The problem asks us to evaluate the definite integral: $A(R) = \int_{-3}^1 (\frac{4-\sqrt{4-4y}}{2} - \frac{y+3}{4}) dy + \int_{0}^1 (\frac{6-y}{2} - \frac{4+\sqrt{4-4y}}{2}) dy + \int_{1}^3 (\frac{6-y}{2} - \frac{y+3}{4}) dy$

1. Problem Description

The problem asks us to evaluate the definite integral:

$A(R) = \int_{-3}^1 (\frac{4-\sqrt{4-4y}}{2} - \frac{y+3}{4}) dy + \int_{0}^1 (\frac{6-y}{2} - \frac{4+\sqrt{4-4y}}{2}) dy + \int_{1}^3 (\frac{6-y}{2} - \frac{y+3}{4}) dy$

2. Solution Steps

Let's evaluate the integrals separately and then add them up.

First Integral:

$\int_{-3}^1 (\frac{4-\sqrt{4-4y}}{2} - \frac{y+3}{4}) dy = \int_{-3}^1 (2 - \frac{\sqrt{4-4y}}{2} - \frac{y}{4} - \frac{3}{4}) dy$

$= \int_{-3}^1 (\frac{5}{4} - \frac{y}{4} - \frac{\sqrt{4-4y}}{2}) dy = \int_{-3}^1 \frac{5}{4}dy - \int_{-3}^1 \frac{y}{4}dy - \int_{-3}^1 \frac{\sqrt{4-4y}}{2}dy$

We evaluate the integrals one by one:

$\int_{-3}^1 \frac{5}{4} dy = \frac{5}{4}y|_{-3}^1 = \frac{5}{4}(1-(-3)) = \frac{5}{4}(4) = 5$

$\int_{-3}^1 \frac{y}{4} dy = \frac{1}{4} \int_{-3}^1 y dy = \frac{1}{4} \cdot \frac{y^2}{2}|_{-3}^1 = \frac{1}{8} (1^2 - (-3)^2) = \frac{1}{8}(1-9) = \frac{-8}{8} = -1$

$\int_{-3}^1 \frac{\sqrt{4-4y}}{2} dy = \frac{1}{2} \int_{-3}^1 \sqrt{4-4y} dy$

Let $u = 4-4y$. Then $du = -4 dy$, so $dy = -\frac{1}{4}du$. When $y=-3$, $u = 4-4(-3) = 4+12=16$. When $y=1$, $u = 4-4(1)=0$.

$\frac{1}{2} \int_{16}^0 \sqrt{u} (-\frac{1}{4}) du = -\frac{1}{8} \int_{16}^0 u^{1/2} du = \frac{1}{8} \int_{0}^{16} u^{1/2} du = \frac{1}{8} [\frac{u^{3/2}}{3/2}]_0^{16} = \frac{1}{8} \cdot \frac{2}{3} [u^{3/2}]_0^{16} = \frac{1}{12} [16^{3/2} - 0^{3/2}] = \frac{1}{12} (16^{1/2})^3 = \frac{1}{12} (4^3) = \frac{64}{12} = \frac{16}{3}$

Therefore, the first integral is $5 - (-1) - \frac{16}{3} = 6 - \frac{16}{3} = \frac{18-16}{3} = \frac{2}{3}$.

Second Integral:

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

$= \int_{0}^1 (1 - \frac{y}{2} - \frac{\sqrt{4-4y}}{2}) dy = \int_{0}^1 1 dy - \int_{0}^1 \frac{y}{2} dy - \int_{0}^1 \frac{\sqrt{4-4y}}{2} dy$

We evaluate the integrals one by one:

$\int_{0}^1 1 dy = y|_0^1 = 1 - 0 = 1$

$\int_{0}^1 \frac{y}{2} dy = \frac{1}{2} \int_{0}^1 y dy = \frac{1}{2} [\frac{y^2}{2}]_0^1 = \frac{1}{4} [1^2 - 0^2] = \frac{1}{4}$

$\int_{0}^1 \frac{\sqrt{4-4y}}{2} dy$

Let $u = 4-4y$. Then $du = -4 dy$, so $dy = -\frac{1}{4}du$. When $y=0$, $u = 4-4(0) = 4$. When $y=1$, $u = 4-4(1)=0$.

$\frac{1}{2} \int_{4}^0 \sqrt{u} (-\frac{1}{4}) du = -\frac{1}{8} \int_{4}^0 u^{1/2} du = \frac{1}{8} \int_{0}^{4} u^{1/2} du = \frac{1}{8} [\frac{u^{3/2}}{3/2}]_0^4 = \frac{1}{8} \cdot \frac{2}{3} [u^{3/2}]_0^4 = \frac{1}{12} [4^{3/2} - 0^{3/2}] = \frac{1}{12} (4^{1/2})^3 = \frac{1}{12} (2^3) = \frac{8}{12} = \frac{2}{3}$

Therefore, the second integral is $1 - \frac{1}{4} - \frac{2}{3} = \frac{12 - 3 - 8}{12} = \frac{1}{12}$.

Third Integral:

$\int_{1}^3 (\frac{6-y}{2} - \frac{y+3}{4}) dy = \int_{1}^3 (\frac{12-2y-y-3}{4}) dy = \int_{1}^3 (\frac{9-3y}{4}) dy = \int_{1}^3 \frac{9}{4} dy - \int_{1}^3 \frac{3y}{4} dy$

$= \frac{9}{4} \int_{1}^3 dy - \frac{3}{4} \int_{1}^3 y dy = \frac{9}{4} [y]_1^3 - \frac{3}{4} [\frac{y^2}{2}]_1^3 = \frac{9}{4}(3-1) - \frac{3}{4} (\frac{9-1}{2}) = \frac{9}{4}(2) - \frac{3}{4}(\frac{8}{2}) = \frac{18}{4} - \frac{3}{4}(4) = \frac{18}{4} - \frac{12}{4} = \frac{6}{4} = \frac{3}{2}$.

Adding the integrals:

$A(R) = \frac{2}{3} + \frac{1}{12} + \frac{3}{2} = \frac{8+1+18}{12} = \frac{27}{12} = \frac{9}{4}$

3. Final Answer

$\frac{9}{4}$