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\centerline{Mathematics 41, section 1 -- Multivariable Calculus}
\centerline{Discussion 8 -- Polar Double Integrals, Centers of Mass}
\centerline{November 18, 1999}
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{\bf Background}
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Here's a joke to try out on your friends:
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{\it Question: What is a polar bear?  Answer:  It's a rectangular
bear after the change of coordinates}
$$x = r\cos\theta, \quad y = r\sin\theta.$$
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Yesterday, we saw that if $f(x,y)$ was any 
function, and $R$ was a region in the plane given in 
polar coordinates by 
$$R = \{(r,\theta) : a\le r\le b, \alpha \le \theta\le \beta\},$$ 
then the double integral
$$\int\int_R f\, dA = \int_\alpha^\beta \int_a^b f(r\cos\theta,r\sin\theta)\, r
dr d\theta.$$
Today we will look at how polar coordinates can simplify
the computations of things like centers of mass {\it when
the geometry of the region $R$ is well adapted to that coordinate system}.  
First, recall the {\it average value of $f$ on $R$} is given by:
$${\int\int_R f(x,y)\, dA \over {\rm area}(R)} = 
  {\int\int_R f(x,y)\, dA \over \int\int_R 1\, dA}$$
In particular, if $f = x$, or $f = y$ we would get 
$$\eqalign{\overline{x} &= {\int\int_R x\, dA \over \int\int_R 1\, dA}\cr
           \overline{y} &= {\int\int_R y\, dA \over \int\int_R 1\, dA}\cr}$$
and the point $(\overline{x},\overline{y})$ can be thought of as the 
{\it center of mass\/} of the region $R$ (it would be the center of mass of 
a thin plate in the shape of $R$ if the material of the plate had 
constant density).
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{\bf Discussion Questions}
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\item{A)}  Compute the center of mass of the region $R$ in the first and
second quadrants bounded by the positive $x$-axis, the line $y = -x$, and
an arc of the circle of radius 2 centered at the origin.  Suggestion:
think polar!  {\it Answer:} 
$$(\overline{x},\overline{y}) = 
\left({8\sqrt{2}\over 9\pi},{8\sqrt{2}+ 16\over 9\pi}\right) 
\doteq (.4001,.9660).$$
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\item{B)}  Suppose you had a thin plate in the shape of the region
$R$ from question A, but now the density of the material at each point
was variable, given by the function $\rho(x,y) = x^2 + y^2$.  
What double integrals would compute total mass of the plate and its 
center of mass?  (Think of $\overline{x}$ and $\overline{y}$ as 
``weighted averages,'' weighted by the density function.)
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\item{C)}  Next, suppose we wanted to determine the center of mass of 
an object that had thickness.  For simplicity, suppose our object
is a rectangular solid $S = [a,b] \times [c,d] \times [e,f]$ in
${\bf R}^3$, but suppose that the object has varying density $\rho(x,y,z)$
at each point $(x,y,z) \in S$.  How could you approximate the 
total mass and the coordinates of the center of mass?
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{\bf Assignment}
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Group write-ups due Monday, November 22.
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