First Quarter Portion Of A Circular Disc

Let $P = (-1,-1)$ denote the center of the circle whose equation is $(x+1)^2 + (y+1)^2 = 3^2$ . Let $Q$ and $R$ be the intersection points between the circle with the $y$ -axis and the $x$ -axis, respectively.

We first find the coordinates of $Q$ and $R$ , respectively.

At point $Q$ , $x = 0 \Rightarrow (0+1)^2 +(y+1)^2 = 3^2 \Rightarrow y = \sqrt8 - 1$ , we take the positive root only. Thus, $Q = (0, \sqrt8 - 1)$ .

Likewise, we get $R = (\sqrt8 - 1, 0)$ .

The area of the orange region can be expressed as the sum of the area of the purple region and the area of the yellow region as shown in the figure on the right.

The area of the purple region represents the area of the right triangle $OQR$ ,

$\text{Area}_\text{purple} = \dfrac12 \times \text{ base } \times \text{ height } = \dfrac12 ( \sqrt8 - 1)^2 = \dfrac92 - 2\sqrt2 .$

The area of the the yellow region represents the area of the segment $QPR$ , let $\theta = QPR$ , then

$\text{Area}_\text{yellow} =\dfrac12 r^2 (\theta - \sin\theta) = \dfrac92 (\theta - \sin\theta) .$

The acute angle $\theta$ can be calculated by applying the cosine rule on the triangle $PQR$ ,

$\cos \theta = \dfrac{(PQ)^2 + (PR)^2 - (QR)^2 }{2 (PQ)(PR) } = \dfrac{(3^2) + (3^2) - ((\sqrt 8 - 1)^2 + (\sqrt 8 - 1)^2 ) }{2 (3)(3) } = \dfrac{4\sqrt2}9 \qquad \Rightarrow \qquad \sin\theta = \sqrt{1- \cos^2\theta} = \dfrac79 .$

Continue where we left off, we have $\text{Area}_\text{yellow} = \dfrac92 \left( \arccos\left( \dfrac{4\sqrt2}9\right)-\dfrac 79\right)$ .

Thus, adding the two areas gives the desired answer

$\begin{aligned} \text{Area}_\text{orange} &= &\text{Area}_\text{purple} + \text{Area}_\text{orange} \\ &= & \dfrac92 - 2\sqrt2 + \dfrac92 \left( \arccos\left( \dfrac{4\sqrt2}9\right)-\dfrac 79\right) \\ &= & \dfrac92 \left( \arccos\left( \dfrac{4\sqrt2}9\right)+\dfrac 29\right) -2\sqrt2 \\ &\approx & \boxed{2.18} \end{aligned}$

The equation of the circle in parametric form is given by

$r(t) = r_c + v_1 \cos t + v_2 \sin t$

where $r_c = [-1, -1, 0]^T , v_1 = [3, 0, 0]^T , v_2 = [ 0, 3, 0 ]$

We need to find values of t, that correspond to the intersection of the circle with the positive x-axis, and the positive y-axis. Using simple geometry, we immediately find that

$t_1 = \sin^{-1} \frac {1}{3} = 0.3398369095$ and $t_2 = \frac{\pi}{2} -\sin^{-1} \frac {1}{3} = 1.230959417341$

Now we use the follwing formula for finding areas of parametric curves:

$A = \frac{1}{2} \| \oint r(t) \times \frac{dr}{dt} dt \|$

This integral reduces to

$A = \frac{1}{2} \| \int_{t_1}^{t_2} r(t) \times \frac{dr}{dt} dt \|$

The integrand is as follows

$r(t) \times \frac{dr}{dt} = (r_c + v_1 \cos t + v_2 \sin t) \times (- v_1 \sin t + v_2 \cos t)$

Evaluating the cross product, results in

$r(t) \times \frac{dr}{dt} = (r_c \times v_1 ) (-\sin t) + (r_c \times v_2) (\cos t) + ( v_1 \times v_2 )$

The cross products have only a z-component that is nonzero. So this will be the magnitude of the vector. In particular,

$r_c \times v_1 = [0, 0, 3]^T, r_c \times v_2 = [0, 0, -3], v_1 \times v_2 = [0,0,9]$

Therefore, the area is given by

$A = \frac{1}{2} \int_{t_1}^{t_2} ( 3 (-\sin t) - 3 \cos t + 9) dt$

$A = \frac{1}{2} \{ 3 (cos(t_2) - cos(t1)) - 3(sin(t_2) - sin(t_1)) + 9 (t_2 - t_1) \}$

Substituting the values of $t_1$ and $t_2$ , we obtain,

$A = \boxed{2.18}$