Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

Answers 43 1
1724. ~jia 2 . 1725. 2ita 2
(2- /2~). 1726. na2 . 1727. Mx=- V
1728. M = ~- 2
; M = & ^.
1729. MX=M K =
^-
J=y = y. 1730. Mx=M K =-~a 2
; x = y = ~a. 1731. 2na 2 . 1732. x =
a 24-sinh2 ._00 asina - < _ . 4 ,__ 4a
^ss- I736t ^^SB- 1737< *~=Jia ;
a - 1738>
^=|-
tion. Divide the hemisphere into elementary spherical slices of area da by
horizontal planes. We have da = 2jiadz, where dz is the altitude of a slice.
a
2ft f az dz
o Ct
Whence z= - = = -?r. Due to symmetry, x = y = Q. 1739. At a dis-
2nfl z
2
tance of -j- altitude from the vertex of the cone. Solution. Partition the
cone into elements by planes parallel
to the base. The mass of an elemen-
z is the distance
tary layer (slice) is dm,- = YftQ 2 dz, wnere Y * s * ne density,
of the cutting plane from the vertex of the cone, Q = -J-Z. Whence
z = _^
h
Jt \ n ;
3 / 3 \
= h. 1740. f 0; 0; a )- Solution. Due to
+-gsymmetry.
5T-= #~=0. To determine 7 we partition the hemisphere into elementary
layers (slices) by planes parallel to the horizontal plane. The mass of such
an elementary layer dm ^nr^dz, where Y * s tne density, a is the distance
of the cutting plane from the base of the hemisphere, r= ^a 2
"33 11 4
15
a
[a 2
z 2
) zdz
11
o
4""
8
JW& ;
2 2
is the
radius of a cross-section. We have: z = =-Q-a. 1741. /==jta 3 .
1742. I a = -z-ab*\ Ib =z-^a s b. 1743. 7=-/?& 8 . 1744. / a = --
1745. /=yjt (#J /?J).
r I = -
b W T na*b.
"4 4
Solution. We partition the ring into elementary
concentric circles. The mass of each such element dm = y2n.rdr and
the moment of inertia 7 = 2nCrdr = y n(/?* #J);(Y ==1). 1746. / = ^nR*Hy.
Solution. We partition the cone into elementary cylindrical tubes parallel
to the axis of the cone. The volume of each such elementary tube is
dV = 2nrhdr, whe r e r is the radius of the tube (the distance to the axis of
the cone), h = H [ \ s - )
is the altitude of the tube; then the moment of