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modafroman

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## 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 :)

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