LINE, SURFACE AND VOLUME INTEGRALS
integral in (11.1). If a loop of wireCcarrying a currentIis placed in a magnetic
fieldBthen the forcedFon a small lengthdrof the wire is given bydF=Idr×B,
and so the total (vector) force on the loop is
F=I∮Cdr×B.11.1.3 Line integrals with respect to a scalarIn addition to those listed in (11.1), we can form other types of line integral,
which depend on a particular curveCbut for which we integrate with respect
to a scalardu, rather than the vector differentialdr. This distinction is somewhat
arbitrary, however, since we can always rewrite line integrals containing the vector
differentialdras a line integral with respect to some scalar parameter. If the path
Calong which the integral is taken is described parametrically byr(u)then
dr=dr
dudu,and the second type of line integral in (11.1), for example, can be written as
∫
Ca·dr=∫Ca·dr
dudu.A similar procedure can be followed for the other types of line integral in (11.1).
Commonly occurring special cases of line integrals with respect to a scalar are
∫Cφds,∫Cads,wheresis the arc length along the curveC. We can always representCparamet-
rically byr(u), and from section 10.3 we have
ds=√
dr
du·dr
dudu.The line integrals can therefore be expressed entirely in terms of the parameteru
and thence evaluated.
Evaluate the line integralI=∫
C(x−y)(^2) ds,whereCis the semicircle of radiusarunning
fromA=(a,0)toB=(−a,0)and for whichy≥ 0.
The semicircular path fromAtoBcan be described in terms of the azimuthal angleφ
(measured from thex-axis) by
r(φ)=acosφi+asinφj,
whereφruns from 0 toπ. Therefore the element of arc length is given, from section 10.3,
by
ds=
√
dr
dφ·
dr
dφdφ=a(cos^2 φ+sin^2 φ)dφ=adφ.