Surface area of rotation of a circle around a tangent Self-studying integral calculus and I got this problem:

The circle $x^2+y^2=a^2$ is rotated around a line tangent to the circle. Find the area of the surface of rotation.

I made the construction:

There were a few hints given alongside this question, namely: "Set up coordinate axes and a convenient parametrization of the circle. What does the polar graph $r=2a\sin(\theta)$ look like?" I understood the first and last hints since under this new coordinate system, the circle's equation becomes:
$${x_n}^2+(y_n-a)^2=a^2$$
Which when converted to polar, gives you the last hint. However I was unable to describe this in terms of parameters, so I decided to take the upper semicircle's surface area of revolution going from $a$ to $-a$ and multiplying that by 2 to account for the lower semicircle. My integral:
$$2\int_{-a}^a 2\pi (\sqrt{a^2-x^2}+a)\sqrt{1+\frac{x^2}{a^2-x^2}} dx$$
Upon simplification:
$$4\pi\int_{-a}^a a+\frac{a^2}{\sqrt{a^2-x^2}}dx$$
Evaluation leads me to:
$$8\pi a^2 + 4\pi^2 a^2$$
However my book (Serge Lang's First Course in Calculus) gives only $4\pi^2 a^2$. Where has my logic gone wrong if I am getting an extraneous term $8\pi a^2$?
EDIT for clarity on integral setup:
I first rearranged for $y$ while taking positive square root as I want to take the upper semicircle into consideration for surface of revolution about x-axis. I'll double this to account for the lower semicircle. This gives:
$$y=\sqrt{a^2-x^2}+a$$
Using the surface of revolution formula with the derivative as
$$\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}$$
Using this into the surface of revolution integral nets me my first integral in this post (also applied $\times$2)
 A: After changing the coordinates, in effect you are rotating $x^2 + (y-a)^2 = a^2$ around x-axis.
The circle is $x^2 + y^2 = 2 ay$
$ \displaystyle y' = \frac{x}{a-y}$
$ \displaystyle ds = \sqrt{1 + (y')^2} ~dx = \frac{a}{|y-a|} ~ dx$
For lower half -
$y = a - \sqrt{a^2-x^2}$
So, $ \displaystyle S_1 = 2 \pi a \int_{-a}^a \frac{a - \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$
$ = 2 \pi a^2 (\pi - 2)$
For upper half -
$y = a + \sqrt{a^2-x^2}$
So, $ \displaystyle S_2 = 2 \pi a \int_{-a}^a \frac{a + \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$
$ = 2 \pi a^2 (\pi + 2)$
Adding both, $S = 4 \pi^2 a^2$
But it is easier in polar coordinates as I mentioned in comments. The circle is,
$r = 2a \sin\theta, 0 \leq \theta \leq a$
$\dfrac{dr}{d\theta} = 2a \cos\theta$
$ \displaystyle ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} ~ d\theta = 2a ~ d\theta$
$y = 2a\sin^2\theta$
So the integral is,
$ \displaystyle S = 8 \pi a^2 \int_0^{\pi} \sin^2\theta ~ d\theta = 4 \pi^2 a^2$
A: In spite of your obfuscating figure, you are asking for the surface area of a torus whose inner radius, $R$ (to the center of the cross-section) and outer radius, $r$ (that of the cross-section) are the same. This is well known to be $S=4\pi^2Rr$ (see, for example the CRC Mathematical Tables). So in your case, $S=4\pi^2a^2$
We can derive this result with Pappus's ($1^{st}$) Centroid Theorem, which states that the surface area $S$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$ and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance $d$ traveled by its geometric centroid. Simply put, $S=2πRL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is the curve length. In your case, $R=a$ and $L$ is the circumference of the circle, i.e., $=2\pi a$, so that $S=4\pi^2a^2.$
A: Comment on the Question
I believe that this is the standard setup for surface of revolution about the $x$-axis
$$
\int_{-a}^a\overbrace{\quad2\pi y\quad\vphantom{\frac{a}{\sqrt{a^2}}}}^{\substack{\text{account for}\\\text{revolution}}}\overbrace{\frac{a}{\sqrt{a^2-x^2}}}^{\mathrm{d}s/\mathrm{d}x}\,\mathrm{d}x
$$
However, the part for the upper arc of the circle does not give the same area as that for the lower arc of the circle, so we need to compute both separately:
$$
\begin{align}
&\int_{-a}^a2\pi\left(a+\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{upper}\\
&+\int_{-a}^a2\pi\left(a-\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{lower}\\
&=2\pi\int_{-a}^a2a\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\[6pt]
&=4\pi^2a^2
\end{align}
$$

Other Approaches
Revolution Around the $\boldsymbol{x}$-axis
As I had posted before I realized the question was revolving around the $y$-axis, if we revolve around the $x$-axis, the upper and lower parts of the surface have the same area, so we can just multiply the upper integral by $2$ in this case. Thus, the formula is
$$
\begin{align}
2\int_0^{2a}2\pi x\,\frac{a}{\sqrt{a^2-(x-a)^2}}\,\mathrm{d}x
&=2\int_{-a}^a2\pi(x+a)\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{1a}\\
&=4\pi a^2\int_{-1}^1(x+1)\frac1{\sqrt{1-x^2}}\,\mathrm{d}x\tag{1b}\\
&=4\pi a^2\int_{-\pi/2}^{\pi/2}(\sin(x)+1)\,\mathrm{d}x\tag{1c}\\[9pt]
&=4\pi^2a^2\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $x\mapsto x+a$
$\text{(1b)}$: substitute $x\mapsto ax$
$\text{(1c)}$: substitute $x\mapsto\sin(x)$
$\text{(1d)}$: integrate
Parametrization
Parametrize the torus as follows: at each point of the circle around the $z$-axis, $a(\cos(\phi),\sin(\phi),0)$ put a circle perpendicular to this circle:
$$
\begin{align}
p(\phi,\theta)
&=\overbrace{a(\cos(\phi),\sin(\phi),0)}^\text{primary circle}+\overbrace{a(\cos(\phi)\cos(\theta),\sin(\phi)\cos(\theta),\sin(\theta))}^\text{secondary circle around the primary circle}\\
&=a(\cos(\phi)(1+\cos(\theta)),\sin(\phi)(1+\cos(\theta)),\sin(\theta))\tag{2a}]\\[6pt]
p_1(\phi,\theta)
&=a(-\sin(\phi)(1+\cos(\theta)),\cos(\phi)(1+\cos(\theta)),0)\\
&=a(1+\cos(\theta))(-\sin(\phi),\cos(\phi),0)\tag{2b}\\[6pt]
p_2(\phi,\theta)
&=a(-\cos(\phi)\sin(\theta),-\sin(\phi)\sin(\theta),\cos(\theta))\tag{2c}
\end{align}
$$
Thus, we get
$$
\begin{align}
|p_1(\phi,\theta)\times p_2(\phi,\theta)|
&=a^2(1+\cos(\theta))\,|(\cos(\theta)\cos(\phi),\cos(\theta)\sin(\phi),\sin(\theta))|\\
&=a^2(1+\cos(\theta))\tag3
\end{align}
$$
and we can compute
$$
\int_0^{2\pi}\int_0^{2\pi}a^2(1+\cos(\theta))\,\mathrm{d}\phi\,\mathrm{d}\theta=4\pi^2a^2\tag4
$$
Theorem of Pappus
As I mentioned in a comment, we can apply the Theorem of Pappus: the primary circle has circumference $2\pi a$ and the secondary circle has circumference $2\pi a$, so the area is
$$
(2\pi a)(2\pi a)=4\pi a^2\tag5
$$
