y = \sqrt{1 - x^2}

Graph of $y = \sqrt{1 - x^2}$ (Upper Semicircle)

$y = \sqrt{1 - x^2}$ is the irrational function describing the upper half of the circle of radius $1$ centred at the origin¹. Squaring both sides produces the equation of the unit circle.

x^2 + y^2 = 1

The full circle is not a function, since a single $x$ would have two values of $y$ . Taking only the positive square root carves out the upper half, and that is this function.

Domain and range

The domain is $-1 \leq x \leq 1$
The range is $0 \leq y \leq 1$
Even function
The domain is a closed bounded interval

The radicand must be non-negative, and that is what confines the domain. A closed interval like this is unusual among the functions met so far.

Monotonicity and the top

The derivative is $y' = \dfrac{-x}{\sqrt{1 - x^2}}$ . It is positive for $x < 0$ and negative for $x > 0$ , so the function rises to a maximum of $1$ at the origin, the top of the semicircle, where the tangent is horizontal.

The endpoints

As $x$ approaches $\pm 1$ the denominator tends to $0$ , so $|y'| \to \infty$ : the tangents at $(-1, 0)$ and $(1, 0)$ stand vertical and the function is not differentiable there. Recalling that the tangent to a circle is perpendicular to the radius, this is exactly what one expects. The curve runs into the $x$ -axis at right angles and stops.

Notable points

$x$	$y$
$0$	$1$
$\pm\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$
$\pm\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$
$\pm 1$	$0$

Every point of the curve is at distance $1$ from the origin, and the curvature is the same everywhere. Unlike a parabola, the bending never varies along the curve.

Other ways to write it

The parametrisation $x = \cos\theta$ , $y = \sin\theta$ with $0 \leq \theta \leq \pi$ traces the same arc. The lower half is $y = -\sqrt{1 - x^2}$ , and together they make the whole circle. For radius $r$ the function becomes $y = \sqrt{r^2 - x^2}$ .

Integration and $\pi$

The area between this curve and the $x$ -axis is the area of a semicircle.

\int_{-1}^{1} \sqrt{1 - x^2}\,dx = \frac{\pi}{2}

It is a classic and beautiful way to extract $\pi$ from an integral. Integrals containing $\sqrt{1 - x^2}$ are also the standard setting for the trigonometric substitution $x = \sin\theta$ .

Circle, Wikipedia

Graph of y=1−x2y = \sqrt{1 - x^2}y=1−x2​ (Upper Semicircle)