Fourier's Heat Law In Integral Form I am having a little trouble with something. Here is a link (wikipedia article) to Fourier's Heat Law in integral form: http://en.wikipedia.org/wiki/Thermal_conduction#Integral_form
What I am trying to do is apply this formula to the one dimensional case. According to my understanding, in this case, rather than integrating over a differential area element, we will be integrating over a length of some object, say a thin rod. Also, I want to integrate this function with respect to time, over the interval $[0,\tau]$, where $ \tau$ is the time after which the rod comes to thermal equilibrium with the environment. 
Here is what I am having trouble with:
$
\displaystyle \frac{\partial Q}{\partial t} = - k \oint \vec{\nabla} T \cdot d \vec{A}
$
$
\displaystyle \frac{\partial Q}{\partial t} = - k \int_{0}^{L} \frac{\partial T(x,t)}{\partial x} dx
$
$
\displaystyle \int_{0}^{\tau} \frac{\partial Q}{\partial t} ~dt = \int_{0}^{\tau} \int_{0}^{L} \frac{\partial T(x,t)}{\partial x} dx ~dt
$
My question is, what is $\frac{\partial T(x,t)}{\partial x} dx$. The definition of a differential of a multivariable function, in this particular case, is 
$
\displaystyle dT = \frac{\partial T}{\partial x}dx + \frac{\partial T}{\partial t} dt
$
If the second term on the l.h.s were zero, then $dT = \frac{\partial T}{\partial x}dx$. 
Am I thinking about this incorrectly?
 A: Assume that the cross-section of the thin rod is constant throughout its length. If the temperature gradient $\frac{\partial T}{\partial x}$ at $x=a$ is positive, then temperature is increasing with $x$, which means that heat is flowing to the left. So, the heat flow to the right is proportional to $-K\frac{\partial T}{\partial x}$. If you look at a section of the rod $a \le x \le b$, the net rate of heat flowing into the bar is
$$
          -K\left[\left.\frac{\partial T}{\partial x}\right|_{x=a}-\left.\frac{\partial T}{\partial x}\right|_{x=b}\right] = K\int_{a}^{b}\frac{\partial^{2}T}{\partial x^{2}}\,dx.
$$
The outside of the rod is assumed to be insulated. So the expression on the left above takes the place of the integral $\int\nabla T\cdot d\vec{A}$. The outward normal points to the left at $x=a$ and to the right at $x=b$, accounting from the subtraction of the values of $\frac{\partial T}{\partial x}$ at the endpoints. No integral over the outside face is needed, because the bar is assumed to be insulated. The balance-of-heat equation is found by equating the above to the time rate of change of the total heat $\int_{a}^{b}CT(x,t)\,dx$, where $C$ is the heat capacity $(heat/temp)/length$ of the material of the rod. The final equation is
$$
           \frac{\partial}{\partial t}\int_{a}^{b}CT(x,t)\,dx=K\int_{a}^{b}\frac{\partial^{2}T}{\partial x^{2}}\,dx.
$$
