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658 CHAPTER 18 The Potential Equation

so
2n + 1 1
a n = f (arccos(x))P n (x)dx.
2R n −1
The steady-state temperature function is
∞ 1 n
2n + 1 ρ

u(ρ,ϕ) = f (arccos(x))P n (x)dx P n (cos(ϕ)).
2 R
n=0 −1
EXAMPLE 18.6
For f (ϕ) = ϕ, the solution is
∞ 1 n
2n + 1 ρ

u(ρ,ϕ) = arccos(x)P n (x)dx P n (cos(ϕ)).
2 R
n=0 −1
We will use numerical integrations to approximate the first six coefficients. P 0 (x),··· , P 5 (x)
were listed in Chapter 15. Using these and MAPLE to perform the computations, we obtain
1
arccos(x)P 0 (x)dx ≈ π,
−1
1
x arccos(x)dx, ≈−0.7854,
−1
1 2
1
(3x − 1)arccos(x)dx = 0,
−1 2
1
1
3
(5x − 3x)arccos(x)dx ≈−.049087,
−1 2
1
1
2
4
(35x − 30x + 3)arccos(x)dx = 0,
−1 8
1
1 5 3
(63x − 70x + 15x)arccos(x)dx ≈−0.012272.
−1 8
Then
π 3 ρ 7 1 ρ
3
3
u(ρ,ϕ) ≈ − (0.7854) cos(ϕ) − (0.049087) (5cos (ϕ) − 3cos(ϕ))
2 2 R 2 2 R
ρ
11 5 1
3
5
− (0.012272) (63cos (ϕ) − 70cos (ϕ) + 15cos(ϕ)).
2 R 8
SECTION 18.7 PROBLEMS
In each of Problems 1 through 4, write a solution for the 4. f (ϕ) = 2 − ϕ 2
steady-state temperature distribution in the sphere if the 5. Solve for the steady-state temperature distribution in a
boundary data is given by f (ϕ). Use numerical integration hollowed-out sphere given in spherical coordinates by
to approximate the first six terms in the solution. R 1 ≤ ρ ≤ R 2 . The inner surface is kept at constant tem-
perature T and the outer surface at temperature zero.
2
1. f (ϕ) = Aϕ ,inwhich A is a positive number. Assume that u is a function of ρ and ϕ only. Approxi-
2. f (ϕ) = sin(ϕ) mate the solution with a sum of the first six terms, using
3. f (ϕ) = ϕ 3 numerical integration to approximate the coefficients.

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October 14, 2010 15:27 THM/NEIL Page-658 27410_18_ch18_p641-666