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rotation about the y-axis (1 Viewer)

crazyymonkey

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heres my question. please help if you know how to do this.


The Interior of the bowl which is to be used to hold the Olympic flame is shaped by rotating the arc of the curve y= loge x from x= 1 to x= 5 about the y-axis.

(i) SHow that the volume is given by V= (pi) S e^2y dy
[with b= ln5 and a= 0 ]

if that makes sense.

(ii) claculate the capacity of the bowl



please help. im stukc ont the first part. i hope it makes sense.
 

tommykins

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y = lnx
x = e^y

Rotation of y axis makes it pi int. x^2 dy

therefore the volume is pi int. e^2y dy

domains are found by subbing x = 5 and x = 1 to find y.

ii) v = pi int. e^2y dy
= pi[(e^2y)/2] sub in ln 5 and 0 and you have your answer.
 
Last edited:

Just.Snaz

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Rotation about the y-axis: make x the subject

Y = lnx

x = e^y

V = pi integrate (from a to b) [ x^2 ] dy

since you'll be integrating in terms of y, you need your limits to be y values

so when x = 1, y = ln1 = 0
x = 5, y = ln5

therefore,

V = pi integrate (from 0 to ln5) [ e^2y ] dy

= pi [ (1/2)e^2y](limits 0 to ln5)
= pi [ (1/2)e^2ln5 - (1/2)e^0)
= pi [ (1/2)x25 - 1/2)
= pi (12)
= 12 pi
 

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