Parametrizing a Surface by Revolving y = lnx Around the y Axis

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SUMMARY

The discussion focuses on parametrizing the surface generated by revolving the curve defined by the equation y = ln(x) around the y-axis, for x values ranging from 1 to 10. The correct parametrization involves using cylindrical coordinates, where y = ln(r) and r = √(x² + z²). Participants emphasize the need for two parameters in the surface representation and clarify that x cannot simply equal ln(x) multiplied by another term. The final parametrization should be structured as x = r cos(θ), y = ln(r), and z = r sin(θ), with θ ranging from 0 to 2π and r from 1 to 10.

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Homework Statement


Parametrize a surface obtained by revolving the curve y = lnx, as x goes from 1 to 10 about the y axis



The Attempt at a Solution



y=lnx
x=lnxcos(theta)
z=lnxsin(theta)

vector r(x,theta) = lnx cos(theta) i + lnx j + lnxsin(theta) k

0\leq (theta) \leq 2pi
1\leq x \leq 10

I'm not really sure how to complete, or if this is correct-any ideas would be helpful, thanks
 
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Hi hils0005! :smile:
hils0005 said:
Parametrize a surface obtained by revolving the curve y = lnx, as x goes from 1 to 10 about the y axis

x=lnxcos(theta)

Nooo … how can x be lnx times something?

Draw a diagram (with pretty circles on :wink:), and try again! :smile:
 
Obviously I 'm confused by Parametrizations
the given function is in the form y= f(x,z)
x=cos(theta)
y=lnx
z=sin(theta)

theta from 0 to 2pi
x from 1 to 10
 
hils0005 said:
Obviously I 'm confused by Parametrizations
the given function is in the form y= f(x,z)
x=cos(theta)
y=lnx
z=sin(theta)

theta from 0 to 2pi
x from 1 to 10

hmm …

i] it's a surface, so it'll need two parameters

ii] how can x=cos(theta) and z=sin(theta)? that means x2 + z2 = 1.

Choose another parameter, and try again! :smile:
 
If y= ln(x) is revolved around the y-axis, then y= ln(r) where r= \sqrt{x^2+ z^2}.

You are going to have "tilt" your head and think about "cylindrical" coordinates where y rather than z is one of the coordinates.
 

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