Proof of the formula for the area of a circle

Hello everyone. Today we will prove the formula for the area of ​​a circle. We will use the method of direct proof as a proof method. 

Theorem: Denote by $P$ the area of ​​a circle and by $r$ the radius of a circle. Then the following equation holds: $P = r ^ 2 \pi$ 

Proof: 

The equation of a circle in Cartesian  coordinate system is $x ^ 2 + y ^ 2 = r ^ 2$. Hence we have that $y = \pm \sqrt {r ^ 2-x ^ 2}$. Based on the geometric interpretation of a certain integral, it follows: $$P = \int \limits _ {- r} ^ r \left(\sqrt {r ^ 2-x ^ 2} - \left (- \sqrt {r ^ 2-x ^ 2} \right) \right) \, dx$$ So, $$P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2-x ^ 2} \, dx$$ $$P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2 \left (1- \frac {x ^ 2} {r ^ 2} \right)} \, dx$$ $$P = \int \limits _ {- r} ^ r 2r \sqrt {1- \frac {x ^ 2} {r ^ 2}} \, dx$$ Let us now introduce the substitution $x = r \cos \theta$. Hence, we have that $dx = -r \sin \theta d \theta$. So we can write the following equation: $$P=-\int\limits_{\pi}^0 2r^2\sqrt{1-\frac{r^2\cos^2\theta}{r^2}}\sin\theta \, d\theta$$ After simplification we get:  $$P=-\int\limits_{\pi}^0 2r^2\sqrt{1-\cos^2\theta}\sin\theta \, d\theta$$$$P=-\int\limits_{\pi}^0 2r^2\sin^2\theta \, d\theta$$$$P=-r^2\int\limits_{\pi}^0 2\sin^2\theta \, d\theta$$$$P=-r^2\int\limits_{\pi}^0 (1-\cos2\theta) \, d\theta$$$$P=-r^2\left(\int\limits_{\pi}^0 d\theta-\int\limits_{\pi}^0 \cos2\theta \, d\theta\right)$$Since $\int\limits_{\pi}^0 \cos2\theta \, d\theta=0$ we have: $$P=-r^2\int\limits_{\pi}^0 d\theta$$$$P=-r^2(0-\pi)$$$$P=r^2\pi$$

$\blacksquare$