91Ó°ÊÓ

The radius of the sphere is 1.

When a conductor reaches a point where no more heat can be absorbed by the rod, it is said to be at a steady-state temperature.

Consider the equation below:

$\begin{array}{c}{u}_{r=1}=\left\{\begin{array}{l}100,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}0<\mathrm{θ}<\frac{\mathrm{Ï€}}{3}\\ 0°,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰otherwise}\end{array}\right.\\ =\left\{\begin{array}{l}100,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰}\frac{1}{2}<x<1\\ 0°, â\P ĉâ\P ĉâ\P ĉâ\P ĉâ\P ĉâ\P ĉâ\P ĉ   \mathrm{otherwise}\end{array}\right.\\ =100f\left(x\right)\end{array}$

Here,

role="math" localid="1664355824351" $f\left(x\right)=\left\{\begin{array}{l}1,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰â€‰â€‰}\frac{1}{2}<x<1\\ 0°,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰o³Ù³ó±ð°ù·É¾±²õ±ð}\end{array}\right.$

Take the equation

$c_m=\frac{2m+1}{2}\underset{−1}{\overset{1}{∫}}(5x^3+3x^2−3)p_m\left(x\right)dx$ ….. (1)

Take m = 0 and put in equation (1).

$\begin{array}{c}{c}_0=\frac{1}{2}\underset{1/2}{\overset{1}{∫}}dx\\ =\frac{1}{2}{\left[x\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\\ =\frac{1}{2}[1−\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.]\\ =\frac{1}{4}\end{array}$

Take m = 1 and put in equation (1).

$\begin{array}{c}{c}_1=\frac{3}{2}\underset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{1}{∫}}xdx\\ =\frac{3}{2}{\left[\frac{x^2}{2}\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\\ =\frac{3}{2}[\frac{1}{2}−\frac{1}{8}]\\ =\frac{9}{16}\end{array}$

Take m = 2 and put in equation (1).

$\begin{array}{c}{c}_2=\frac{5}{4}\underset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{1}{∫}}(3x^2−1)dx\\ =\frac{5}{4}{(3\frac{x^3}{3}−x)}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\\ =\frac{5}{4}(\frac{3}{3}−1−\frac{3}{12}+\frac{1}{2})\\ =\frac{15}{32}\end{array}$

Use the equation

$u=\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l{r}^l{p}_l\left(\cos \mathrm{θ}\right)$

$u=100(\frac{1}{4}{p}_0\left(\cos \mathrm{θ}\right)+\frac{9}{16}r{p}_1\left(\cos \mathrm{θ}\right)+\frac{15}{32}{r}^2{p}_2\left(\cos \mathrm{θ}\right)+……)$