Surface Integrals and Average Surface Temperature of a Torus

[modafroman]

Homework Statement

A torus is a surface obtained by rotating a circle about a straight line. (It looks like a
doughnut.) If the z-axis is the axis of rotation and the circle has radius b, centre (0, a, 0)
with a > b, and lies in y − z plane, the torus obtained has the parametric form
r(u, v) = (a + b cos v) cos u i + (a + b cos v) sin u j + b sin v k
with 0 <= u, v < 2pi. Consider such a torus with the surface temperature given by
T(x, y, z) = 1 + z^2.
Calculate the average surface temperature.

Homework Equations

Average Value over surface = (1/area of s) * double int (s) f(x,y,z) dS

The Attempt at a Solution

Area of Torus = 4pi^2 R r = 4pi^2 ab (in given notation)

Surface integral for temperature => double int (s) t(x,y,z) dS, since T(x,y,z) = 1 + z^2, and z = b sin(v)

limits
0 < v < 2pi
0 < b < a (since b > 0 and a > 0

dS => rdrdtheta

=> double int (s) 1 + (b sin(v))^2 dS
=> int 0 to 2pi, int a to 0, b + b^(3)cos^(2)v db dv

which I solve and get 0.785398a^(2)(a^(2) + 4)

and then

Average Value over surface = (1/area of s) * double int (s) f(x,y,z) dS

=> (1/(4pi^2 ab)) * 0.785398a^(2)(a^(2) + 4)
=> 0.0198944a(a^(2) + 4) / b

Now, I'm not sure I've done the right thing for the surface integral, I wasn't sure what the limits of the surface integral were supposed to be, and if I was supposed to substitute the temperature equation for the z part of the area of the torus equation...

Edit: Wait, do I integrate with respect to u and v? If I'm supposed to, then what are the limits for u and v? I gather its 0 < v < 2pi, but what's the upper limit for u, since it only gives that u > 0?

And if that's the case, do I still substitute t(x,y,z) = 1 + z^2, for z = b sin v, in which case b becomes a constant?

Help please?

Thanks guys :)

 

[HallsofIvy]

Homework Statement

A torus is a surface obtained by rotating a circle about a straight line. (It looks like a
doughnut.) If the z-axis is the axis of rotation and the circle has radius b, centre (0, a, 0)
with a > b, and lies in y − z plane, the torus obtained has the parametric form
r(u, v) = (a + b cos v) cos u i + (a + b cos v) sin u j + b sin v k

So a and b are constants, u and v are the parameters determining a specific point on the torus.

with 0 <= u, v < 2pi. Consider such a torus with the surface temperature given by
T(x, y, z) = 1 + z^2.
Calculate the average surface temperature.

Homework Equations

Average Value over surface = (1/area of s) * double int (s) f(x,y,z) dS

The Attempt at a Solution

Area of Torus = 4pi^2 R r = 4pi^2 ab (in given notation)

Surface integral for temperature => double int (s) t(x,y,z) dS, since T(x,y,z) = 1 + z^2, and z = b sin(v)

limits
0 < v < 2pi
0 < b < a (since b > 0 and a > 0

b is not a parameter, it is a constant. The parameter u goes from 0 to $2\pi/$ also.

dS => rdrdtheta

NO! For one thing, there is no "r" in your formulas nor is there any "$\theta$! For another, that is the differential of area in polar coordinates which has nothing to do with this problem. You want the differential of surface area for a torus. Do you know how to find the differential of surface area for a given surface?

=> double int (s) 1 + (b sin(v))^2 dS
=> int 0 to 2pi, int a to 0, b + b^(3)cos^(2)v db dv

What, exactly is "dS"? It certainly cannot involve "db"- b is a constant, not a parameter.

which I solve and get 0.785398a^(2)(a^(2) + 4)

and then

Average Value over surface = (1/area of s) * double int (s) f(x,y,z) dS

=> (1/(4pi^2 ab)) * 0.785398a^(2)(a^(2) + 4)
=> 0.0198944a(a^(2) + 4) / b

Now, I'm not sure I've done the right thing for the surface integral, I wasn't sure what the limits of the surface integral were supposed to be, and if I was supposed to substitute the temperature equation for the z part of the area of the torus equation...

Edit: Wait, do I integrate with respect to u and v? If I'm supposed to, then what are the limits for u and v? I gather its 0 < v < 2pi, but what's the upper limit for u, since it only gives that u > 0?

And if that's the case, do I still substitute t(x,y,z) = 1 + z^2, for z = b sin v, in which case b becomes a constant?

Help please?

Thanks guys :)

 

[modafroman]

So a and b are constants, u and v are the parameters determining a specific point on the torus.What, exactly is "dS"? It certainly cannot involve "db"- b is a constant, not a parameter.

I was reading my notes and there was a similar question where he used dS, but I think that was just because it was in polar co-ords, which confused me.

I think I ended up getting it out, where I did:

int int T(u,v) du dv => int 0->2pi int 0->2pi 1 + b^2 sin^2 (v) du dv

where T(u,v) was T(x,y,z) with the parameterised form substituted in, so it was 1 + b^2 sin^2 v, and using the limits 0 < u < 2pi and 0 < v < 2pi, I ended up getting a nice answer of

2 pi^2(b^2 + 2)

so my answer for the average surface temperature came out to be:
(b^2 + 2)/2ab.

Sounds right, yea?

 

[HallsofIvy]

No, the differential of surface area is NOT just "dudv".

 

[modafroman]

No, the differential of surface area is NOT just "dudv".

Well then I'm lost as to what else it could be. None of my notes say anything differently than what I have done (in post 3, I see now what I did in the OP is crazy wrong :p)(even tho the example given uses polar co-ordinates)...

The only other thing I can find after some googling is this page: http://math.etsu.edu/multicalc/Chap5/Chap5-5/index.htm

that says that it also involves ||ru x rv || in the integral, but where do these come from? and is that what I'm supposed to be using?

Thanks for you help so far :)