Circular-Reasoning

First step, find point of intersection. Quit easy to do since we have

$\begin{array}{l} x = \frac{1-t^2}{1+t^2} \\ y = \frac{2t}{1+t^2} \\ \end{array}$

We can plug into the equation $y^2 = x - 1$ to get:

$\Bigg( \frac{2t}{1+t^2} \Bigg)^2 = \frac{1-t^2}{1+t^2}-1$

This can be simplified to yield:

$\frac{4t^2}{(1+t^2)^2} = \frac{-2t^2}{1+t^2}$ $2t^2(1+t^2)(3+t^2) = 0$

This can only happen when $t = 0$ which then means that

$\begin{array}{l} x = \frac{1-(0)^2}{1+(0)^2} = 1 \\ y = \frac{2(0)}{1+(0)^2} = 0\\ \end{array}$

The point of intersection is then $(x,y) = (1,0)$

Now, we can calculate the area using basic integration:

$A = \int_{0}^{1} y dx$

But the complicated function is expressed in parametric form, so we can convert the integral as follows:

$y = \frac{2t}{1+t^2}$

$\begin{array}{l} \frac{dx}{dt} = \frac{d}{dt} \Bigg(\frac{1-t^2}{1+t^2} \Bigg) = \frac{-4t}{(1+t^2)^2} \\ dx = \frac{-4t}{(1+t^2)^2}dt \\ \end{array}$

Clearly, when $x = 0$ we have $t = 1$ and when $x = 1$ we have $t = 0$ so the integral becomes:

$\begin{array}{l} A = \int_{1}^{0} \frac{2t}{1+t^2} \frac{-4t}{(1+t^2)^2}dt \\ = -8 \int_{1}^{0} \frac{t^2}{(1+t^2)^3} dt \\ \end{array}$

Now using a trig substitution, mainly $t = \tan\theta$ : $t=1 \rightarrow \theta = \pi /4$ and $t=0 \rightarrow \theta=0$ :

$\begin{array}{l} A = -8 \int_{\pi/4}^{0} \frac{\tan^2{\theta}}{(\sec^2(\theta))^3} \sec^2{\theta} d \theta \\ = -8 \int_{\pi/4}^{0} \sin^2(\theta) \cos^2(\theta) d \theta \\ = -8 \int_{\pi/4}^{0} \frac{\sin^2(2 \theta)}{4} d \theta \\ = - \frac{8}{4} \int_{\pi/4}^{0} \frac{1- \cos(4 \theta)}{2} d \theta \\ = - \Bigg[ \theta - \frac{1}{4} \sin(4 \theta) \Bigg]_{\pi/4}^{0} \\ = - \Bigg[ 0 - \big( \frac{\pi}{4} - 0 \big) \Bigg] \\ = \frac{\pi}{4} \approx \boxed{0.785} \\ \end{array}$

Rewrite the parametric curve a bit: $x=\frac{1-t^2}{1+t^2},\, y=\frac{2t}{1+t^2}$

Therefore, $x^2+y^2=\frac{1-2t^2+t^4+4t^2}{(1+t^2)^2}=\frac{1+2t^2+t^4}{(1+t^2)^2}=\frac{(1+t^2)^2}{(1+t^2)^2}=1$

Thus, the parametric curve is just a unit circle $x^2+y^2=1$

Solve both $x^2+y^2=1$ and $y^2=x-1$ , we find out $x=-2$ or $x=1$

Since $y^2=x-1$ is not defined at $x=-2$ , we take $x=1$

Notice that we are now finding the area of part of unit circle that lies above x-axis for $0\le x\le 1$ , which is a quarter circle.

So the area $=\pi (1)^2(\frac{1}{4})=\frac{\pi}{4}=\boxed{0.785398163}$

Let $t = \tan\Big(\frac{x}{2}\Big)$ . Then we have $\begin{aligned} \frac{2t}{1+t^2} = & \frac{2\tan\Big(\frac{x}{2}\Big)}{1+\tan^2\Big(\frac{x}{2} \Big)} \\ = & \frac{2\tan\Big(\frac{x}{2}\Big)}{\sec^2\Big(\frac{x}{2}\Big)}\\ = & 2\tan\Big(\frac{x}{2}\Big) \cos^2\Big(\frac{x}{2}\Big) \\ = & \sin(x). \end{aligned}$ Similar reasoning shows that the parametric equation for $x(t)$ in terms of $x$ is $\cos(x)$ . Therefore the parametric equation represents a circle with radius one. We know that the point (1,0) belongs to this circle, and we can easily see by inspection it also belongs to the curve $y^2=x-1$ . Therefore the area requested is $\int_0^1 \sqrt{1-x^2}\,dx \, \, \, \, \color{#3D99F6} \text{via } x^2+y^2 =1 \Rightarrow y = \sqrt{1-x^2}$ but we know this is also the area of a quarter circle, so the area can be solved conceptually: $\int_0^1 \sqrt{1-x^2}\,dx = \frac{\pi}{4} \approx \boxed{0.785}$

Given that $\textbf r(t) = \left(\frac {(1-t)(1+t)}{1+t^2}, \frac {2t}{1+t^2}\right) = \left(\frac {1-t^2}{1+t^2}, \frac {2t}{1+t^2}\right) = (\cos \theta, \sin \theta )$ , where $t = \tan \frac \theta 2$ . Therefore, $x=\cos \theta$ and $y = \sin \theta$ $\implies x^2+y^2=1$ which is a circle centered at the origin $O(0,0)$ and with a radius of 1.

The point of intersection between the circle $\textbf r(x)$ and curve $y^2=x-1$ is given by:

$\begin{aligned} \color{#3D99F6} y^2 & = x - 1 & \small \color{#3D99F6} \text{Since }x^2+y^2 = 1 \\ 1-x^2 & = x-1 \\ x^2 + x - 2 & = 0 \\ (x-1)(x+2) & = 0 \\ \implies x & = 1 & \small \color{#3D99F6} \text{Note that }x=-2 \text{ is outside the circle }\textbf r(x). \end{aligned}$

When $x=1$ , $y^2 = 1-1 = 0$ $\implies y = 0$ . The area under $\textbf r(x)$ and above the $x$ -axis from $x=0$ to $x=1$ is a quarter of a circle of radius 1 or $\frac \pi 4 \approx \boxed{0.785}$ .