The shaded region enclosed by the circile x^2+y^2=a^2 and the the line
y=-a+h,where 0<h<=a ,is revolution about the y-axis.Prove that the volume of the solid of revolution is π/3h^2(3a-h) (Proved)
A hemispherical bowl of inner radius 6 cm contains water to depth of 2 cm.
Q1.) When a sphere of radius 3 cm is now placed in the bowl so that it sinks into the water. Find the rise in the water level.
The answer is 2/3 cm.
Application Definite Integral
2010-01-28 5:11 am
回答 (1)
2010-01-28 6:13 am
✔ 最佳答案
Q0:By disk method, V=∫[-a~-a+h] π[√(a^2-y^2)]^2 dy
=∫[-a~-a+h] π(a^2-y^2) dy
= π(a^2 y - y^3/3) for y=-a~ -a+h
= π{a^2*h-[(-a+h)^3+a^3]/3 } = (π/3)*h^2(3a-h)
Note: 0<= h <= 2a
Q1:
Suppose that the water level rise h, then the water level is 2+h, thus
(π/3)*4*(3*6-2)+(π/3)(2+h)^2*[3*3-(2+h)]=(π/3)(2+h)^2*[3*6-(2+h)]
64+(2+h)^2*(7-h)=(2+h)^2*(16-h)
64=(2+h)^2*9
8=3(2+h), so h=2/3 (cm)
收錄日期: 2021-04-30 14:14:16
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