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

[hils0005]

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

 

[tiny-tim]

Hi hils0005!

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 ), and try again!

 

[hils0005]

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

 

[tiny-tim]

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! ​

 

[HallsofIvy]

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.