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

27.2 Line integrals 869
Solution
The work done by gravity isfound by evaluating the line integral
W=mg
∫ s=10
s=0
ds =mg[s] 10
0 = 10mg
Note thatthisis alsothe potential energy lostbyminfallingadistance of 10 units.
Inthe previous example the path along which weintegrated was a straightline, butthis
need not always bethe case.Consider the following example.
Engineeringapplication27.2
Theworkdonebyanelectricfield
Figure 27.2 shows a unit charge moving along a curveC from point A to point B in
an electric fieldE.
At the particular point of interest, M, we have resolved the electric field vector
into two components. Resolving a vector into perpendicular components has been
described in Example 7.3. One component is tangential toC, namelyE t
, and one is
normaltoC,namelyE n
.AsaunitchargemovesfromMtoN,adistance δsalongthe
curve, the work done by the electric field isE t
δs. The componentE n
does no work
sincethereisnomotionperpendiculartothecurveC.Tofindthetotalworkdonewe
mustadd up all contributions, resulting inthe integral
total work done =
∫ B
A
E t
ds
Thisisasecondexampleofalineintegral,thelinebeingthecurve,C,joiningAand
B. Itisusual todenote thisby
∫
total work done = E t
ds
∫
C
where the symbol tells us tointegrate along the curveC.
C
C
B
M
E t
ds
N
E n
A
E
Figure27.2
Asacharge movesfrom Mto N,
the fieldEdoesworkE t δs.