Hello I try to present what I know the best.
We know that heat flow takes place in three forms :
- Conduction.
- Convection.
- Radiation.
For one dimensional heat flow, we have $q= k \dfrac{T_2-T_1}{L}$ , where $T_1$ and $T_2$ are terminal temperatures and $L$ is the length of the material. Switching to $3$-D spaces as you said, we need to look at the fourier system of the heat transfer before proceeding further. $\dfrac{q_x}{A}= - K \dfrac{dT}{dx}$. So integrating we obtain $\dfrac{q_x}{A}\large \int_{0}^{L} \ dx= -k \int_{T_1}^{T_2} \ dT.$
So given that the heat flow ( rate of conduction ) in the $3$-D space $(x,y,z)$ is given by the following equation $ q= - k \nabla T = - k \bigg( \hat{i} \dfrac{\partial T}{\partial x}+\hat{j} \dfrac{\partial T}{\partial y}+\hat{k} \dfrac{\partial T}{\partial y}\bigg).$The negative sign, there indicates the transfer of heat from one place to another. So you can simply substitute the heat equation in the place of $T$ and compute the partial derivatives to get the rate of flow of heat.
I can write up the cylindrical and spherical coordinate version too . It can be like this .
- For cylindrical coordinate system $(r,\phi)$ we have $ \large q= - k \nabla T = - k \bigg(\hat{i} \dfrac{\partial T}{\partial r}+\hat{j} \dfrac{1}{r} \dfrac{\partial T}{\partial y}+\hat{k} \dfrac{\partial T}{\partial Z}\bigg).$ and individual heat flows are given by $\large q_r = - K \dfrac{dT}{dx}, \\ \large q_{\phi} = - \dfrac{K}{r}\dfrac{dT}{d \phi},\\ \large q_{Z} = - K\dfrac{dT}{dZ}$ where $\large q_r$ represents the flow in $r$-th direction. Here we include an extra dimension $Z$ to make $2$-D into $3$-D.
- For Spherical coordinate system $(r,\theta,\phi)$ we have $ \large q= - k \nabla T = - k \bigg(\hat{i} \dfrac{\partial T}{\partial r}+\hat{j} \dfrac{1}{r} \dfrac{\partial T}{\partial \theta}+\dfrac{\hat{k}}{r\sin \theta} \dfrac{\partial T}{\partial \phi}\bigg).$
I hope this will be of some use to you. Feel free to ask if you want to hear more on anything.
Thank you.