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\centerline{Mathematics 173 -- Applied Mathematics 1}
\centerline{Problem Set 2 -- Line and Surface Integrals}
\centerline{September 14, 1999}
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\noindent
I.  Evaluate the line integrals of each of the following
vector fields along the indicated curves, and answer the 
accompanying questions.
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\item{A)} ${\bf F} = (x,x+y)$, $\alpha(t) = (t-t^2,t^2+t^3)$, $t \in [0,1]$.
{\it Answer: 139/60}.
\item{B)} ${\bf F} = (y^2,2xy)$, $\alpha$ the path from 
$(-1,2)$ to $(1,5)$ made up of two straight line segments, the
first parallel to the $x$-axis, the second parallel to the $y$-axis.
\item{C)} ${\bf F} =$ same vector field as in part B, but now
$\alpha$ the line segment directly from $(-1,2)$ to $(1,5)$.
\item{D)}  Your answers for B and C should be equal.  Explain why this 
is true.  {\it Hint:}  Is ${\bf F}$ a gradient vector field?
\item{E)}  The vector field
$${\bf F} = \left( {-y \over x^2 + y^2}, 
{x \over x^2+y^2}\right)$$
$\alpha$ the circle of radius 1 centered at $(0,0)$ in ${\bf R}^2$,
traversed counterclockwise.  
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\noindent
II.  Use Green's Theorem to evaluate each of the following
line integrals around the specified simple closed curve, and 
answer the other questions.
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\item{A)}  ${\bf F} = (\sin(\pi xy),\cos(\pi xy))$, $\alpha$ the 
positively-oriented boundary of the square
$[0,1]\times [0,1]$.  {\it Answer: $-{2+ \pi \over \pi}$}.
\item{B)}  ${\bf F} = (y^2,xy)$, $\alpha$ the ellipse with equation
$x^2/9 + y^2/4 = 1$.
\item{C)}  Compute 
$$\int\!\!\!\int_D {\partial Q\over \partial x} - {\partial P\over \partial y} dA$$
for the vector field ${\bf F}$ and the region $D$ inside $\alpha$ 
from part E of question I.  Your answer should be different from what
you saw in that part.  Is this a contradiction to Green's Theorem?? Explain.
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\noindent
III.  The statement of Green's Theorem is also valid in more general
situations.
For example, consider the {\it circular ring\/} $D$ in the plane between the 
two circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.  The positively
oriented boundary of $D$ consists of the the circle of radius 2 traversed 
counterclockwise, and the circle of radius 1 traversed clockwise. 
Let ${\bf F} = (P,Q)$ be a vector field whose component
functions have continous first order partial derivatives 
on some open set in ${\bf R}^2$ containing the whole ring. 
Show that the line integral of ${\bf F}$ over the positively oriented
boundary is equal to the double integral of 
${\partial Q\over \partial x} - {\partial P\over \partial y}$ over the 
ring.  {\it Hint: Consider the curve $\alpha_1$ which consists of 
the line segment from $(1,0)$ to $(2,0)$, followed by the upper semicircle 
of radius 2 traversed counterclockwise, followed the line segment from 
$(-2,0)$ to $(-1,0)$, followed by the upper semicircle of radius 1 traversed 
clockwise. Similarly let $\alpha_2$ be the curve consisting
of the line segment from 
$(-1,0)$ to $(-2,0)$, followed by the lower semicircle of radius 2 traversed 
counterclockwise, followed by the line segment from $(2,0)$ to $(1,0)$, followed
by the lower semicircle of radius 1 traversed clockwise.}
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\noindent
IV.  For each of the following vector fields and surfaces, verify
the Divergence Theorem by direct computation of the surface integral
and the triple integral:
\item{A)}  ${\bf F} = (z,y,x)$, $S$ the sphere of radius 1 centered at 
$(0,0,0)$.
\item{B)}  ${\bf F} = (x+y,y+z,x+z)$, $S$ the surface of the tetrahedron
bounded by the coordinate planes and the plane $x + y + z = 1$.  (For
the surface integral, you will need to parametrize each of the four
faces of the tetrahedron separately, and add the corresponding surface
integrals.)
{\it Answer: 1/2 (both values)}.
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{\it Assignment}
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Individual solutions, due Tuesday, September 21.
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