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 :)