Question: I have trouble relating two forms of finding the volume of a cylinder. First of all you can imagine integrating the area of a disk along the height. Something like this, $V=\int\_{0}^{h} \pi r^2 dx = \pi r^2h$. But also you can think of it as integrating the area of a rectangle ($hr$) along the "angles", so the volume would be $V=\int\_{0}^{2\pi} hrd\theta=2\pi hr$. I know the last one is wrong. But I really don't see why. Hope you can help

Answer: 
All you need is to choose a positive epsilon, say $\epsilon = 1$, then there is a delta $\delta > 0$ such that $|f(x) - f(c)| < \epsilon = 1$ if $0 < |x-c| < \delta\implies - 1 < f(x) - f(c) < 1 \implies -1 + f(c) < f(x) < 1 + f(c) \implies |f(x)| < 1 + |f(c)|= M$ which shows $f$ is bounded.
From the definition

$A\subset \mathbb{R}, c\in A$

$\forall \varepsilon>0,\exists \delta >0,\forall x\in A \quad |x-c|<\delta \implies |f(x)-f(c)|<\varepsilon$

Let $\varepsilon >0$, so according to the definition, there is $\delta >0$ such as $\forall x\in (c-\delta, c+\delta)$ this inequality ($|f(x)-f(c)|<\varepsilon$) is satisfied.

then

$|f(x)-f(c)|<\varepsilon \iff f(c)-\varepsilon<f(x)<f(c)+\varepsilon$

$\implies f$ is bounded