Parameterization of a torus problem

[mcafej]

Homework Statement

Consider the parametrization of the torus given by:
x = x(s, t) = (3 + cos(s)) cos(t)
y = y(s, t) = (3 + cos(s)) sin(t)
z = z(s, t) = sin(s),
for 0 ≤ s, t ≤ 2π.

(a). What is the radius of the circle that runs though the center of the tube, and what is
the radius of the tube, measured from the central circle?
(b). Compute the rate of coverage of the torus by the parametrization. (Look for simplifi-
cations using trig identitites.)
(c). The parametrization is 1-1. Compute the area of the torus.

Homework Equations

Rate of coverage
|<∂x/∂s, ∂y/∂s, ∂z/∂s>x<∂x/∂t, ∂y/∂t, ∂z/∂t>|
(if this is hard to read, its basically the magnitude of the cross product of the velocity vectors)

Given that the transformation is 1-1 on a finite region D, the area on the s-t plane covered by the parameterized surface is

∫∫|<∂x/∂s, ∂y/∂s, ∂z/∂s>x<∂x/∂t, ∂y/∂t, ∂z/∂t>| dsdt

(in other words, the area covered by the torus is going to be the double integral of the rate of coverage with respect to s and t)

The Attempt at a Solution

Part b and c are pretty straight forward. For part B, I use the equation he gave us, for rate of coverage (listed above). Finding the partials, the cross product and the magnitude is pretty straight forward (although they are very confusing). I ended up with

Cross product of the velocity vectors is
<-cos(s)cos(t)(3+cos(s)), -sin(t)cos(s)(3+cos(s)), (-3-cos(s))(sin(s)cos²(t)+sin²(t)cos(s)>

Then I need to take the magnitude of that, but I can't figure out how to simplify it using trig identities (taking the magnitude of that mess would be a nightmare, but I may need to just sit down and do it, is there any easier way?). The last one looks like I should be able to simplify more cause I have the sin² and cos², but I can't figure out how to because they are both attached to a sin(s) and cos(s).

For part c, I just take my answer from part b and plug it into the integral. The hard part about this question is the bounds of the integral. I know that 0≤s and t≤2π, but I have know idea what the upper bound for s should be and what the lower bound for t should be.

As for part a, I really have no clue how to even get started on this one. I have a feeling it has something to do with the 3 in the equation, but I really don't know. I'm not sure which of s or t relates to the circle that runs through the center of the tube, and which is related to the radius of the tube itself. Sorry if this is a little confusing. If you have any questions about anything, just let me know.

 

[LCKurtz]

Homework Statement

Consider the parametrization of the torus given by:
x = x(s, t) = (3 + cos(s)) cos(t)
y = y(s, t) = (3 + cos(s)) sin(t)
z = z(s, t) = sin(s),
for 0 ≤ s, t ≤ 2π.

(a). What is the radius of the circle that runs though the center of the tube, and what is
the radius of the tube, measured from the central circle?
As for part a, I really have no clue how to even get started on this one. I have a feeling it has something to do with the 3 in the equation, but I really don't know. I'm not sure which of s or t relates to the circle that runs through the center of the tube, and which is related to the radius of the tube itself. Sorry if this is a little confusing. If you have any questions about anything, just let me know.

Look at ##x^2+y^2## and see how it varies as a function of ##s## and a function of ##t## to get an idea about the two radii.

 

[mcafej]

Ok, so I just looked up the formula for the radii of a torus, and I just explained how I got 1 as the radius of the circle of the tube, and 3 as the radius of the torus itself. As for the Area of coverage, I was able to simplify it down and for the Area you end up just getting the double integral from 0 to 2pi (for both s and t) of 3+cost. I ended up getting 12pi^2 for the integral, which is the surface area of the torus (which is correct because the surface area of a torus and it is 4rRpi^2, with r being the radius of the tube and R being the radius of the torus, so 4*3*1*pi^2=12pi^2 as needed). The response before was helpful about deriving the radii of the torus, thank you.