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\(first lecture of wk10 is contained in wk9.pdf\)

\(second lecture, covering volumes of revolution and averages, was taught for
me by Dr. Haskell - see scans2.pdf\)

## Work

Warning:
    I largely follow Stewart here. This is very much a ``Just-So'' account of
    the physics involved. I will leave it to your physics lecturers to give
    you the full story.

Definition:
    Suppose a particle moves along a line from $x=a$ to $x=b$ while a force
    acts on the particle along the line of motion, and suppose the force in
    the direction of motion is a continuous function $F(x)$ of position. Then
    the /work done/ by the force is
	\[ W = \int_a^b F(x)dx .\]

    If $x$ and $F$ are measured in metres and Newtons respectively, then
    $W$ is measured in Joules. 

Example:
    I hold a 5000kg anvil 10m above the ground, then let it drop. As it falls,
    it is acted on by gravity with a force of $g=9.8$ Newtons. So the work
    done by gravity as the anvil falls is
    \[ \int_{10}^0 -mg dh = \int_{10}^0 -490000 dh = 4900000 \]
    Joules, i.e. 4.9MJ.

    \(Enough to power a 15W light bulb for nearly 4 days\)

Generally:
    for a #constant# force in the direction of motion, work is force times
    distance travelled
    \[ W = Fd ;\]
    integration is needed when the force is varying.

Example:
    How much work do I have to do to reset the previous experiment - i.e. lift
    the anvil back up 10m?

    In lifting it, I have to counteract the force of gravity - but can apply
    arbitrarily small forces beyond that. So the work I have to do is
    \[ \int_0^{10} mg dh = \int_{0}^{10} 490000 dh = 4900000 .\]

    \(Fact: it actually doesn't matter what forces I apply in lifting the
    anvil; as long as it ends up at rest 10m above the ground, the work done
    by the forces will be the same, 4.9MJ.\)

Example:
    The force exerted by a compressed or stretched spring is approximately
    proportional to the difference between its length and its relaxed length
	\[ F=-kx \]
    \(Hooke's law\).

    I pull back the spring on a pinball table by 0.05m and then let it go. If
    $k=1000$ for this spring, what is the work done by the spring as it springs
    back?

    \(Again, it turns out that this is the same as the work I had to do in
    pulling back the spring.\)

    Solution:
    *\[a* \int_{-0.05}^0 -kx dx &= \int_{-0.05}^0 -1000x dx \\
    &= -1000 \evalat{\frac{x^2}{2}}{-0.05}{0} \\
    &= -1000 * (0-0.0025/2) = 1.25 J *a\]*

Example:
    The CN tower collapses. How much energy is released by the falling tower?

    Solution: for simplicity, let's assume that the entirety of the tower
    falls to ground level.

    We take the height of the tower to be $H=550$ metres and its total mass
    to be $120*10^6$ kg.

    First, suppose we know that how the mass of the tower is distributed along
    its height. Suppose that at height $h$, the mass per unit height is $M(h)
    kgm^-1$.

    Then the energy is an integral: if we divide the tower into $n$ segments
    of equal length $\Delta_n=\frac{H}{n}$, and estimate the mass of the
    $i^{th}$ as $M(h_i^*)\Delta_n$, then as in the anvil example we estimate
    the energy released by the falling of the $i^{th}$ segment as
    $M(h_i^*)\Delta_ngh_i^*$, and hence the entire energy as
    \[ \sum_{i=1}^n M(h_i^*)\Delta_ngh_i^* .\]

    Taking the limit as $n->\infty$, we get

    \[ g\int_0^H hM(h)dh .\]
    
    Now let's estimate $M(h)$: for simplicity, let's assume a linear
    relationship. \(So in particular we entirely ignore the SkyPod...\)
    So if the mass per unit height at height $h$ is $M(h)=(500-h)k$, then
      \[ 120*10^6 = \int_0^{500} (500-h)k dh =
	k\evalat{500h-\frac{h^2}{2}}{0}{500} = k\frac{500^2}{2} \]
    So
      \[ k = \frac{120*10^6}{\frac{500^2}{2}} = 960 kgm^{-1}.\]

    So the work done by gravity as the tower falls is
    *\[a* g\int_0^{500} h(500-h)k dh &=
    kg\evalat{\frac{500h^2}{2}-\frac{h^3}{3}}{0}{500} \\
    &= kg\frac{500^3}{3} \\
    &= 960*9.8*\frac{500^3}{3} \\
    &= 0.39*10^{12} J \\
    &= 0.39 TJ *a\]*

    \(Again, this is also the energy spent fighting gravity to construct the
    tower... for comparison, the Hiroshima bomb is said to have released
    67TJ.\)

## Integration by parts

\[ \ddx f(x)g(x) = f'(x)g(x) + f(x)g'(x) \]
so
\[ f(x)g(x) = \int (f'(x)g(x) + f(x)g'(x))dx \]
so
\[ \int f'(x)g(x)dx = f(x)g(x) - \int f(x)g'(x)dx .\]

Example:
    \[ \int x\cos(x) dx = \int \sin'(x)x dx = \sin(x)x - \int \sin(x) 1 dx =
    x\sin(x) + \cos(x) + C .\]

Remark:
  Setting $u=f(x)$ and $v=g(x)$ \(and assuming $u$ is a function of $v$ and
  $v$ is a function of $u$\) we can use the Substitution rule to rewrite this
  as \[ \int vdu = uv - \int udv .\]

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