1.3 Solutions 77
The volume:V=π∫a0y^2 dxTherefore, dropping off the constant factors
K=y^2 +λy(1+y′^2 )^1 /^2
which must satisfy the Euler’s equation
∂K
∂x−
d
dx(K−y′∂K
∂y′)= 0
It is convenient to use the above form asKdoes not explicitly containx, and
∂K
∂x=0. Therefore,K−y′∂K
∂y′=y^2 +λy(1+λy′^2 )(^12)
−λyy′^2 (1+y′^2 )−
(^12)
= 0
Nowy=0atx=0 and atx=awhich can be true ifC=0. Hence
y^2 +λy(1+y′^2 )−^1 /^2 = 0
Ory=−λ(1+y′^2 )−^1 /^2
Solving fory′,
dy
dx
=
1
y(λ^2 −y^2 )^1 /^2Integrating,
−(λ^2 −y^2 )1(^2) =x−x 0
Or (x−x 0 )^2 +y^2 =λ^2
This is the equation to a sphere with the centre on thex-axis atx 0 , and of
radiusλ.
1.3.13 StatisticalDistribution..............................
1.93 (a)∑∞
x= 0
Px=∑∞
x= 0e−mmx
x!=e−m(
1 +
m
1!+
m^2
2!+···
)
=e−m×e+m= 1
Thus the distribution is normalized.
(b)<x>=∑∞
x= 0
xPx=∑∞
x= 0xe−mmx
x!=
∑∞
x= 0e−mmx
(x−1)!=e−m(
m+m^2
1!+
m^3
2!+···
)
(∵(−1)!=∞)
=me−m(
1 +
m
1!+
m^2
2!+···
)
=me−m×em=m