Laplacian with Integral BC(s) I want to solve the three-dimensional laplacian 
$$\nabla^{2} T = 0$$ where $\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$
defined on $x\in[0,L],y\in[0,l],z\in[-w,0]$ with the following boundary conditions 
$$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 \rightarrow Neumann$$
$$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0\rightarrow Neumann$$
$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) -  \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg) \rightarrow Convection$$
$$\frac{\partial T(x,y,0)}{\partial z}=p_c\bigg(\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y -T(x,y,0) \bigg)\rightarrow Convection$$
$b_h,b_c,p_h,p_c,L,l$ are constants $>0$
Addendum The physical situation that this problem describes has two additional pieces of information
$\rightarrow$ At $(0,y,-w)$ $\rightarrow$ $\frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x = T_{h,i}$ a constant
and
At $(x,0,0)$ $\rightarrow$ $\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y = T_{c,i}$ a constant 
Attempt
I know that for the 3D Laplacian,, when separation of variables is applied the following holds
$k_x^{2} + k_y^{2} = k_z^{2}$
So if $X(x)Y(y)Z(z)$ is substituted as $T$ we can say $X \sim \cos(n\pi x)$, $Y \sim \cos(m \pi y)$. I know that Z can be  some hyperbolic function like $[A_{m,n}\cosh(\pi \sqrt{n^2 + m^2}) + B_{m,n}\cosh(\pi \sqrt{n^2 + m^2})]$
But when the solution form is substituted onto the $z$ bc, what i get is just two homogeneous linear equations in $A_{m,n}$ and $B_{m,n}$. From this point, I cannot proceed. 
Any help or guidance on how this problem can be tackled will be appreciated. 
Attempt 2 
As suggested by @DisintegratingByParts I need to divide the problem into two parts. 
Let us take Problem 1.
As already mentioned,I have the additional info that:
At $(0,y,-w)$ $\rightarrow$ $\frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x = T_{h,i}$. 
So we can write 
$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) -  \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg)  $$
as
$$\frac{\partial T(0,y,-w)}{\partial z}=p_h\bigg( T(0,y,-w) -  T_{h,i} \bigg)$$
So now i have the boundary condition (more like an edge condition in 3D) without the integral on the RHS. 
Can this BC be used along with the others to form a solution of Problem 1 or am i wrong in some way ?

On applying the third BC from Problem 1 but at $(x,y,-w)$ I arrive at
$$
\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\sinh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) = p_h\bigg(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{m\pi y}{l})\cos(\frac{n\pi x}{L})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) - \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{A_{n,m}b_h}{b_h^2 + n^2 \pi^2 }\cos(\frac{m\pi y}{l})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg)\bigg[b_h\cos(\frac{n\pi x}{L}) + n\pi\sin(\frac{n\pi x}{L}) \bigg]\bigg)
$$
This needs to be now solved for Fourier coefficients $A_{n,m}$. I know that we have to somehow use the property of orthogonality of the Eigen functions, but the problem pertains from the fact that $A_{n,m}$ exist on both sides of the equation.
 A: You can solve two problems and sum the two solutions:
Problem 1
$$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 \rightarrow Neumann$$
$$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0\rightarrow Neumann$$
$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) -  \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg) \rightarrow Convection$$

$$
\frac{\partial T(x,y,0)}{\partial z} = 0.
$$
Problem 2
$$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 \rightarrow Neumann$$
$$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0\rightarrow Neumann$$

$$
   \frac{\partial T(x,y,-w)}{\partial z} = 0.
$$
$$\frac{\partial T(x,y,0)}{\partial z}=p_c\bigg(\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y -T(x,y,0) \bigg)\rightarrow Convection$$
The Problem 1 Form of solution is
$$
        T(x,y,z)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\cosh\left(\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}z\right).
$$
The constants $A_{n,m}$ are determined as Fourier coefficients through the third condition at $z=-w$.
