Exercise 9a

1.
Stewart, Section 14.8, Exercises 1, 2, 10, 19.

2.
Show by direct calculation that the conclusion of Stokes Theorem does not hold for $ \mathbf{F}= \frac{1}{x^2 + y^2} (-y \mathbf{i}+ x \mathbf{j})$.

3.
Let $ r = \sqrt{x^2 + y^2 + z^2}$, $ \mathbf{r}= (x\mathbf{i}+ y \mathbf{j}+
z\mathbf{k})$
(a)
Find the gradient of $ \frac{1}{r(z-r)}$. Find the curl of $ -y \mathbf{i}+ x \mathbf{j}$.
(b)
Show that $ \nabla \times \mathbf{F}= \mathbf{E}$, everywhere except on the non-negative part of the $ z$-axis, where

\begin{displaymath}
\begin{split}
\mathbf{F}& = \frac{Q}{4 \pi \epsilon_0} \frac...
...frac{Q}{4 \pi \epsilon_0} \frac{1}{r^3} \mathbf{r},
\end{split}\end{displaymath}

and $ Q, \epsilon_0$ are strictly positive constants.
(c)
Let $ C$ be the curve $ x^2 + y^2 =1, z=0$ oriented in the anti-clockwise direction when viewed from above. Use Stokes theorem to show that if $ S$ is any oriented surface with oriented boundary $ C$, and $ S$ does not intersect the non-negative part of the $ z-axis$, then

$\displaystyle \iint_S \mathbf{E}\cdot d\mathbf{S}= -\frac{Q}{2 \epsilon_0}
$

4.
(Optional) Suppose $ \mathbf{G}$ is a vector field with $ \nabla \cdot \mathbf{G}=
0$, everywhere in $ 3$-space. Define $ \mathbf{H}$ in terms of $ \mathbf{G}$ by

\begin{displaymath}
\begin{split}
H_1 & = 0\\
H_2 & = \int_a^x G_3(u,y,z) du\\...
... - \int_a^x G_2(u,y,z) du + \int_b^y G_1(a,v,z) dv,
\end{split}\end{displaymath}

where $ a,b$ are arbitrary constants. Show by direct calculation that $ \nabla \times \mathbf{H}= \mathbf{G}$ everywhere in $ 3$-space.



Math 13 Winter 1999
1999-03-03