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!
