4 02/06 F

1. 1st Ed: p103, q44, 2nd Ed: p123, q44. Do it using Stokes.

2. 1st Ed: p104, q62, 2nd Ed: p124, q62. Do the surface integrals both directly and using the divergence theorem. Make sure to include the flat circle of the cone. Note: in doing the surface integrals directly, you are required to write them down in Cartesian coordinates using the expression for $\vec{n} {\rm d}{}S$ given in class when $F(x,y,z)=0$. After that, switch to polar coordinates to actually do the integration.

3. MODIFIED version of 1st Ed: p132, q50, 2nd Ed: p154, q50. Given
   
   $$\vec v = \frac{(-y,x)}{x^2+y^2}$$
   
   
   1. Evaluate $\nabla\times\vec{v}$.
   2. Also evaluate, presumably using polar coordinates,
      
      $$\int_{\rm I} \vec v \cdot {\rm d}\vec r \qquad \int_{\rm II} \vec v \cdot {\rm d}\vec r$$
      
      
      where path I is the semi circle of radius $r$ going clockwise from $(r,0)$ to $(-r,0)$, and path II is the semi circle of radius $r$ going counter-clockwise from $(r,0)$ to $(-r,0)$.
   3. Explain why the integral over II minus the integral over I is the integral over the closed circle.
   4. Explain why Stokes implies that the closed contour integral should be the integral of the $z$-component of $\nabla\times\vec{v}$ over the inside of the circle.
   5. Then explain why you would then normally expect the contour integral to be zero. That means that the two integrals I and II should be equal, but they are not.
   6. Explain what the problem is.
   7. Do you expect integrals over closed circles of different radii to be equal? Why?
   8. Are they actually equal?
   Now assume that you allow singular functions to be OK, like Heaviside step functions and Dirac delta functions say. Then figure out in what part of the interior of the circle, $\int\!\int\nabla\times\vec{v}\cdot{\hat k} {\rm d}{x}{\rm d}{y}$ is not zero. So how would you describe $\nabla\times\vec{v}$ for this vector field in terms of singular functions?

4. 1st Ed: p133, q56, 2nd Ed: p155, q56.

5. 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 $u$ and $v$. 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?

6. 1st Ed: p160, q38, 2nd Ed: p183, q38. Simplify as much as possible. Sketch each surface, taking the $z$-axis upwards.

7. 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,
   
   $$\frac{\partial {\hat\imath}_i}{\partial u_i} = \frac{1}{h_... ...frac{1}{h_i} \frac{\partial h_j}{\partial u_i} {\hat\imath}_j$$
   
   

8. Express the acceleration in terms of the spherical velocity components $v_r,v_\theta,v_\phi$ and their first time derivatives, instead of time derivatives of position coordinates. Like $a_r = \dot v_r + \ldots$, etc. This is how you do it in fluid mechanics, where time-derivatives of particle position coordinates are normally not used. (So, get rid of all position coordinates with dots on them in favor of the velocity components and position coordinates without dots.) Hint: you may want to differentiate the expressions for the velocity components with respect to time to get expressions for the second order derivatives of the position coordinates. Then get rid of the second order derivatives first.