# Homework Help: Parametrize a surface

1. Dec 8, 2008

### hils0005

1. The problem statement, all variables and given/known data
Parametrize a surface obtained by revolving the curve y = lnx, as x goes from 1 to 10 about the y axis

3. 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

2. Dec 8, 2008

### tiny-tim

Hi hils0005!
Nooo … how can x be lnx times something?

Draw a diagram (with pretty circles on ), and try again!

3. Dec 8, 2008

### 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

4. Dec 8, 2008

### tiny-tim

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!

5. Dec 8, 2008

### 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.