Irrational Integral

The curve $y= \sqrt{1-x^2}$ represents the potion of the circle^with unit radius $x^2+y^2=1$ first quadrant.

The area: $\int_{0}^{1} \sqrt{1-x^2} \ dx$

Represents the area of the circle in the first quadrant which is essentially the area of a quarter circle.

The area: $\int_{-1}^{1} \sqrt{1-x^2} \ dx$

Represents the area of the circle with unit radius above the X-axis which is essentially the area of a semi-circle.

Therefore the area: $2\int_{-1}^{1} \sqrt{1-x^2} \ dx$

Is twice the area of a semicircle of radius $1$ which is the area of a complete circle. Therefore, the answer is $\pi r^2 = \boxed{\pi}$

$\int_{-1}^1 \sqrt{1-x^2}dx$ Take $x^2=t$ , the integral becomes $2\int_0^1 t^{-1/2} \sqrt{1-t} dt$ $=2\Beta{(\dfrac{1}{2},\dfrac{3}{2})}=2\dfrac{\Gamma{(\dfrac{1}{2})}\Gamma{(\dfrac{3}{2})}}{\Gamma{(2)}}$

$= 2×\sqrt{π}×\dfrac{\sqrt{π}}{2} =\boxed{π}$ Relevant wiki

- Beta function

$I= \displaystyle 2\cdot \int_{-1}^1 \sqrt{1-x^2}\, dx = \int_{-1}^1 \sqrt{1-x^2}\, dx + \int_{1}^{-1} -\sqrt{1-x^2}\, dx$

Note that if $S= \{ (x,y) \in \mathbb{R}^2 : x^2+y^2\leq1\}$ , this sum of integrals is exactly the area integral:

$I=\iint_S\, dA = \iint_S \vec{\nabla}\times (0,x,0)\, d\vec{A}$

And, now, using Stokes' Theorem , we find out:

$I=\oint_{\partial S} (0,x,0) \cdot d\vec{l}$

As $\partial S$ is the unit circle, we can use polar coordinates:

$I=\int_{0}^{2\pi} (0,\cos(\theta), 0) \cdot (-\sin(\theta), \cos(\theta), 0)\, d\theta = \int_{0}^{2\pi} \cos^2(\theta)\, d\theta$

$\Leftrightarrow I= \int_{0}^{2\pi} \frac{1+\cos(2\theta)}{2}\, d\theta = \int_{0}^{2\pi} \frac{1}{2}\, d\theta + \frac{1}{2}\cdot \int_{0}^{2\pi} \cos(2\theta)\, d\theta$

$\Leftrightarrow I = \pi + \frac{1}{2}\cdot \frac{1}{2}\cdot (\sin(4\pi) - \sin(0)) = \boxed{\pi}$