Find the value of the line integral. F · dr (Hint: If F is conservative, the integration may be easier on an alternative path.) (2x − 8y + 6) dx − (8x + y − 6) dy?

Question

part c:

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Answer 1 · 2018-07-28T12:50:09

(c) #qquad (21 - 20 e^2 - e^4)/2#

(d) #qquad 0#

If you mean that the line integral is:

#int_C (2x − 8y + 6) dx − (8x + y − 6) dy qquad = triangle#

Then #bbF = (:2x − 8y + 6,-(8x + y − 6) :)#

And #"curl " bbF = det[(del_x, del_y),(2x − 8y + 6,-(8x + y − 6))] = 0#

So #bbF# is a conservative vector field.

You can only see (c) and (d) in the screengrab

Along #y = e^x qquad :. dy = e^x dx#:

#triangle = int_(0,2) (2x − 8bbe^x + 6) dx − (8x + bbe^x − 6) bb(e^x dx)#

That looks ugly so just use a different 2-step path:

#{(bbbA: qquad dx = 0, x = 0, y: \ 1 to e^2 ),(bbbB: qquad dy = 0, y = e^2, x: \ 0 to 2 ):}#
Path #bbbA#

#triangle = int_C cancel((2x − 8y + 6) dx) − (8*0 + y − 6) dy#

#= int_(1,e^2) - y + 6 \ dy = [6y- y^2/2]_1^(e^2)#

# = 6e^2- (e^4+11)/2 #

#triangle = int_C (2x − 8e^2 + 6) dx − cancel((8*0 + y − 6) dy)#

# = int_(0,2) 2x − 8e^2 + 6 \ dx = [ x^2 − (8e^2 - 6)x]_0^2 #

# = 4 − 16e^2 + 12 #

#C_3 = bbbA + bbbB#

# = (21 - 20 e^2 - e^4)/2#

Zero, because this is a closed path and the field is conservative