Area of Row

For the letter $R$ above:

$A_{R} = -\int_{3}^{6} y^2 - 9y + 18 dy + 3 = -(\dfrac{y^3}{3} -\dfrac{9y^2}{2} + 18y)|_{3}^{6} + 3 = 3 + \dfrac{9}{2} = \boxed{\dfrac{15}{2}}$ .

For the letter $O$ using the ellipse above:

$A_{e} = 6\int_{3}^{5} \sqrt{1 - (x - 4)^2} + 1 dx$

Let $x - 4 - \sin(\theta) \implies dx = \cos(\theta) d\theta \implies A_{e} = 6(\dfrac{1}{2}(\theta + \dfrac{1}{2}\sin(2\theta))|_{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}} + x|_{3}^{5}) = \boxed{3(\pi + 4)}$

For the letter $W$ with the line $y = 6$ :

Using the symmetry about the line $x = 8$ :

On the left side of the line $x = 8$ we have a right triangle and a trapezoid $\implies A_{W} = 2(3 + \dfrac{9}{2}) = \boxed{15}$

$\therefore$ The sum of the areas $A_{ROW} = \dfrac{6\pi + 69}{2} \approx \boxed{43.924777}$ .

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Note: Using the definite integral for the letter $W$ with the line $y = 6$ we obtain:

Again, using the symmetry about the line $x = 8 \implies$ $A_{W} = 2(\int_{0}^{6} \dfrac{y + 21}{3} - (\dfrac{42 - y}{6}) dy - \int_{3}^{6} 8 - (\dfrac{y + 21}{3}) dy) =$

$2(\dfrac{1}{4}y^2|_{0}^{6} - \dfrac{1}{3}(3y-\dfrac{y^2}{2})|_{3}^{6} = 2(9 - \dfrac{3}{2}) = \boxed{15}.$