Finding surface area - integral of $\sqrt{1+z^2}$ Sorry about this, this is more of a "am I going the right way" question, there's a surface it goes:
$$x^2+y^2-z^2=1$$
Now this is nice because
$x^2+y^2=r^2=1+z^2$ thus $r=\sqrt{1+z^2}$ (I want the positive side)
(BORING PART - I am sure this is right)
So the shape of this thing has rotational symmetry about z (the vertical axis) some sort of hyperbola comes to mind because it has (for r) two parts (the + and -) and it "tends to linear" as z gets large.
That's not analysis reasoning, just for large $z$ $z^2$ will be pretty big compared to one, so $\sqrt{1+z^2}$ will be slightly bigger than $z$.
Anyway, my task is to find the surface area between the planes z=1 and z=0.
So.... the circumfrence of a circle is $2\pi r$and $r$ is as above. The height of each 'lil cylinder is $dz$ (This was once fine, Analysis makes me feel bad for saying this!) so $dA = 2\pi\sqrt{1+z^2}dz$ It makes sense to sum from $z=0$ to 1, hence:
ACTUAL QUESTION
$$A = 2\pi\int^1_0\sqrt{1+z^2}dz$$
I am pretty sure I am being dippy by not seeing it, I can see that:
using $\cosh(\theta)^2-\sinh(\theta)^2=1$ one can easily get to:
$\cosh(\sinh^{-1}(z)) = \sqrt{1+z^2}$
The question is supposed to be easy, just wondering if there's a quicker way, someone to go "yes that's right!" would be great.
 A: Your analysis is incorrect, although your approach is certainly in the right spirit.  The "height" of a cylinder is not $dz$, but $ds$, or the arc length element along the surface.  In this case
$$ds = \sqrt{1+\left ( \frac{\partial r}{\partial z}\right)^2} dz = \sqrt{1+\frac{z^2}{1+z^2}}dz = \sqrt{\frac{1+2 z^2}{1+z^2}}dz$$
Then the integral for the surface area is instead
$$2 \pi \int_0^1 dz \, \sqrt{1+2 z^2}$$
You can then make a substitution $u=z \sqrt{2}$ to get
$$\sqrt{2}\pi \int_0^{\sqrt{2}} du \, \sqrt{1+u^2}$$
Substituting $u=\sinh{v}$ produces a simpler integral to evaluate.  The result I get for the surface area is
$$S =  \frac{\pi}{\sqrt{2}} \left [\sqrt{6} + \log{(\sqrt{2}+\sqrt{3})}\right]$$
A: For future people looking for an answer, the integral in the question is totally wrong. I'd gone the wrong direction.
To see an answer that uses the rotational symmetry about the z axis of the question (ie elementary .... tapered disks (NOT cylinders)) please see Ron Gordon's answer. 
To have your memory jogged (subtly getting my excuse in there) about surface area, read on.
It ought to be known that a curve in $\mathbb{R}^3$ can be parameterised by a map $\boldsymbol{r}:\mathbb{R}\rightarrow\mathbb{R}^3$
and by analogy a surface $S$ in $\mathbb{R}^3$ by:
$\boldsymbol{r}:\mathbb{R}^2\rightarrow\mathbb{R}^3$
The surface $S =\lbrace\boldsymbol{r}((u,v))|(u,v)\in\Omega\rbrace$
From now on I'll write r(u,v) as r of the vector (u,v), save some brackets.
and: $\Omega=\lbrace(u,v)|u\in U, v\in V\rbrace$
tl-dr parametric surface.
Now neglecting questions of existence (a smooth way of pretending I'm not sidestepping answering those questions) the following all happens:
with the substitution involving root 2, it should be 1 over root 2du not root2du
consider the parallelogram (with curvy edges maybe, but for small enough delta u and v will exist such that the four 'edges' do not cross, which is what is important):
$\boldsymbol{r}(u,v)$
$\boldsymbol{r}(u+\delta u,v)$
$\boldsymbol{r}(u,v+\delta v)$
and
$\boldsymbol{r}(u+\delta u,v+\delta v)$
the area of a parallelogram can be found quite easily by showing that (looking at it so one of the sets of parallel lines is 'an x axis' and reflecting is (isometry) as required so the slanty edge is to the right, ie the top now horizontal line ends to the right of the base) you can 'cut off' a triangle that starts at the right end of the base, defined by going vertically upwards from that, then right to the end of the top edge,
and that it fits with the bottom right vertex being the top vertex of a similarly constructed triangle on the other side.
This sounds so much harder than my simple picture, but just use an angle "theta" from the base to the left slanty edge, if you continue the base, there's another angle theta from that base to the right slanty edge. cos defines the 'overhang' it's very easy to show they're similar triangles.
Thus you can turn the parallelogram into a rectangle, of width |b| (b for the base vector) and height |a|sin(theta) where a is the vector that is the slant, and |a| the length of it 
So the area is |a||b|sin(theta)
Now you should know that a x b = |a||b|sin(theta)n where n is a unit normal vector to a, b (perpendicular to them both)
so |a x b| = | |a||b|sin(theta)n | = |a||b|sin(theta) |n| = |a||b|sin(theta)
this means the magnitude of the cross product of the (direction) vectors that represent the edges of the parallelogram is the area of that parallelogram.
So going back to our parallelogram, we can see that:
$\boldsymbol{r}(u+\delta u,v)-\boldsymbol{r}(u,v)\approx\frac{\partial\boldsymbol{r}}{\partial{u}}\delta u$ 
and 
$\boldsymbol{r}(u,v+\delta v)-\boldsymbol{r}(u,v)\approx\frac{\partial\boldsymbol{r}}{\partial{v}}\delta v$
Which are the edges of the 'parallelogram' 
the area of which is ($\approx$):
$\lVert\frac{\partial\boldsymbol{r}}{\partial{u}}\delta u\times\frac{\partial\boldsymbol{r}}{\partial{v}}\delta v\rVert=\lVert\frac{\partial\boldsymbol{r}}{\partial{u}}\times\frac{\partial\boldsymbol{r}}{\partial{v}}\rVert\delta u\delta v$
Thus: $\int\int_S ds=\int\int_\Omega \lVert\frac{\partial\boldsymbol{r}}{\partial{u}}\times\frac{\partial\boldsymbol{r}}{\partial{v}}\rVert dudv$
I really really .... want to stop typing now, so skipping working out all that you get:
$A=\int^{u=1}_{u=0}{\sqrt{1+2u^2}\int^{v=2\pi}_{v=0}{dv}du}$
Clearly the integral involving v is just 2pi, the second one, I just said, let sinh(t)=root(2)u and got a new integral with limits arcsinh(root(2)) and 0, blah blah blah, answer is:
$A=\pi[\sqrt{3}+\frac{1}{\sqrt{2}}\ln(\sqrt{2}+\sqrt{3})]$
