# How to apply Leibniz's Rule to a Metal Pipe Temperature's Partial Derivative in this Example

Refer to this image showing the temperature of a metal pipe at the inlet and the outlet: The temperature $$T(z,t)$$ is a function of the length $$z$$ and time $$t$$. Let the average temperature be
$$T_mathrm{avg}(t)=frac{1}{2}big(T(a,t)+T(b,t)big).$$
Integrating the partial derivative of $$T$$ with respect to $$t$$, $$frac{partial T}{partial t}$$ over the entire length of the pipe (from $$z=a$$ to $$z=b$$) and applying Leibniz’s rule, we should get
$$int_{z=a}^{z=b}frac{partial}{partial t}T(z,t)dz=frac{d}{dt}int_{z=a}^{z=b}T(z,t)dz-T(b,t)frac{db}{dt}+T(a,t)frac{da}{at}.$$
In this paper (eqs. (21) and (22)), the result is reported as follows:
$$int_{z=a}^{z=b}frac{partial}{partial t}T(z,t)dz=(b-a)frac{d}{dt}T_mathrm{avg}(t)+big(T(a,t)-T_mathrm{avg}(t)big)frac{da}{dt}+big(T_mathrm{avg}(t)-T(b,t)big)frac{db}{dt}.$$

Are the two equivalent? How to correctly solve this integral? Thanks!

Note: The integral in this question is only a portion of a larger integration problem in the energy balance, not reported for brevity. Equations (21) and (22) in the reference describe the complete energy balance.

This answer is based on the original answer here. Some supplemental details relevant to the problem are added.

Referring to the original problem, let $$T(z,t)$$ vary linearly with $$z$$ as shown in the following figure: As proposed in the original solution, integrating integral of the partial derivative of $$T$$ from $$a$$ to $$b$$ and applying Leibniz's Rule: begin{align} int_{a}^{b}frac{partial}{partial t}T(z,t)dz&=int_{a}^{b}frac{partial}{partial t}big(T(z,t)-T_mathrm{avg}(t)+T_mathrm{avg}(t)big)dz\ &=int_{a}^{b}frac{partial}{partial t}T_mathrm{avg}(t)dz + underbrace{int_{a}^{b}frac{partial}{partial t}big(T(z,t)-T_mathrm{avg}(t)big)dz}_text{apply Leibniz Rule}\ &=(b-a)frac{d}{dt}T_mathrm{avg}(t)\ &quad+underbrace{frac{d}{dt}int_{a}^{b}big(T(z,t)-T_mathrm{avg}(t)big)dz+big(T(a,t)-T_mathrm{avg}(t)big)frac{da}{dt}+big(T_mathrm{avg}(t)-T(b,t)big)frac{db}{dt}.}_text{Leibniz's Rule applied} end{align}

The area under the curve in the figure (shaded green) can be found using the integral $$int_{a}^{b}T(z,t)dz$$. Approximating the curve with a straight line, we can write

begin{align} int_{a}^{b}T(z,t)dz&approxunderbrace{frac{1}{2}(b-a)big(T(b,t)-T(a,t)big)}_text{area of top triangle}+underbrace{T(a,t)(b-a)}_text{area of bottom rectangle}\ &=frac{1}{2}big(T(a,t)+T(b,t)big)(b-a)=(b-a)T_mathrm{avg}(t). end{align}

Thus, the second term in the third equality of the first equation vanishes, and the desired result is obtained: $$int_{a}^{b}frac{partial}{partial t}T(z,t)dzapprox(b-a)frac{d}{dt}T_mathrm{avg}(t)+big(T(a,t)-T_mathrm{avg}(t)big)frac{da}{dt}+big(T_mathrm{avg}(t)-T(b,t)big)frac{db}{dt}.$$

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