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Putnam
1940 Putnam
B2
B2
Part of
1940 Putnam
Problems
(1)
Putnam 1940 B2
Source: Putnam 1940
2/22/2022
A cylindrical hole of radius
r
r
r
is bored through a cylinder of radiues
R
R
R
(
r
≤
R
r\leq R
r
≤
R
) so that the axes intersect at right angles. i) Show that the area of the larger cylinder which is inside the smaller one can be expressed in the form
S
=
8
r
2
∫
0
1
1
−
v
2
(
1
−
v
2
)
(
1
−
m
2
v
2
)
d
v
,
where
m
=
r
R
.
S=8r^2\int_{0}^{1} \frac{1-v^{2}}{\sqrt{(1-v^2)(1-m^2 v^2)}}\;dv,\;\; \text{where} \;\; m=\frac{r}{R}.
S
=
8
r
2
∫
0
1
(
1
−
v
2
)
(
1
−
m
2
v
2
)
1
−
v
2
d
v
,
where
m
=
R
r
.
ii) If
K
=
∫
0
1
1
(
1
−
v
2
)
(
1
−
m
2
v
2
)
d
v
K=\int_{0}^{1} \frac{1}{\sqrt{(1-v^2)(1-m^2 v^2)}}\;dv
K
=
∫
0
1
(
1
−
v
2
)
(
1
−
m
2
v
2
)
1
d
v
and
E
=
∫
0
1
1
−
m
2
v
2
1
−
v
2
d
v
E=\int_{0}^{1} \sqrt{\frac{1-m^2 v^2}{1-v^2 }}dv
E
=
∫
0
1
1
−
v
2
1
−
m
2
v
2
d
v
. show that
S
=
8
[
R
2
E
−
(
R
2
−
r
2
)
K
]
.
S=8[R^2 E - (R^2 - r^2 )K].
S
=
8
[
R
2
E
−
(
R
2
−
r
2
)
K
]
.
Putnam
geometry