1. **Stating the problem:** Find the area inside both polar curves $r=3+2\cos(\theta)$ and $r=3+2\sin(\theta)$.\n\n2. **Formula for area inside a polar curve:** The area enclosed by a polar curve $r(\theta)$ from $\theta=a$ to $\theta=b$ is given by $$\text{Area} = \frac{1}{2} \int_a^b r(\theta)^2 \, d\theta.$$\n\n3. **Important rules:** To find the area common to both curves, we need to find the intersection points (limits of integration) and integrate the minimum of the two $r^2$ values over the appropriate intervals.\n\n4. **Find intersection points:** Set $3+2\cos(\theta) = 3+2\sin(\theta)$ which simplifies to $\cos(\theta) = \sin(\theta)$. This occurs at $\theta = \frac{\pi}{4}$ and $\theta = \frac{5\pi}{4}$.\n\n5. **Determine which curve is inside where:** For $\theta$ in $[0, \frac{\pi}{4}]$, $\cos(\theta) > \sin(\theta)$ so $r=3+2\cos(\theta)$ is larger. For $\theta$ in $[\frac{\pi}{4}, \frac{5\pi}{4}]$, $\sin(\theta) > \cos(\theta)$ so $r=3+2\sin(\theta)$ is larger.\n\n6. **Set up the integral for the common area:** The common area is the integral of the smaller radius squared over $\theta$ from $0$ to $2\pi$. By symmetry and the intersection points, the common area is twice the integral from $\frac{\pi}{4}$ to $\frac{5\pi}{4}$ of the smaller radius squared.\n\n7. **Calculate the area:**\n$$\text{Area} = 2 \times \frac{1}{2} \int_{\pi/4}^{5\pi/4} (3+2\cos(\theta))^2 \, d\theta = \int_{\pi/4}^{5\pi/4} (3+2\cos(\theta))^2 \, d\theta.$$\n\n8. **Expand the integrand:**\n$$(3+2\cos(\theta))^2 = 9 + 12\cos(\theta) + 4\cos^2(\theta).$$\n\n9. **Use identity:**\n$$\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}.$$\n\n10. **Rewrite integrand:**\n$$9 + 12\cos(\theta) + 4 \times \frac{1+\cos(2\theta)}{2} = 9 + 12\cos(\theta) + 2 + 2\cos(2\theta) = 11 + 12\cos(\theta) + 2\cos(2\theta).$$\n\n11. **Integrate term-by-term:**\n$$\int_{\pi/4}^{5\pi/4} 11 \, d\theta = 11 \times \left(\frac{5\pi}{4} - \frac{\pi}{4}\right) = 11 \times \pi = 11\pi.$$\n$$\int_{\pi/4}^{5\pi/4} 12\cos(\theta) \, d\theta = 12 \times (\sin(5\pi/4) - \sin(\pi/4)) = 12 \times \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = 12 \times (-\sqrt{2}) = -12\sqrt{2}.$$\n$$\int_{\pi/4}^{5\pi/4} 2\cos(2\theta) \, d\theta = 2 \times \frac{\sin(2\theta)}{2} \Big|_{\pi/4}^{5\pi/4} = \sin(2 \times \frac{5\pi}{4}) - \sin(2 \times \frac{\pi}{4}) = \sin(\frac{5\pi}{2}) - \sin(\frac{\pi}{2}) = 1 - 1 = 0.$$\n\n12. **Sum the integrals:**\n$$11\pi - 12\sqrt{2} + 0 = 11\pi - 12\sqrt{2}.$$\n\n**Final answer:** The area inside both curves is $$\boxed{11\pi - 12\sqrt{2}}.$$