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# Week 7 Tuesday 8/1 brief notes.
Exam 2 corrections + all HW due by the end of last day of course.
Agenda of things to go over.
(0) Gabriel's horn: Infinity is weird. (arclength, surface area and volume) 8.1 and 8.2
(1) Powerseries are nice.
(2) How to manipulate geometric series to obtain representations of certain functions.
(3) Finding interesting/curious representation of numbers as series.
(4) Powerseries are only as good as on their interval of convergence. Pay attention to it (at least the open interval of convergence/radius of convergence)
(5) Taylor and Maclaurin series; Taylor inequality.
(6) Important basic examples of Taylor series; binomial series.
(7) How to get new series from old ones.
(8) Application: Integration.
(9) Application: Limits without L'Hospital rule.
(10) Computing derivatives.
(11) Basic differential equation with powerseries.
We may or may not have time to get to the following two topics (Ch.10) So you want to read them yourselves:
(13) Parametric equation
(14) Polar equation
Class photo / pizza on Wednesday 8/9 (day before final exam?)
Final exam Thursday 8/10 !!
## Surface area of solid of revolution (8.1 and 8.2).
Recall from Math 7 calculus 1, we learned to find the volume of a solid of revolution. One way is by the disk method:
Given a positive function $f(x)$ on the interval $[a,b]$, take its graph $y=f(x)$ on $[a,b]$ and revolve it around the $x$-axis. This generates a solid of revolution, whose volume is $$
\int_{a}^{b} \pi f(x)^{2}dx
$$
This volume is essentially adding a lot of tiny disks of volume $\pi R^{2}\Delta x$ where $R=f(x)$. That is,$$
\text{volume}=\int_{a}^{b}\pi f(x)^{2}dx \approx\sum_{i=1}^{N}\pi f(x_{i})^{2}\Delta x
$$
We can also ask what is the **surface area** of this solid, that is, what is the area of the sides generated by rotating the graph $y=f(x)$ about the $x$-axis?
To answer this, we first ask what is even the arclength of this graph $y=f(x)$ on the interval $[a,b]$. If we use this idea of chopping up the graph of $y=f(x)$ into small pieces, we see that we need to add up a lot of tiny ilne segments, each is in general the hypotenuse of a small right triangle. These line segments have length $$
\Delta s=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}=\sqrt{1+\left( \frac{\Delta y}{\Delta x} \right)^{2}} \Delta x
$$
So the arclength of $y=f(x)$ on $[a,b]$ is approximately $$
\text{arclength} \approx \sum_{i=1}^{N} \sqrt{1+\left( \frac{\Delta y}{\Delta x} \right)^{2}} \Delta x
$$
Now as $\Delta x \to 0$, the ratio $\frac{\Delta x}{\Delta y}\approx f'(x)$. So in the limit, we have the arclength formula: $$
\text{arclength} = \int_{a}^{b} \sqrt{1+(f'(x))^{2}}dx
$$for the graph $y=f(x)$ over the interval $[a,b]$.
Now how do we get surface area of the surface obtained by revolving the graph of $y=f(x)$?
Notice if we chop up the surface, each piece is a ribbon of width $\Delta s$ (because it is slanted), and length (circumference) $2\pi R=2\pi f(x)$. So the area is approximately $$
\text{surface area} \approx \sum_{i=1}^{N} 2\pi f(x_{i})\Delta s
$$
And we know $\Delta s = \sqrt{1+\left( \frac{\Delta y}{\Delta x} \right)^{2}} \Delta x$. So in the limit as $\Delta x \to 0$, we have $$
\text{surface area} = \int_{a}^{b} 2\pi f(x) \sqrt{1+(f'(x))^{2}}dx
$$
(Note, this surface area does not include the two tops! Just the part revolved by the graph $y=f(x)$)
Remark. Although one can set these formulas up, quite often it's not easy to compute them exactly if $f(x)$ is a wild function. However one can always perform integral estimations.
**Example.**