Here's a derivation without trigonometry.
There is also a short discussion of the geometric significance
of the exponents in the answer after the derivation.
Let $L$ be the length of a line touching the inner corner of the
two corridors and extending to the outer wall of each corridor.
Where this line meets the outer wall of each corridor, draw a line
perpendicular to the corridor.
Let $x$ be the distance from the inner corner to the perpendicular line
across corridor $B$
and let $y$ be the distance from the inner corner to the perpendicular line
across corridor $A$.
The result is the figure shown above.
Applying the Pythagorean Theorem to the obvious
right triangle with hypotenuse $L$,
$$ L^2 = (A + x)^2 + (B + y)^2.$$
The longest pipe that can fit around the corner is the smallest
value of $L$ for any value $x > 0$, so let's minimize $L$ as a function of $x$.
But $L$ is minimized when $L^2$ is minimized, so we would like to set
$$ \frac{d}{dx} L^2 = 0. $$
That is,
$$\begin{eqnarray}
0 = \frac{d}{dx} L^2 &=& \frac{d}{dx}\left((A + x)^2 + (B + y)^2\right) \\
&=& 2(A + x) + 2(B + y)\frac{dy}{dx}.
\end{eqnarray}$$
Now, by similar triangles,
$ \dfrac Bx = \dfrac yA .$
That is, $xy = AB.$ Differentiating both sides of this by $x$,
$$ x\frac{dy}{dx} + y = 0,$$
$$ \frac{dy}{dx} = -\frac yx.$$
Substituting for $\frac{dy}{dx}$ in our earlier equation for $\frac{d}{dx}L^2,$
$$ 0 = 2(A + x) + 2(B + y)\left(-\frac yx\right),$$
from which it follows that
$$ (A + x)x = (B + y)y,$$
But (again by similar triangles)
$$ \frac Bx = \frac{B+y}{A+x},$$
$$ B+y = \frac Bx (A+x),$$
and combining this with the fact that $y = \dfrac{AB}{x},$
$$ (B + y)y = \left(\frac Bx (A+x)\right) \frac{AB}{x}
= \frac{AB^2}{x^2}(A + x).$$
That is,
$$ (A + x)x = \frac{AB^2}{x^2}(A + x),$$
$$ x^3 = AB^2,$$
$$ x = A^{1/3}B^{2/3},$$
$$ y = \frac{AB}{x} = A^{2/3}B^{1/3},$$
$$A + x = A + A^{1/3}B^{2/3} = \left(A^{2/3} + B^{2/3} \right)A^{1/3},$$
and $$B + y = B + A^{2/3}B^{1/3} = \left(A^{2/3} + B^{2/3} \right)B^{1/3}.$$
Therefore at the value of $x$ that minimizes $L$,
$$\begin{eqnarray}
L^2 &=& (A + x)^2 + (B + y)^2 \\
&=& \left(A^{2/3} + B^{2/3} \right)^2 A^{2/3} +
\left(A^{2/3} + B^{2/3} \right)^2 B^{2/3} \\
&=& \left(A^{2/3} + B^{2/3} \right)^2 \left(A^{2/3} + B^{2/3}\right) \\
&=& \left(A^{2/3} + B^{2/3} \right)^3
\end{eqnarray}$$
and
$$ L = \left(A^{2/3} + B^{2/3} \right)^{3/2}.$$
Regarding the symmetry of the solution, clearly the length of the pipe
does not change if we swap the labels $A$ and $B$ on the two corridors.
We should suspect from this that there might be some expression for
$L$ in which $A$ and $B$ appear in completely symmetric roles.
Regarding the geometric interpretation of the exponents, the key facts
that lead to the $\frac23$ exponents are the findings that $x^3 = AB^2$
and (by symmetry) $y^3 = BA^2.$
This is what leads to the conclusion that $L$ is the hypotenuse of a
right triangle with legs $A + A^{1/3}B^{2/3}$ and $B + B^{1/3}A^{2/3}$;
it is from the squares of those two lengths that we get terms
in $A^{2/3}$ and $B^{2/3}$.
Finally, note that if we double both $A$ and $B$ simultaneously,
we should expect the entire figure showing the "tightest" angle
of the pipe around the corner to scale up by a factor of $2$,
hence $L$ also doubles.
But we see that $A^{2/3}$ and $B^{2/3}$ scale up only by a factor
of $2^{2/3}$ when we double $A$ and $B$; it is the exponent of $\frac32$
outside the parentheses that gives us a scaling factor
of $\left(2^{2/3}\right)^{3/2} = 2.$
So it is not a mere coincidence that the exponents $\frac23$ and $\frac32$
are multiplicative inverses.
Another way to approach this is to apply the fact that
$B+y = \dfrac Bx (A+x)$
before differentiating $L^2$ rather than afterwards.
That is, we simply substitute for $B+y$ in the formula for $L^2$,
obtaining
$$\begin{eqnarray}
L^2 &=& (A + x)^2 + \frac{B^2}{x^2}(A + x)^2 \\
&=& \left(1 + \frac{B^2}{x^2}\right) (A + x)^2.
\end{eqnarray}$$
We can differentiate this with respect to $x$ to obtain an analytic solution,
or apply numeric methods using $x$ as an independent variable.
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$
Best Answer
From what I can see from the parametrization you've written, you have axis $x$ directed down, and axis $y$ directed left, is that right? That means you have the axes oriented in a way opposite than it's normaly done ($y$ counterclockwise to $x$). This is the origin of your opposite sign.
I suggest using axis $x$ oriented right and axis $y$ oriented up, as that's how they are usually drawn, so it's harder to make a mistake. The center of the coordinate system will be the center of the base of the silo.
You can paramterize the border of the grazing area by $$ x(\phi)= -R\sin\phi + \phi R\cos\phi, \qquad y(\phi) = -R\cos\phi - \phi R\sin\phi$$ where $R=10 {\rm m}$ and $\phi$ is the angle showing how much of the chain is unwinded and does not follow the circumference of the silo. We have $\phi\in[0,2\pi]$.
We can note that for $x$ initially decreases from $0$ for $\phi=0$ to some $x_{\rm min}<0$ for some $\phi=\phi_0$ (as it will turn out, the value of $\phi_0$ is not important) before it starts icreasing and reaches $x=2\pi R$ for $\phi=2\pi$. For $x<0$ we have two possible values of $y$ one related to $\phi<\phi_0$ and the other from $\phi>\phi_0$. Let's denote them $y_1(x)$ and $y_2(x)$. For $x>0$ we have only $y_2(x)$, but the lower boundary of this region is the wall for $y=-R$.
The full area surrounded by the chain, including the base of the silo, is given by \begin{align} S &= \int_{x_{min}}^0 (y_2(x)-y_1(x))dx + \int_0^{2\pi R} (y_2(x)-(-R)) dx =\\ &= - \int_{x_{min}}^0 y_1(x) dx + \int_{x_{min}}^{2\pi R} y_2(x) dx + 2\pi R^2=^\text{changing variables} \\ &= - \int_{\phi_0}^0 y(\phi) \frac{dx}{d\phi}(\phi) d\phi + \int_{\phi_0}^{2\pi} y(\phi) \frac{dx}{d\phi}(\phi) d\phi + 2\pi R^2= \\ &= \int_0^{\phi_0} y(\phi) \frac{dx}{d\phi}(\phi) d\phi + \int_{\phi_0}^{2\pi} y(\phi) \frac{dx}{d\phi}(\phi) d\phi + 2\pi R^2= \\ &= \int_0^{2\pi} y(\phi) \frac{dx}{d\phi}(\phi) d\phi + 2\pi R^2 = \\ &= (-\pi + \frac43\pi^3)R^2 + 2\pi R^2 = \\ &= (\pi + \frac43\pi^3)R^2 \end{align}
At the end we need to subtract $\pi R^2$, which is the area of the silo's base to get the answer: $$ A = \frac43 \pi^3 R^2 \approx 4134 {\rm m}^2$$