Spiral doodle

Slip the blue region into three rings $A_1$ , $A_2$ , and $A_3$ as shown. As the areas are symmetrical about the $y$ -axis, we need only to consider the positive side of $x$ -axis then multiplied by $2$ . We note that:

$\begin{aligned} A_1 & = 2 \left(\int_\frac \pi 2^{\frac {3\pi}2} \frac {r^2}2 d\theta - \int_0^\frac \pi 2 \frac {r^2}2 d \theta \right) = \int_\frac \pi 2^{\frac {3\pi}2} \theta^2 \ d\theta - \int_0^\frac \pi 2 \theta^2 \ d \theta \\ & = \frac {\theta^3}3 \ \bigg|_\frac \pi 2^\frac {3\pi}2 - \frac {\theta^3}3 \ \bigg|_0^\frac \pi 2 = \frac {27}{24} \pi^3 - \frac 1{24} \pi^3 -\frac 1{24} \pi^3 - 0 = \frac {25}{24} \pi^3 \end{aligned}$

Similarly, $A_2 = \dfrac {\theta^3}3 \ \bigg|_\frac {5\pi}2^\frac {7\pi}2 - \dfrac {\theta^3}3 \ \bigg|_\frac {3\pi}2^\frac {5\pi}2 = \frac {120}{24} \pi^3$ and $A_3 = \dfrac {\theta^3}3 \ \bigg|^\frac {11\pi}2_\frac {9\pi}2 - \dfrac {\theta^3}3 \ \bigg|^\frac {9\pi}2_\frac {7\pi}2 = \frac {216}{24} \pi^3$ .

Therefore, $A_1+A_2+A_3 = \dfrac {25+120+216}{24} \pi^3 = \dfrac {361}{24}\pi^3$ $\implies a+b+c = 361 + 24 + 3 = \boxed{388}$ .

The area of the central white region is $W_1 = 2 \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^2 d\theta = \frac{1}{24}\pi^3$ .

The area of the central blue region is $B_1 = 2 \cdot \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \theta^2 d\theta - W_1 = \frac{26}{24}\pi^3 - \frac{1}{24}\pi^3 = \frac{25}{24}\pi^3$ .

The area of the next white region $W_2$ with $B_1$ and $W_1$ is $B_1 + W_1 + W_2 = 2 \cdot \frac{1}{2} \int_{\frac{3\pi}{2}}^{\frac{5\pi}{2}} \theta^2 d\theta = \frac{98}{24}\pi^3$ .

The area of the next blue region is $B_2 = 2 \cdot \frac{1}{2} \int_{\frac{7\pi}{2}}^{\frac{5\pi}{2}} \theta^2 d\theta - (B_1 + W_1 + W_2) = \frac{218}{24}\pi^3 - \frac{98}{24}\pi^3 = \frac{120}{24}\pi^3$ .

The area of the third white region $W_3$ with $B_1$ , $B_2$ , $W_1$ , and $W_2$ is $B_1 + B_2 + W_1 + W_2 + W_3 = 2 \cdot \frac{1}{2} \int_{\frac{7\pi}{2}}^{\frac{9\pi}{2}} \theta^2 d\theta = \frac{386}{24}\pi^3$ .

The area of the third blue region is $B_3 = 2 \cdot \frac{1}{2} \int_{\frac{9\pi}{2}}^{\frac{11\pi}{2}} \theta^2 d\theta - (B_1 + B_2 + W_1 + W_2 + W_3) = \frac{602}{24}\pi^3 - \frac{386}{24}\pi^3 = \frac{216}{24}\pi^3$ .

The total area of the blue regions is therefore $B = B_1 + B_2 + B_3 = \frac{25}{24}\pi^3 + \frac{120}{24}\pi^3 + \frac{216}{24}\pi^3 = \frac{361}{24}\pi^3$ , so $a = 361$ , $b = 24$ , $c = 3$ , and $a + b + c = \boxed{388}$ .