Surface area of this helicoid? What is the surface area of the paramatized helicoid 
$$X(r,s)=(r\cos(s), r\sin(s), s)$$ 
with $0\leq r \leq 1$ and $0\leq s \leq 2n\pi$, where $n$ is a positive integer? 
I attempted to take the standard normal vector with the two tangent vectors 
$$T_r=(\cos(s),\sin(s),0) \text{ and } Ts=(-r\sin(s),r\cos(s), 1).$$
and I got the cross product of these two as $\sin(s)i+\cos(s)j+ rk$. Then when I took the length of this I got $\sqrt{r^2+1}$. I don't know how to integrate this function and I also wanted to be sure that I did the rest of the set up correctly.
 A: I would use surface integrals. That is 
$\displaystyle \iint_S 1 \ dS =$ Surface area of S.
We can rewrite this as
$\displaystyle \iint_R \|\mathbf Q_r \times \mathbf Q_s \| \ dA$
where R is the domain of the parameters r and s and Q(r,s) is the parametrization of the surface. 
For $\mathbf Q(r,s) = \big <r\cos s,r\sin s,s \big>$ we get
$ \mathbf Q_r = \big <\cos s,\sin s,0 \big>$
$ \mathbf Q_s = \big <-r\sin s,r\cos s,1 \big>$ as you said. 
Now crossing and taking the magnitude gives $\sqrt{1+r^2}$ , which we can now plug into the integral.
$\displaystyle \iint_R \sqrt{1+r^2} \ dA$ = $\displaystyle \int_0^{2n\pi} \int_0^1 \sqrt{1+r^2}  \ dr \ ds$
First let's do the inner integral (the outer integral will just multiply the answer by $2n\pi$). We can do a trig substitution by letting $r = \tan\theta$ with $dr = \sec^2\theta \ d\theta$. 
So we get $\displaystyle \int \sqrt{1+r^2} dr = \int \sqrt{1+\tan^2\theta} \sec^2\theta \ d\theta = \int \sec^3\theta \ d\theta = \int \sec\theta\sec^2\theta \ d\theta$
We can calculate this via integration by parts. Let $u= \sec\theta ;\ du= \sec\theta\tan\theta ;\ dv = \sec^2\theta \ d\theta$ and $v=\tan\theta$. 
Substituting and skipping a couple steps we get:
$\displaystyle \int \sec^3\theta \ d\theta = \sec\theta\tan\theta - \int\sec^3\theta \ d\theta +\int\sec\theta \ d\theta$. We bring the integral with like terms to the left side and perform some more laborious but straightforward integration to get:
$\displaystyle \int\sec^3\theta \ d\theta = \frac12 \tan\theta\sqrt{1 + \tan^2\theta} + \frac12 \ln |\tan \theta +\sqrt{1 + \tan^2\theta} | + C $
Substituting back $\tan\theta = r$
$\displaystyle \int \sqrt{1+r^2}  \ dr =\frac12 r\sqrt{1 + r^2} + \frac12 \ln | r +\sqrt{1 + r^2}| + C$.
Evaluating over $[0,1]$ gives $\displaystyle \frac{\sqrt2}{2} + \frac12 \ln(1+\sqrt2)$ and multiplying by $2n\pi$ gives the surface area: 
$\displaystyle n\pi(\sqrt2 + \ln(1+\sqrt2))$
