1. **State the problem:** We want to evaluate the double integral $$\iint_D x^2 \, dx \, dy$$ where $D$ 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, it is convenient 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 $$u^2 + v^2 \leq 1,$$ which is the unit disk $D'$ in the $uv$-plane.\n\n3. **Jacobian of the transformation:** The area element transforms as $$dx \, dy = |J| \, du \, dv,$$ where the Jacobian determinant is $$J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} a & 0 \\ 0 & b \end{vmatrix} = ab.$$\n\n4. **Rewrite the integral:** Substitute $x = au$, $y = bv$, and $dx \, dy = ab \, du \, dv$ into the integral:\n$$\iint_D x^2 \, dx \, dy = \iint_{D'} (au)^2 (ab) \, du \, dv = a^2 b \iint_{D'} u^2 \, du \, dv.$$\n\n5. **Evaluate the integral over the unit disk:** The region $D'$ is the unit disk $u^2 + v^2 \leq 1$. Use polar coordinates:\n$$u = r \cos \theta, \quad v = r \sin \theta, \quad r \in [0,1], \theta \in [0, 2\pi].$$\nThe area element is $$du \, dv = r \, dr \, d\theta.$$\nThe integral becomes:\n$$a^2 b \int_0^{2\pi} \int_0^1 (r \cos \theta)^2 r \, dr \, d\theta = a^2 b \int_0^{2\pi} \cos^2 \theta \, d\theta \int_0^1 r^3 \, dr.$$\n\n6. **Compute the integrals:**\n- $$\int_0^{2\pi} \cos^2 \theta \, d\theta = \pi$$ (since average value of $\cos^2 \theta$ over $0$ to $2\pi$ is $1/2$ and $2\pi \times 1/2 = \pi$).\n- $$\int_0^1 r^3 \, dr = \frac{1}{4}.$$\n\n7. **Combine results:**\n$$a^2 b \times \pi \times \frac{1}{4} = \frac{\pi a^2 b}{4}.$$\n\n**Final answer:** $$\boxed{\frac{\pi a^2 b}{4}}.$$