4 HW 4

1. Derive $\vec n {\rm d}{}S$ in terms of ${\rm d}\theta$ and ${\rm d}\phi$, where $(r,\theta,\phi)$ are spherical coordinates. Assume that the surface is described as $r=f(\theta,\phi)$ for some given function $f$. Use the formulae given earlier in class for $\vec n {\rm d}{}S$ in terms of two parameters $u=\theta$ and $v=\phi$. The formula requires you to differentiate $\vec r$ with respect to the parameters. Now in spherical,
   
   $$\vec r = r \hat\imath_r$$
   
   
   From class, the derivatives of $\hat\imath_r$ are
   
   $$\frac{\partial \hat\imath_r}{\partial r} = 0 \qquad \fra... ...al \hat\imath_r}{\partial\phi} = \sin\theta \hat\imath_\phi$$
   
   
   So you can now write $\vec{n} {\rm d}{}S$ in terms of $r$ and the derivatives $f_\theta$ and $f_\phi$ of function $f$.

   Next assume that the surface is not given as $r=f(\theta,\phi)$, but as $F(r,\theta,\phi)$ = constant. Rewrite your expression for $\vec{n} {\rm d}{}S$ in terms of $F$ instead of $f$. Hint: To get the derivatives of $f$ in terms of those of $F$, look at the total differential of $F$ at a point on the surface:
   

   $${\rm d}F \equiv \frac{\partial F}{\partial r} {\rm d}{}r +... ... {\rm d}\theta + \frac{\partial F}{\partial\phi} {\rm d}\phi$$
   
   
   Now if you take ${\rm d}{}r=f_\theta{\rm d}\theta+f_\phi{\rm d}\phi$, you stay on the surface, so ${\rm d}{}F$ will then be zero:
   
   $$0 = \frac{\partial F}{\partial r} (f_\theta{\rm d}\theta+f... ... {\rm d}\theta + \frac{\partial F}{\partial\phi} {\rm d}\phi$$
   
   
   From this you can find $f_\theta$ and $f_\phi$ in terms of the derivatives of $F$, by taking ${\rm d}\phi$, respectively ${\rm d}\theta$ zero. Plug that into the earlier expression for $\vec{n} {\rm d}{}S$ in terms of $f$ and you have $\vec{n} {\rm d}{}S$ in terms of $F$. Write this expression in terms of the gradient of F in spherical coordinates, as given by the expression in your notes, or in any mathematical handbook. Compare with the Eulerian expression
   
   $$\vec n {\rm d}S = \frac{\nabla F}{F_z} {\rm d}x {\rm d}y$$
   
   
   as derived in class. Here ${\rm d}x{\rm d}y$ can be denoted symbolically as ${\rm d}S_z$: it is the area of a surface of constant $z$ of dimensions ${\rm d}x\times{\rm d}y$. (In other words, it is the projection of surface element ${\rm d}S$ on a surface of constant $z$.) What is the equivalent to ${\rm d}S_z$ in your spherical coordinates expression?

2. 1st Ed: p133, q56, 2nd Ed: p155, q56.

3. 1st Ed: p160, q38, 2nd Ed: p183, q38. Simplify as much as possible. Sketch each surface, taking the $z$-axis upwards.

4. Finish finding the derivatives of the unit vectors of the spherical coordinate system using the class formulae. Then finish 1st Ed p160 q47, 2nd Ed p183 q47, as started in class, by finding the acceleration. As noted in class,
   
   $$\frac{\partial {\hat\imath}_i}{\partial u_i} = \frac{1}{h_... ...frac{1}{h_i} \frac{\partial h_j}{\partial u_i} {\hat\imath}_j$$
   
   

5. Derive the heat equation for conduction in a bar physically. To do so, look at a small segment of length $\Delta x$ of the bar. Fourier's law says that the heat going into the segment at its left hand surface at position $x$ is given by
   
   $$q_1 = - k u_x(x,t) A \Delta t$$
   
   
   where $k$ is the material's heat conduction coefficient in kW/m K, $u$ the temperature, so $u_x(x,t)$ the partial derivative of the temperature with respect to $x$ at the considered position and time (x,t), $A$ is the cross sectional area of the bar, and $\Delta t$ a small increase in time. Similarly, the heat going out the segment at its right hand surface at position $x+\Delta x$ is given by
   
   $$q_2 = - k u_x(x + \Delta x,t) A \Delta t$$
   
   
   Now apply that the net heat going into the segment must be the increase in temperature of the segment in time interval $\Delta t$, which can be written as $u_t(x,t)\Delta t$ times its volume, times the specific heat capacity of the segment, which can be written in terms of the specific heat per unit mass $C_p$ of the material in kJ/kg K, its density $\rho$ in kg/m$^3$, and the volume of the segment. From that, obtain the heat equation. In particular, identify the value and units of $\kappa$ in terms of the material properties.