Evaluate the integral $\iint\limits_R \sqrt{1-x^2} {d}A$ If $R = \{(x, y) | -1 \leqslant x \leqslant 1, -2 \leqslant y \leqslant 2 \}$, evaluate the integral
$$
\iint\limits_R \sqrt{1-x^2} {d}A
$$
The author of the book ('Multivariable Calculus' by James Stewart) explains that
$$
\iint\limits_R \sqrt{1-x^2} {d}A = \frac{1}{2}\pi(1)^2\times4=2\pi
$$
And I'm failing to understand why
$$
\iint\limits_R \sqrt{1-x^2} {d}A
$$
is equal to
$$
\frac{1}{2}\pi(1)^2\times4
$$
How does $\pi$ come into play?
Could you please show me the steps and assumptions that the author is making? I really tried to figure this out by my own, but as a last resource I need to ask you guys.
 A: Hint:
note that $z=\sqrt{1-x^2}$ is a semicircle of radius $1$. So
$$
\mbox{ Area semicircle=}\int_{-1}^1\sqrt{1-x^2} dx=\frac{\pi}{2} (1)^2
$$
A: I would say
$\displaystyle I=\int_{-2}^{2}\int_{-1}^{1} \sqrt{1-x^2}\,dx\,dy=\int_{-2}^{2}\left(\int_{-1}^{1} \sqrt{1-x^2}\,dx\right)\,dy$
and 
$\int_{-1}^{1} \sqrt{1-x^2}\,dx$ = the area of semicircle of radius  1 = $\pi/2$
$\displaystyle \Rightarrow I = \int_{-2}^{2}\left(\frac{\pi}{2}\right)\,dy=\cdots$
A: Applying Fubini theorem $$I = \int_{-1}^{1} \int_{-2}^{2} {\sqrt{1-x^{2}} dx dy} = \int_{-2}^{2} (\int_{-1}^{1} {\sqrt{1-x^{2}} dx}) dy$$ In order to calculate $I_{1} = \int_{-1}^{1}{\sqrt{1-x^{2}}dx}$ it's possible to  change of variable in the following way: $x = \cos{\varphi}, 0 \leq \varphi \leq \pi$, then $dx = -\sin{\varphi} d\varphi$. So, $I_{1} = -\int_{0}^{\pi}{\sin^{2}{\varphi} d\varphi}$, which can be evaluated by the power reduction, for example.
A: The reason for the $\pi$ is the use of trigonometric substitution.
Working in a Cartesian coordinate system,
$$\mathrm{d}A=\mathrm{d}x\mathrm{d}y$$
Therefore, we can change in the integral - and its bounds. We are given that
$$-1\leq x\leq 1,\quad -2\leq y\leq 2$$
Therefore, we have
$$\int_{-2}^2\int_{-1}^1\sqrt{1-x^2}\mathrm{d}x\mathrm{d}y$$
We can start with trigonometric substitution:
$$x=\sin\theta,\quad \mathrm{d}x=\cos\theta \mathrm{d}\theta$$
Given the property that
$$1-\sin^2\theta=\cos^2\theta$$
we have
$$\int_{-2}^2\int_{\theta_a}^{\theta_b}\cos^2\theta \mathrm{d}\theta \mathrm{d}y$$
That trigonometric substitution shows you that $\pi$ will be involved. All you have to do is integrate the $\cos^2\theta$, substitute back in $x$ (or directly solve it using the bounds for $\theta$) and then integrate over the bounds for $y$.
A: As we see, the ranges of $x$ and $y$ shows that these two variabels are independently acting here, so, just to think of solving $$\int_{-1}^{+1}\sqrt{1-x^2}dx$$
