

See below

Explanation:

#bbF(bbr) = y bbi − x bbj #

Testing for conservative field:

#"curl " bbF = det[(del_x, del_y),(y, -x)] = del_x(-x) - del_y(y) = -2 bbk color(red)(ne bb0)#

This is not conservative , which is fine.

To use the parameterisation in #t#, note that:

  • #int_C bbF(bbr) * d bb r equiv int_(Delta t)bbF(bbr(t)) * bbr'(t) \ dt#

(a)

  • #bbr_1(t) = underbrace(t)_(x)bbi + underbrace(t)\_(y)bbj, qquad 0 ≤ t ≤ 1 qquad bbr_1^'(t) = bbi + bbj#

#implies int_(Delta t)bbF(bbr(t)) * bbr'(t) \ dt#

#= int_(0,1) (t bbi − t bbj) * (bbi + bbj) \ dt = 0#

(b)

  • #bbr_2(t) = underbrace(t)_(x)bbi + underbrace(t^2)\_(y)bbj, qquad 0 ≤ t ≤ 1 qquad bbr_2^'(t) = bbi + 2t bbj#

#implies int_(0,1) (t^2 bbi − t bbj) * (bbi + 2t bbj) \ dt#

#=- int_(0,1) t^2 \ dt = - 1 /3#

(c)

  • #bbr_3(t) = underbrace(t)_(x)bbi + underbrace(t^3)\_(y)bbj, qquad 0 ≤ t ≤ 1 qquad bbr_3^'(t) = bbi + 3t^2 bbj#

#implies int_(0,1) (t^3 bbi − t bbj) * (bbi + 3t^2 bbj) \ dt#

#= -2 int_(0,1) t^3 \ dt = - 2 /3 = - 1/2#