1.3 Solutions
77
The volume:
V = π
a
0
y
2 dx
Therefore, 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 as K does not explicitly contain x, and
∂ K
∂ x
= 0. Therefore,
K − y
∂ K
∂ y = y
2
+ λy(1 + λy
2 )
1
2 − λyy
2 (1 + y
2 )
−
1
2 = 0
Now y = 0 at x = 0 and at x = a which can be true if C = 0. Hence
y
2
+ λy(1 + y
2 )
−1/2
= 0
Or y = −λ(1 + y
2 )
−1/2
Solving for y
,
dy
dx
=
1
y
(λ
2
− y
2 )
1/2
Integrating,
− (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 the x-axis at x 0 , and of
radius λ.
1.3.13 Statistical Distribution
1.93 (a)
∞
x=0
P x =
∞
x=0
e
−m m
x
x!
= e
−m
1 +
m
1!
+
m
2
2!
+ · · ·
= e
−m
× e
+m
= 1
Thus the distribution is normalized.
(b) < x >=
∞
x=0
x P x =
∞
x=0
xe
−m m
x
x!
=
∞
x=0
e
−m m
x
(x − 1)!
= e
−m
m +
m
2
1!
+
m
3
2!
+ · · ·
(∵ (−1)! = ∞)
= m e
−m
1 +
m
1!
+
m
2
2!
+ · · ·
= m e
−m
× e
m
= m