Find the area bounded by the arc of the curve y = ± x

Find the area bounded by the arc of the curve $y = \pm x\sqrt{1 - x^2}$.

We will use the curve's point symmetry about the origin to reduce the area to a single elementary integral.

Step 1: Understand the curve

The curve is defined for $|x|\leq1$ (so that $1-x^2\geq0$), and consists of two branches: the upper part $y=x\sqrt{1-x^2}$ and the lower part $y=-x\sqrt{1-x^2}$.

Step 2: Find key points

At $x=0$: $y=0$. At $x=\pm1$: $y=0$.

Maximum of the upper curve: Differentiating $y=x\sqrt{1-x^2}$:

\[\frac{dy}{dx} = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} = \frac{1-2x^2}{\sqrt{1-x^2}}\]

Setting equal to zero: $x^2=\frac12 \implies x=\frac{1}{\sqrt2}$, giving $y=\frac12$. Maximum point: $\left(\frac{1}{\sqrt2},\frac12\right)$; by symmetry the lower branch has a minimum at $\left(\frac{1}{\sqrt2},-\frac12\right)$.

Step 3: Analyze symmetry

The curve is symmetric about the origin (if $(x,y)$ lies on it, so does $(-x,-y)$), and the upper and lower branches are mirror images across the $x$-axis.

Step 4: Set up the integral

For $x \in [0,1]$, the upper boundary of the enclosed region is $y=x\sqrt{1-x^2}$ and the lower boundary is $y=-x\sqrt{1-x^2}$, a vertical gap of $2x\sqrt{1-x^2}$. By the origin symmetry, the region for $x<0$ contributes an identical area, so:

\[A = 2\int_0^1 2x\sqrt{1-x^2} \, dx = 4\int_0^1 x\sqrt{1-x^2} \, dx\]

Step 5: Evaluate the integral

Let $u=1-x^2$, so $du=-2x\,dx$, i.e. $x\,dx=-\frac12du$. When $x=0$, $u=1$; when $x=1$, $u=0$.

\[\int_0^1 x\sqrt{1-x^2} \, dx = \frac{1}{2}\int_0^1 u^{1/2}\,du = \frac12\left[\frac{2u^{3/2}}{3}\right]_0^1 = \frac13\]

Step 6: Calculate the total area

\[A = 4\cdot\frac13 = \frac43\]

\[\boxed{A = \frac{4}{3} \text{ square units}}\]

Step 7: Geometric picture

The curve traces a symmetric two-lobed figure passing through $(-1,0)$, $(0,0)$, and $(1,0)$, with maximum height $\frac12$ and minimum height $-\frac12$.

Summary

\[\boxed{A = \frac{4}{3} \text{ square units}}\]

Key insight: Even though the curve is given as $y=\pm(\ldots)$ rather than a single function, its point symmetry about the origin lets the whole enclosed area be reduced to four times a single elementary integral on $[0,1]$.