Consider a concentric spherical shell of radius $r$ and thickness $\mathrm{d}r$ as shown in diagram. The radial rate of flow of heat through this shell in steady state will be,



$\mathit{\text{H}}=\frac{\text{d}\mathit{\text{Q}}}{\text{d}\mathit{\text{t}}}=-\mathit{\text{KA}}\frac{\text{d}\mathit{\text{θ}}}{\text{d}\mathit{\text{r}}}$

[Negative sign is used because, with an increase in $r$, $\theta$ decreases.]

Now as for spherical shell $A=4\pi r^2$

So$\mathit{\text{H}}=-4\text{π}{\mathit{\text{r}}}^2\mathit{\text{K}}\frac{\text{d}\mathit{\text{θ}}}{\text{d}\mathit{\text{r}}}\text{or}{\int }_a^b\frac{\text{d}\mathit{\text{r}}}{{\mathit{\text{r}}}^2}=-\frac{4\text{π}\mathit{\text{K}}}{\mathit{\text{H}}}{\int }_{{\mathit{\text{θ}}}_1}^{{\mathit{\text{θ}}}_2}\text{d}\mathit{\text{θ}}$

which on integration and simplification gives

$\mathit{\text{H}}=\frac{\text{d}\mathit{\text{Q}}}{\text{d}\mathit{\text{t}}}=\mathit{\text{K}}\frac{4\text{π}ab\left({\mathit{\text{θ}}}_1-{\mathit{\text{θ}}}_2\right)}{\left(b-a\right)}\text{...(i)}$

Now in steady state, as no heat is absorbed, the rate of loss of heat by conduction is equal to that of supply, i.e., $H = P$, and here

          ${\theta }_1-{\theta }_2=T$   and   $\mathit{\text{a}}\simeq \mathit{\text{b}}=\mathit{\text{R}}$

So Equation (i) becomes,

          $\mathit{\text{P}}=\frac{4\pi {\mathit{\text{R}}}^2\mathit{\text{KT}}}{\left(\mathit{\text{b}}-\mathit{\text{a}}\right)}$

i.e thickness of shell is $\left(b-a\right)=\frac{4\pi {\mathit{\text{R}}}^2\mathit{\text{KT}}}{\mathit{\text{P}}}$