082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

Substitutingy =x 2 and dy = 2xdx we have
∫ ∫
F·ds= x 4 dx + (2x)x 2 (2xdx)
C 1
C 1
=
∫ x=1
x=0
27.5 Conservative fields and potential functions 879
5x 4 dx = [x 5 ] 1 0 = 1
(b) C 2
hasequationy =xandsody = dx.NotethatthistimeweareintegratingfromB
toA.Therefore
∫
F·ds=
C 2
∫
(y 2 i +2xyj)·(dxi +dyj) =
C 2
∫
y 2 dx +2xydy
C 2
Substitutingy =xand dy = dx we have
∫ ∫ x=0
F·ds= x 2 dx+2x 2 dx=
C 2
x=1
∫ 0
(c) Now
∮ ∫ ∫
F·ds= F·ds+ F·ds=1−1=0
C 1
C 2
1
3x 2 dx = [x 3 ] 0 1 = −1
The line integral around the closed path is zero because the field is conservative.
You should check that ∇ ×F = 0and thatasuitablepotential function is φ =y 2 x.
EXERCISES27.5
1 Verifythat the vector field
F = (4x 3 +y)i +xj is conservative.
2 Forthe vector field
F =ycosxyi +xcosxyj +2zkfind ∇ ×Fandverify
that the field isconservative.
3 Considerthe fieldF =yi +xj.
(a) ShowthatFis aconservative field.
(b) Find afunction φ such that ∂φ
∂x =yand∂φ ∂y =x.
(c) Find asuitable potential function φ for F.
(d) IfAisthe point (2,1)and Bis the point
(5,8)evaluate ∫ B
A F·ds.
(e) Find φ at Band φ atAandshow that the value of
the line integralcalculated in part (d)is equal to
thedifferencebetweenthevaluesof φ atBandA.
4 Showthat the vectorfield,
F = (2xy −y 4 +3)i + (x 2 −4xy 3 )j,ofQuestion5in
Exercises27.3, isconservative andfindasuitable
potential function φ from which Fcan be derived.
Showthat the difference between φ evaluated at
B(2, 1)andatA(1,0)is equalto the value ofthe line
integral ∫ B
A F·ds.
5 Show that
∫ (1,1)
I = 3x 2 y 2 dx +2x 3 ydy
(−1,−1)
is independentofthe pathofintegration,andevaluate
the integral.
6 The function φ = 4xy is apotential functionfor
whichF = ∇φ.
(a) Find F.
(b) Evaluate ∫ F ·dsalong the curvey =x 3 from
A(0, 0)to B(2, 8).
(c) Evaluate φ at BandatAandshowthat the
difference between these values is equalto the
value ofthe lineintegral obtainedin part (b).
7 IfF =P(x,y)i
(
+Q(x,y)j
)
+0k,showthat
∂Q
∇×F=
∂x − ∂P k.Deduce that ifFis
∂y
conservative then ∂Q
∂x = ∂P
∂y .