Question

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

Answer 1

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`