Grazing area for a goat around a circle.

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: $$ A = \frac43 \pi^3 R^2 \approx 4134 {\rm m}^2$$

I have found another, much simpler derivation of the result. Let us divide the area of grazing into infinitesimaly thin triangles, as below:

They all approximately isoceles. The triangle with the apex at point $(R\sin\phi,R-R\cos\phi)$ has the legs of the length $L(\phi) = (2\pi-\phi)R$ (the length of unwinded chain). they are supposed to be very thin, with the angle at the apex being $\Delta\phi$. Therefore, the area of each triangle is

$$\Delta S \approx \frac12 L(\phi)^2 \sin( \Delta\phi) \approx \frac12 (2\pi-\phi)^2R^2 \Delta\phi$$

The full area is therefore equal to

$$ S = \int_0^{2\pi} \frac12 (2\pi-\phi)^2 R^2d\phi = -\frac{(2\pi-\phi)^3R^2}{6}\Big|_{\phi=0}^{2\pi} = \frac43\pi^3R^2$$

The previous method I've given is more general: it can be used for any area with a boundery with known parametrization. This answer is specific to this problem.