answer:
hope this , this took me literally 2 hours to answer..
step-by-step explanation:
ah, im really good at this : )
if the region is multiply-connected we render it simply connected by
drawing diaphragms* when we fall back on case (a) if q happens to be
many-valued. a diaphragm corresponds to a surface of discontinuity, and
qi - q^ in (v.) becomes np where p is the cyclic increment of q and where n is
an integer.
considering now the similar cases of exception for the circuit integral, we
shall suppose
(a') that a surface of discontinuity cuts the given circuit in two points
a and b. let the surface containing the mesh- work be drawn through an
arbitrary curve acb on the surface of discontinuity. on adding the results
of integration for the two circuits consisting of the part on one side of the
surface of discontinuity and the curve acb, and of the part on the other side
of the surface and the curve bca, we have exactly as in (v.)
jdp . g + jdpi2 . (ji - 5-2) = jtdvv . q (xii.)
it follows from (iv.) that we get exactly the same result had any other
curve adb been taken on the sunace of discontinuity.
(b') if q is not single-valued over the continuous net, its value is definite
if a definite value is chosen at some one point of the net, or else q is inde-
terminate at some point of the net. such a point may be surrounded by a
small closed curve joined by a barrier to the circuit. the barrier must be
treated as a line of discontinuity and the value of the integral round the
closed curve must be taken account of.
(o') when q becomes infinite at a point on the surface of the mesh- work,
let the point be surrounded by a small circle of radius r. then the relation
becomes, when we exclude the point from the surface integral,
jydvv. q^jdp . 2'- jrdup . (qq+q^r-^-^q^r-^+ (xiii.)
the second line integral bein^ taken round the circle.t this integral
vanishes unless there are negative powers of r. the part depending on q^ is
jdup . gi=|dtjp . (satjp+^up)
*the interior of a hollow curtain ring becomes simply connected when a
diaphragm is drawn across one normal section.
t the two line integrals are taken in the same sense of rotation round the axis
of the small circle. if we choose the minus sign may be placed on the right of
the sign of integration, and then we shall have the surface integral equal to the
sum of two line integr^s taken in opposite directions.
s18
the opeeator v.
[chap. xvi.
suppose where ^ is a linear vector function, the terms not linear in up
leading to a vanishing integral round the circle. putting up = ^ cos w +^‘ sin-w?
where i and / are in the plane of the small circle, the integral easily reduces
to tr(jsai-isaj and to 7 r(va^-x'^+ 2 s€^) where x' and € have
the same signification as in the chapter on linear vector functions.
p.t. 1, if /(v) is any linear function of the operator v with constant
■coefficients,
|/(dv). g=j/(v) .q.dv, |/(dp) . ? =f/(td.^v) . q,
and j ? •/(d'') = j s' -/(v) • di> , | =\q ./(v dvv).
[no step in the proof of the simpler case need be modified. in the
second set of relations the operator is placed in front of the operand. see
art. 57, ex. 1 1, and m^aulay’s utility of quaternions in physics^
ex. 2. in general if /(a) is a linear function of an arbitrary vector a
while the variable vector p is involved in the constitution of the function,
show that
i/(dv)=j/(v).d«, l/(d/,)=|/(vdvv),
where /(v) means that v operates in situ on the variable vector p as involved
in the structure of the function.
• ex. 3. prove that | - jsdvv. vtp-i,
where no infinites occur.
[see tait’s quat& miom^ art. 504. here the line integral is jvdpvtp”^,
which transforms into
jv.vdvy.vr/5-i or jdvv^tp-isjsdi/v.vt/j-i.]
ex. 4. prove that
j jdv = j i p . vg . d« - j j pd vy.
[this is an example of an extensive class of transformations depending on
the invariantal properties of v. transforming the surface integral, we
have ^ pdvq—\ p(^)qdv, where v operates both on p and on q. but
pv=vp= -3. see art. 132, p 235.]
