New trend volume 2
Application of definite Intefrals
Revision Exercise
Q25
In Fig. 23(a),the shad3ed region is bounded by the circle x^2+y^2=r^2,the x-axis ,the y-axis and the lind Y=-h where h>0.,If the shaded region is revolved aobut the y-axis ,show that the volume of the solid generated is (r^2-h^3/3)pai cubic units
(b) In Fig.23(b),A and C are points on the x-axis and y-axis respectively,AB is an arc of the circle 3x^2+3y^2=89 and BC is a segment of the liine y=-1.A mould is formed by revolving AB and BC about the y-axis. Using (a). or otgherwise ,show that the capacity of the mould is 88pai/3 cubic units.
(C)A hemispherical pot of inner radius 4 units is completely filled with molten gold>(See Fig 23(c).)The molten gold is then poured into the mould mentioned in (b) by steadily tilting the pot . Suppose the pot is tilted through an angle X and G is the centre of the rim of the pot.
(i)find ,in trms of x.
(1)the distance between G and the surface of the molten gold remaining in the pot
(2)the volume of gold pouted into the mould
(ii)When the mould is completely filled with molten gold,show that
8(sin x)^3-24sinx +11=0
Hence fin the value of x
Amath(application of definite
2009-01-29 8:17 pm
回答 (1)
2009-01-30 1:05 am
✔ 最佳答案
(a)x^2+y^2=r^2
x^2=(r^2-y^2)
the volume of the solid generated
=π∫(r^2-y^2) dy [from 0 to -h]
= π(r^2y-y^3/3)
=(r^2h-h^3/3)π cubic units
(b)
Substitute h=1 and r^2=89/3
the capacity of the mould
=(89/3-1/3)π
=88π/3 cubic units
(c)(i)
The distance between G and the surface of the molten gold remaining in the pot
=4 sinx
(ii)
the volume of gold pouted into the mould
=(64sin x^2-64 sin^3 x/3)π cubic units
(iii)
Let
(64sin x^2-64 sin^3 x/3)π= (88/3)π
(192sin x^2-64 sin^3 x)= 88
8(sin x)^3-24sinx +11=0
(2sinx-1)(4sin^2x+2sinx-11)=0
sinx=1/2
x=π/6
收錄日期: 2021-04-26 13:04:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090129000051KK00579