## Essential University Physics: Volume 1 (3rd Edition)

We start with the given equation: $H = -kA \frac{dT}{dr}$ We know that the area of a cylinder is $2\pi rL$, so we find: $\frac{H}{2\pi R} dr = -kLdT$ Applying the integral on both sides, it follows: $\frac{HlnR_2 - HlnR_1}{2\pi}= -kL(T_1-T_2)$ $H = \frac{2\pi kL(T_1-T_2)}{ln(R_2/R_1)}$