1. **Problem statement:** We want to evaluate the double integral $$\iint_\mathcal{D} x^2 \, dx \, dy$$ where $\mathcal{D}$ is the interior of the ellipse given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$\n\n2. **Formula and approach:** To integrate over an ellipse, a common method is to use a change of variables that transforms the ellipse into a unit circle. Define new variables $$u = \frac{x}{a}, \quad v = \frac{y}{b}.$$ Then the ellipse becomes $$u^2 + v^2 \leq 1.$$\n\n3. **Jacobian of the transformation:** The area element transforms as $$dx \, dy = |J| \, du \, dv,$$ where the Jacobian determinant is $$|J| = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| = ab.$$\n\n4. **Rewrite the integral:** Substitute $x = au$, $y = bv$, and $dx \, dy = ab \, du \, dv$ into the integral:\n$$\iint_\mathcal{D} x^2 \, dx \, dy = \iint_{u^2+v^2 \leq 1} (au)^2 (ab) \, du \, dv = a^2 b \iint_{u^2+v^2 \leq 1} u^2 \, du \, dv.$$\n\n5. **Symmetry and integration over the unit disk:** The region is the unit disk in $uv$-plane. Because of symmetry, the integral of $u^2$ over the unit disk equals the integral of $v^2$. Also, $$\iint_{u^2+v^2 \leq 1} (u^2 + v^2) \, du \, dv = \pi \cdot \frac{1^4}{2} = \frac{\pi}{2}$$ (using polar coordinates). Since $u^2$ and $v^2$ contribute equally,\n$$\iint_{u^2+v^2 \leq 1} u^2 \, du \, dv = \frac{1}{2} \iint_{u^2+v^2 \leq 1} (u^2 + v^2) \, du \, dv = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.$$\n\n6. **Final evaluation:** Substitute back to get\n$$\iint_\mathcal{D} x^2 \, dx \, dy = a^2 b \cdot \frac{\pi}{4} = \frac{\pi a^2 b}{4}.$$\n\n**Answer:** $$\boxed{\frac{\pi a^2 b}{4}}.$$