644 Geometry and Trigonometry
cost=
c−xcosB
√
x^2 +c^2 − 2 xccosB
.
The integral from the statement is
∫a
0
cost(x)dx=
∫a
0
c−xcosB
√
x^2 +c^2 − 2 xccosB
dx.
Using the standard integration formulas
∫
dx
√
x^2 +αx+β
=ln
(
2 x+α+ 2
√
x^2 +αx+β
)
,
∫
xdx
√
x^2 +αx+β
=
√
x^2 +αx+β−
α
2
ln
(
2 x+α+ 2
√
x^2 +αx+β
)
,
we obtain
∫a
0
cost(x)dx=csin^2 Bln
(
2 x+ 2 ccosB+ 2
√
x^2 − 2 cxcosB+c^2
)∣∣
∣
a
0
−cosB
√
x^2 − 2 cxcosB+c^2
∣
∣
∣
a
0
=csin^2 Bln
a−ccosB+b
c( 1 −cosB)
+cosB(c−b).
642.It is equivalent to ask that the volume of the dish be half of that of the solid of
revolution obtained by rotating the rectangle 0≤x≤aand 0≤y≤f(a). Specifically,
this condition is
∫a
0
2 πxf(x)dx=
1
2
πa^2 f(a).
Because the left-hand side is differentiable with respect toafor alla>0, the right-hand
side is differentiable, too. Differentiating, we obtain
2 πaf(a)=πaf(a)+
1
2
πa^2 f′(a).
This is a differential equation inf, which can be written asf′(a)/f (a)=^2 a. Integrating,
we obtain lnf(a)=2lna,orf(a)=ca^2 for some constantc>0. This solves the
problem.
(Math Horizons)
643.Parametrize the curve by its length as(x(s), y(s), z(s)),0≤ s≤ L. Then the
coordinates(ξ,η,ζ)of its spherical image are given by