Graph of this integral. Is it a half circle or quarter circle? I'm reading this text:

$$\int_0^1 \sqrt{1 - x^2} \cdot dx$$
Why isn't the above integral the right half of a circle? Why did they show the equation of $y^2$? Was the point of that just to show that it's a circle? But that just confuses people because they don't really explain why it's just 1 quarter of a circle. I guess y cannot be negative? But $y$ can be negative in the $y^2$ equation.
 A: The question is asking you to evaluate the integral
$$\int_{0}^{1} \sqrt{1-x^2}\,\mathrm{d}x$$
by interpreting this integral as an area.  In order to do this, we need to understand the shape of the region being enclosed.  To do this, we might reason as follows:


*

*Since the limits of the integral are $0$ and $1$, the area we are interested lives on the interval $[0,1]$.  That is the area pictured here:


*Next, the integral is the area enclosed between the curve $y = \sqrt{1+x^2}$ and the $x$-axis.  Note that the square root function is, by definition, the positive square root.  This implies that when $\sqrt{x}$ is defined, it is nonnegative.  In other words, the curve that bounds the region of interest, i.e. $y = \sqrt{1-x^2}$, lives only above the $x$-axis.  It doesn't extend below the $x$-axis at all.  Hence we are interested in the region that is above the $x$-axis and below the curve.

*Finally, using a graphing utility, this looks like a semicircle above the $x$-axis:

If we are being careful (as we ought to be), we should check this:  recall that that a circle centered at the point $(h,k)$ with radius $r$ is described by the equation
$$ (x-h)^2 + (y-k)^2 = r^2. $$
Since the region that we are interested in looks like a semicircle of radius 1 centered at the origin, this should be the region bounded by the curve
$$ x^2 + y^2 = 1. $$
Solving for $y$, we get
$$ y^2 = 1-x^2
\implies y = \pm \sqrt{1-x^2}.$$
We are interested in the region above the $x$-axis and below the line $y = \sqrt{1-x^2}$.  The above argument shows that this upper bound is a semicircle.  Yay!
Intersecting the blue and red regions above, we get the quarter circle, which has area $\frac{\pi}{4}$.

A: If $x^2 + y^2 = 1$ then $y = \pm \sqrt{1-x^2}.$ You've got only the "plus" part; not the "minus" part.
A: Remember that $\sqrt{1-x^2} \geq 0$ for all $x$ because $\sqrt{x} \geq 0$ for all $x$.
Perhaps you are thinking of the following derivation:
$$\begin{align}x^2+y^2&=1 \\
y^2&=1-x^2\\
y&=\sqrt{1-x^2}\\
\end{align}$$
But remember that $x^2+y^2=1$ is an implicit curve, not a function. For us to end up with a function we restrict $\sqrt{x}$ to principal square roots. In reality, to get 'both quarters', you need:
$$y=\pm \sqrt{1-x^2}$$
