Area calculating part 3

area of green portion is equal to :area of rectangle - (area of circular arc $BEF$ + $\frac{1}{2}$ area of square having EC & BC as two sides ) $2-(\frac{45}{360} ×π×(\sqrt{2})^2+0.5)$

which upon solving gives $\frac{6-π}{4}$ so the correct answer is $6+4=\boxed{10}$

Solution:

Rectangle's area - (area of $\triangle BCE$ + area of sector $BEF$ ) = green region.

$\triangle BCE$ is isosceles and its area is $\frac{1}{2}$ .

$BE = \sqrt{2}$ and that is $r$ for the sector $BEF$ . Sector $BEF$ is $\frac{1}{8}$ of the whole circle, so its area is $\frac{r^{2}}{8}\pi = \frac{\pi}{4}$ .

Rectangle area is $2$ . So, the green region is $\frac{3}{2} - \frac{\pi}{4}$ [1]. In $\frac{1}{a}(b - \pi)$ , it can be seen that coefficient of $\pi$ is $1$ . After we transform [1] into appropriate term, we get that $\frac{1}{4}(6 - \pi)$ is the green area. Therefore, $a + b = 10$ .

First I want to say that my solution will not be the easiest--and maybe it's a bit unnecessary--but I just finished multi-variable calculus and so I see things as the Cartesian Coordinate Plane, and double integrals, so here goes...

First, we have a rectangle that can be described by the combination of these four equations

$\textrm{Rectangle}\left\{\begin{matrix} x=0, \: 0 \le y \le1\\ y=0, \: 0 \le x \le 2 \\y=1, \: 0 \le x \le 2 \\ x=2, \: 0 \le y \le 1 \end{matrix}\right.$

Then we have an arc which I figured to be a segment of the circle $(x-2)^2+y^2=2$ . This arc goes through the points $(2-\sqrt{2}, \; 0)$ and $(1, \; 1)$ . Let's call the green shaded area AFED a region "D". Here we have

$D= \left \{ (x, \; y) \; : \; 0 \le y \le 1, \: 0 \le x \le - \sqrt{2-y^2}+2 \right \}$

Note that $(x-2)^2+y^2=2 \: \Rightarrow x= \pm \sqrt{2-y^2}+2 \:$

So, in order to find the area of the shaded region we can evaluate the following iterated integral:

$\textrm{Area}= \int_{0}^{1} \int_{0}^{- \sqrt{2-y^2}+2} \; dx \; dy = \frac{1}{4} (6-\pi)$

We see that $a=4$ and $b=6$ and $6+4=\boxed{10}$

A calculus based solution It looks more difficult than the geometrical solution but gives the same result.

Assuming that AD forms the x-axis, you need to evaluate the area under the $f(x) = 2 - \sqrt{2 - (x-1)^2}$ where the limits of x are between 0 and 1 (the length of AD = 1).

Now let us evaluate, $\int_{0}^{1} f(x)dx$ .

$\int_{0}^{1} \big(2 - \sqrt{2 - (x-1)^2}\big) dx = 2 - \int_{0}^{1} \big(\sqrt{2 - (x-1)^2}\big) dx$

Assuming $x = 1 + \sqrt{2} sin(\theta )$ and $\theta$ varies from $- \dfrac{\pi}{4}$ to $0$ .

$\int_{0}^{1} \big(\sqrt{2 - (x-1)^2}\big) dx = \int_{\frac{-\pi}{4}}^{0} 2 \cos^2{ \theta} d\theta = \int_{\frac{-\pi}{4}}^{0} (1 + \cos{2\theta }) d\theta = \dfrac{\pi}{4} + \dfrac{\sin{2 \theta}}{2} \bigg|_{-\pi /4}^{0} = \dfrac{\pi}{4} + 0 - (-\dfrac{1}{2}) =$

$\dfrac{\pi}{4} + \dfrac{1}{2}$

Substituting the integral in the original equation gives us $\int_{0}^{1} f(x)dx = 2 - \dfrac{\pi}{4} - \dfrac{1}{2}= -\dfrac{\pi}{4} + \dfrac{3}{2} = \dfrac{1}{4} (6 - \pi) = \dfrac{1}{a} (b - \pi)$ . Hence, $a = 4$ and $b = 6$