area of points in a square closer to the center than the edge I'm hoping someone can confirm my solution to the following problem or perhaps provide better alternatives. Cheers.
A shape is defined by consisting of all the points closer to the center than the edge of a 2 inch square. Find the area of the shape.
Use symmetry is simplify the integral. Placing the center at the origin, we can find $\frac{1}{8}$ of the area by considering an angle sweep of $\frac{\pi}{4}$ radians. With that condition we know: the left and upper bounds are given by $y=x$ and the lower bound is the $x$-axis. We need to find the right bound.
Find the equation that defines the right bound or the rounded edge from the $x$-axis to the intersection of the line $y=x$.
Let $r$ be the distance from the center, the origin, to a point the edge of the shape $(x,y)$. Then
\begin{align}
    r&=\sqrt{x^2+y^2}\\
\text{and for the first $\frac{\pi}{4}$ radians $r$ is restricted to}
    r&=1-x\\
\text{$\therefore$}    
    \sqrt{x^2+y^2}&=1-x\\
    x^2+y^2&=(1-x)^2\\
    y^2&=1-2x+x^2-x^2\\
    y&=\sqrt{1-2x}
\end{align}
Find intersection of $y=x$ and $y=\sqrt{1-2x}$
\begin{align}
    x&=\sqrt{1-2x}\\
    x^2&=\sqrt{1-2x}^2\\
    0&=1-2x-x^2\\
    0&=x^2+2x-1\\
    1&=x^2+2x\\
    1+1&=x^2+2x+1\\
    2&=(x+1)^2\\
    \sqrt{2}&=x+1\\
    \sqrt{2}-1&=x
\end{align}
Find the $x$-intercept of $\sqrt{1-2x}$
\begin{align}
    0&=\sqrt{1-2x}\\
    0&=1-2x\\
    -1&=-2x\\
    \frac{1}{2}&=x
\end{align}
Now find the area under the curve $y=x$ from $0$ to $\sqrt{2}-1$ and the area under the curve $y=\sqrt{1-2x}$ from $\sqrt{2}-1$ to $\frac{1}{2}$
\begin{align}
    \frac{1}{8}A&=\int_0^{\sqrt{2}-1}xdx+\int_{\sqrt{2}-1}^{\frac{1}{2}}\sqrt{(1-2x)}dx\\
    A&=8\bigg(\int_0^{\sqrt{2}-1}xdx+\int_{\sqrt{2}-1}^{\frac{1}{2}}\sqrt{(1-2x)}dx\bigg)\\
    A&\approx 0.876in^2
\end{align}

Since our square is $[{-1},1]^2$ the arc you considered is not characterized by $\sqrt{x^2+y^2}=2-x$, but by $\sqrt{x^2+y^2}=1-x$. It is a parabolic arc given by $$x={1\over2}(1-y^2)\qquad\bigl(0\leq y\leq\sqrt{2}-1\bigr)\ .$$
