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

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Answers
inertia /=Y 2jttf 1 r * dr = ' ia /=Y 2jttf ( 1
\ ^) r * dr = '
^lr^
where Y is the densit y
of the
2
cone. 1747. / = - Ma 2 . Solution. We partition the sphere into elementary
5
cylindrical tubes, the axis of which is the given diameter. An elementary
/-r 2
"
volume dV 2nrhdr, where r is the radius of a tube, h = 2a I/ 1 _ --a
f a2 is its altitude. Then the moment of inertia I = 4nay I I/ 1 L./-*dr = A na'v,
4
where y is t e density of the sphere, and since the mass M = ita'y, it folo
lows that y = |- Ma 2 . 1748. V--=2n 2 a 2
b\ S--=4n 2 ab. 1749. a) 7-=//~=4 a;
o o
9 4 r
b) x==(/ = T7 -n. 1750. a) x^=0, f/^-rr Hint. The coordinate axes are cho-
1U o 71
sen so that the jc-axis coincides with the diameter and the origin is the
centre of the circle; b) ^=4o Solution. The volume of the solid a double
cone obtained from rotating a triangle about its base, is equal to V -- rft/i 2
o
,
where b is the base, h is the altitude of the triangle. By the Guldm theo-
rem, the same volume V 2Jtx -z-b'i. where x is the distance of tlu> centre
H vt' 2
of gravity trom the base. \\hence x = -
1751. v t ^ .
i 2
1752. -lnl+--. 1753. x=t: D flt,= -f 1754.
= ~^/? 2 // 2
Hint. The elementary force (force of gravity) is equal to the
weight of water in the volume of a layer cf thickness dx, that is, dF =
2 = ynR dx, where y is the weight of unit volume of water. Hence, the elementary
work of a force dA ynR'* (H x} dx, where x is the water level.
1757. A=ryR 2H 2 . 1758. A
---^ R*TM ^Q 79-10* -0 79- 10 7
l^& tn0h
1759. A = ynR*H. 1760. ^ = ; ^
on a mass m is equal to F = k
kgm.
/lo.-mg/?. Solution. The force acting
j- , where r is the distance from the centre
of the earth. Since for r = R we have F=mg f it follows that kM=gR 2 .
R + h.
. The
- =
, )
sought-for work will have the form A= \ kr-dr kmM { -^-
J * \KK-\-nJ
m^- When/i = oo we have A = mgR. 1761. 1.8-10 4
ergs. Solution.