Find the volume of the described solid.
The base of a solid is the region between the curve y=4cos x and the x-axis from x=0 to x= pi/2. The cross sections perpendicular to the x-axis are squares with bases running from the x-axis to the curve.
Please show work.
Calculus: Find the volume of the described solid.?
2014-04-11 12:49 pm
回答 (1)
2014-04-11 7:12 pm
Firstly make a sketch of the base region, and draw squares on the area, of thickness dx
The limits are 0 to π/2 and the side of the square is (cos x - 0)
so the volume element is :
dV = (4cos x )^2 dx = 16 ( cosx)^2
so the total volume is V = ∫dV = 16∫ (cosx)^2 dx from 0 to π/2
the integral of ( is [x/2 +sin 2x /4 ]
from a table, or can also be found by "parts" ...
( best to memorize this and sin x)^2 as they are so very common )
so V = 16[x/2 +sin2x4] from 0 to π/2
V = 16(π/4 +sin2(π/2) /4 -0 -0
and sin π = 0 so
V = 16.π/4 = 4π = 12.56633... <--- answers
==============
Confirmation ..
Using something like a TI-84, the built in integrator gives
ans = fnInt (4cos x)^2 ,x , 0 , π/2) = 12.566....
The limits are 0 to π/2 and the side of the square is (cos x - 0)
so the volume element is :
dV = (4cos x )^2 dx = 16 ( cosx)^2
so the total volume is V = ∫dV = 16∫ (cosx)^2 dx from 0 to π/2
the integral of ( is [x/2 +sin 2x /4 ]
from a table, or can also be found by "parts" ...
( best to memorize this and sin x)^2 as they are so very common )
so V = 16[x/2 +sin2x4] from 0 to π/2
V = 16(π/4 +sin2(π/2) /4 -0 -0
and sin π = 0 so
V = 16.π/4 = 4π = 12.56633... <--- answers
==============
Confirmation ..
Using something like a TI-84, the built in integrator gives
ans = fnInt (4cos x)^2 ,x , 0 , π/2) = 12.566....
參考: Retired Calculus Teacher
收錄日期: 2021-04-13 21:17:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140411044928AA3MN6N