1. **State the problem:** Calculate the double integral $$\int_0^1 \int_0^y x y \cdot 8 x y \, dy \, dx$$. 2. **Simplify the integrand:** The integrand is $$x y \cdot 8 x y = 8 x^2 y^2$$. 3. **Rewrite the integral:** $$\int_0^1 \int_0^y 8 x^2 y^2 \, dy \, dx$$. 4. **Integrate with respect to $y$ first:** Treat $x$ as constant. $$\int_0^y 8 x^2 y^2 \, dy = 8 x^2 \int_0^y y^2 \, dy = 8 x^2 \left[ \frac{y^3}{3} \right]_0^y = \frac{8 x^2 y^3}{3}$$. 5. **Substitute back into the outer integral:** $$\int_0^1 \frac{8 x^2 y^3}{3} \, dx$$. 6. **Note the limits of the outer integral are from 0 to 1 for $x$, but the inner integral's upper limit is $y$, which depends on $x$.** 7. **Re-examine the limits:** The original integral is $$\int_0^1 \int_0^y 8 x^2 y^2 \, dy \, dx$$, but the inner integral is with respect to $y$ from 0 to $y$, which is confusing because $y$ is the variable of integration. 8. **Assuming the integral is $$\int_0^1 \int_0^x 8 x^2 y^2 \, dy \, dx$$ (switching the inner limit to $x$), proceed:** 9. **Integrate inner integral:** $$\int_0^x 8 x^2 y^2 \, dy = 8 x^2 \left[ \frac{y^3}{3} \right]_0^x = \frac{8 x^2 x^3}{3} = \frac{8 x^5}{3}$$. 10. **Integrate outer integral:** $$\int_0^1 \frac{8 x^5}{3} \, dx = \frac{8}{3} \int_0^1 x^5 \, dx = \frac{8}{3} \left[ \frac{x^6}{6} \right]_0^1 = \frac{8}{3} \cdot \frac{1}{6} = \frac{8}{18} = \frac{4}{9}$$. **Final answer:** $$\frac{4}{9}$$.