Find the value of the line integral.F · dr (Hint: If F is conservative, the integration may be easier on an alternative path.) F(x,y) = ye^xyi + xe^xyj (a) r1(t) = ti − (t − 8)j, 0 ≤ t ≤ 8 ?

(b) the closed path consisting of line segments from (0, 8) to (0, 0), from (0, 0) to (8, 0), and then from (8, 0) to (0, 8)

1 Answer
Jul 27, 2018

Answer:

Zero in both cases

Explanation:

#bbF(bbr) = ye^(xy) bbi + xe^(xy) bbj #

A field #bbF(bbr)# is conservtive if there exists a potential function #phi(bbr)# such that:

  • #bbF = - nabla phi#

Or, in 2-D:

  • #bbF = - (: phi_x, phi_y:) #

This can be solved directly here, because:

#{(phi_x = - ye^(xy)),(phi_y = - xe^(xy) ):} implies phi = - e^(xy) + phi_o#

So for:

Path a)

  • #bbr_1(t) = t bbi − (t − 8) bbj, qquad 0 le t le 8#

This is along line #( 0, 8 ) to ( 8, 0 )#

#Delta phi = - e^(0xx8 ) + e^(8xx0 ) = 0#

Path b)

Regardless of path, this is closed path from #( 0, 8 ) to ( 0, 8 )#

#:. Delta phi = 0#

Going instead with the hint , the curl of a conserved vector field is zero. This follows from the existence of a potential function and the gradient theorem.

Curl of the Field: #= bbnabla times bb F = det ((del_x, del_y),(ye^(xy), xe^(xy))) #

#=( del_x(xe^(xy)) - del_y(ye^(xy))) bbk#

#=( e^(xy) + xye^(xy) - (e^(xy) + xye^(xy) ) )bbk= bb0#

The field is conservative / path-independent

Path a)

#bbr_1(t) = t bbi − (t − 8) bbj, qquad 0 le t le 8#

This is along line #( 0, 8 ) to ( 8, 0 )# ie #y = 8 - x#

Following the hint, break this into #bb 1# followed by path #bb2#:

  • #{(bb1, x = 0, dx = 0, y in [8,0]),(bb2, y = 0, dy = 0, x in [0,8]):}#

Path #bb1#:

#implies int_C bbF * d bbr #

#= int_C (ye^(xy) bbi + xe^(xy) bbj )* (bb i dx + bb j dy)#

#= int_C (y bbi + 0 bbj )* (bb i 0 + bb j dy) = 0#

Path #bb2#:

#= int_C (0 bbi + x bbj )* (bb i dx + bb j 0) = 0#

#bb1 + bb 2 = 0#

Path b)

This is clearly zero as it is a closed path in a conserved field.

FINALLY:

Noting that #bbF * dbbr equiv bbF * bbr' \ dt#, the full integral using the parameterisation is:

  • #bbr(t) : {(x = t),(y = 8 - t):} qquad bbr'(t) : {( x' = 1),( y' = -1):}#

#bbF(bbr(t)) * bbr'(t) = ( (8 - t)e^(t(8 - t)) bbi + te^(t ( 8 - t)) bbj ) *(bbi - bbj)#

#= (8 - t)e^(t(8 - t))- te^(t ( 8 - t)) #

#= (8 - 2t)e^(t(8 - t)) #

# implies int_0^8 (8 - 2t)e^(8t- t^2) \ dt#