Derivation of Heat Equation for frustum-shaped rod

[telercoi]

Homework Statement

Derive the Heat Equation for a rod in the shape of a frustum. Assume the specific heat c and density p are all constant. Use the "exact" method (through an integral) to derive the heat equation. Also, there is no heat source in the rod.

Homework Equations

The cross-section area for this rod is not constant, thus A=A(x)

The Attempt at a Solution

\frac{\mathrm{d} }{\mathrm{d} t} \int_{a}^{b} e(x,t)A(x)dx

= \Phi(a,t)A(a) - \Phi(b,t)A(b)

=-\int_{a}^{b}\frac{\mathrm{d} }{\mathrm{d} x}(\Phi(x,t)A(x))dx

=-\int_{a}^{b}\frac{\mathrm{d} \Phi(x,t)A(x))}{\mathrm{d} x} + \Phi(x,t)A'(x)dx

And this is as far as I've gotten. I think the last step may be incorrect as well; it's quite different from anything we've done in class. Can anyone help please?

 

[LCKurtz]

Homework Statement

Derive the Heat Equation for a rod in the shape of a frustum. Assume the specific heat c and density p are all constant. Use the "exact" method (through an integral) to derive the heat equation. Also, there is no heat source in the rod.

Homework Equations

The cross-section area for this rod is not constant, thus A=A(x)

The Attempt at a Solution

\frac{\mathrm{d} }{\mathrm{d} t} \int_{a}^{b} e(x,t)A(x)dx

= \Phi(a,t)A(a) - \Phi(b,t)A(b)

=-\int_{a}^{b}\frac{\mathrm{d} }{\mathrm{d} x}(\Phi(x,t)A(x))dx

=-\int_{a}^{b}\frac{\mathrm{d} \Phi(x,t)A(x))}{\mathrm{d} x} + \Phi(x,t)A'(x)dx

And this is as far as I've gotten. I think the last step may be incorrect as well; it's quite different from anything we've done in class. Can anyone help please?

$$\frac{\mathrm{d} }{\mathrm{d} t} \int_{a}^{b} e(x,t)A(x)dx<br /> <br /> = \Phi(a,t)A(a) - \Phi(b,t)A(b)<br /> <br /> =-\int_{a}^{b}\frac{\mathrm{d} }{\mathrm{d} x}(\Phi(x,t)A(x))dx<br /> <br /> =-\int_{a}^{b}\frac{\mathrm{d} \Phi(x,t)A(x))}{\mathrm{d} x} + \Phi(x,t)A'(x)dx$$

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