The question goes as follows: Let S be that portion of the cylinder $x^2+y^2=a^2$ bounded below by the $xy$-plane and above by the cone $z=\sqrt{(x-a)^2+y^2}$. Find the area of S.
It's not exactly clear to me if S being referred to is the top of the cylinder curved by the cone or the cylinder up to said top. In either case both are interesting.
I've gone as far as finding that if I used cylindrical coordinates $z\in[0, a\sqrt{2}\sin{\frac{\theta}{2}}]$ and $\theta\in[0,\pi]$. $dS$ would be $adzd\theta$ I think that would be the area of the cylinder's sides minus the curved top(?).
But I'm not sure how to go about finding the top's area.
For a surface parametrised by ${\bf r}(u,v)=(x(u,v),y(u,v),z(u,v))$ we have that the area of the surface is given by $$ A = \iint_S dS = \int_D \|{\bf r}_u\times {\bf r}_v\|dudv $$ where $(u,v)$ range over $D$.
We are given that the surface is that bounded by the cylinder $x^2+y^2=a^2$, the cone $z=\sqrt{(x-a)^2+y^2}$ and the plane $z=0$.
TOP SURFACE:
The height of the top surface is determined by the cone. The portion of the cone considered is determined by the cylinder. Hence,
${\bf r}=(u,v,\sqrt{(u-a)^2+v^2})$ where $D$ is given by $u^2+v^2\leq a^2$.
${\bf r}_u=(1,0,(u-a)((u-a)^2+v^2)^{-\frac{1}{2}})$
${\bf r}_v=(0,1,v((u-a)^2+v^2)^{-\frac{1}{2}})$
${\bf r}_u\times {\bf r}_v = (-(u-a)((u-a)^2+v^2)^{-\frac{1}{2}},-v((u-a)^2+v^2)^{-\frac{1}{2}},1)$
$\|{\bf r}_u\times {\bf r}_v\| = \sqrt{2}$
\begin{eqnarray*} A_{top} &=& \iint_D \sqrt{2}dudv = \sqrt{2}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=a}rdrd\theta = 2\sqrt{2}\pi\int_{r=0}^{r=a}rdr = \sqrt{2}\pi r^2|_{r=0}^{r=a} = \sqrt{2}\pi a^2 \end{eqnarray*}
SIDE SURFACE
The height of the sides is determined by the cone. Hence,
${\bf r}=(a\cos u, a\sin u, v)$ where $D$ is given by $0\leq u\leq 2\pi$ and $0\leq v \leq \sqrt{(x-a)^2+y^2}$. We rewrite the upper bound for $v$ in terms of $u$: \begin{eqnarray*} \sqrt{(x-a)^2+y^2} &=&\sqrt{(a\cos u-a)^2+a^2\sin^2 u} =\sqrt{a^2\cos^2 u-2a^2\cos u+a^2+a^2\sin^2 u}\\ &=&\sqrt{2a^2-2a^2\cos u} =\sqrt{2}a\sqrt{1-\cos u} \end{eqnarray*}
${\bf r}_u=(-a\sin u,a\cos u,0)$
${\bf r}_v=(0,0,1)$
${\bf r}_u\times {\bf r}_v = (a\cos u, a\sin u, 0)$
$\|{\bf r}_u\times {\bf r}_v\|=a$
\begin{eqnarray*} A_{side} &=& \iint_D adudv = \int_{u=0}^{u=2\pi}\int_{v=0}^{v=\sqrt{2}a\sqrt{1-\cos u}}adudv = 2\int_{u=0}^{u=\pi}\int_{v=0}^{v=\sqrt{2}a\sqrt{1-\cos u}}adudv\\ &=& 2\sqrt{2}a^2\int_{u=0}^{u=\pi}\sqrt{1-\cos u}du = 2\sqrt{2}a^2\int_{w=0}^{w=2}(2-w)^{-\frac{1}{2}}dw = 2\sqrt{2}a^2\Big(-2(2-w)^{\frac{1}{2}} \Big)|_{w=0}^{w=2}\\ &=& 2\sqrt{2}a^2 2\sqrt{2} = 8a^2 \end{eqnarray*} where we used the substitution $w=1-\cos u$.
BOTTOM SURFACE
It is simply the area of a circle of radius $a$, namely $A_{bottom}=\pi a^2$.
TOTAL AREA
$A_{total}=A_{top}+A_{side}+A_{bottom}=\sqrt{2}\pi a^2+8a^2+\pi a^2$
Hopefully, there aren't too many typos in this!