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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)

\section*{Work}

\paragraph{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.

\paragraph{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 $F\left(x\right)$ of position. Then
    the {\em work done} by the force is
	\[ W = \int_a^b F\left(x\right)dx .\]

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

\paragraph{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)

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

\paragraph{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.)

\paragraph{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:
    \begin{align*} \int_{-0.05}^0 -kx dx &= \int_{-0.05}^0 -1000x dx \\
    &= -1000 \evalat{\frac{x^2}{2}}{-0.05}{0} \\
    &= -1000 * \left(0-0.0025/2\right) = 1.25 J \end{align*}

\paragraph{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 $M\left(h\right)
    kgm^-1$.

    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 $M\left(h_i^*\right)\Delta_n$, then as in the anvil example we estimate
    the energy released by the falling of the $i^{th}$ segment as
    $M\left(h_i^*\right)\Delta_ngh_i^*$, and hence the entire energy as
    \[ \sum_{i=1}^n M\left(h_i^*\right)\Delta_ngh_i^* .\]

    Taking the limit as $n\rightarrow \infty$, we get

    \[ g\int_0^H hM\left(h\right)dh .\]
    
    Now let's estimate $M\left(h\right)$: 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 $M\left(h\right)=\left(500-h\right)k$, then
      \[ 120*10^6 = \int_0^{500} \left(500-h\right)k dh =
	k\evalat{500h-\frac{h^2}{2}}{0}{500} = k\frac{500^2}{2} \]
    So
      \[ k = \frac{120*10^6}{\frac{500^2}{2}} = 960 kgm^{-1}.\]

    So the work done by gravity as the tower falls is
    \begin{align*} g\int_0^{500} h\left(500-h\right)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 \end{align*}

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

\section*{Integration by parts}

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

\paragraph{Example:}
    \[ \int x\cos\left(x\right) dx = \int \sin'\left(x\right)x dx = \sin\left(x\right)x - \int \sin\left(x\right) 1 dx =
    x\sin\left(x\right) + \cos\left(x\right) + C .\]

\paragraph{Remark:}
  Setting $u=f\left(x\right)$ and $v=g\left(x\right)$ (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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