1. **Problem statement:** Analyze the function $$f(a,b) = (2a + 2b)^2 + (2a - 2b)^2$$ over variables $a$ and $b$. 2. **Simplify the function:** $$f(a,b) = (2a + 2b)^2 + (2a - 2b)^2 = 4(a+b)^2 + 4(a-b)^2.$$ Expanding both: $$4(a^2 + 2ab + b^2) + 4(a^2 - 2ab + b^2) = 4a^2 + 8ab + 4b^2 + 4a^2 - 8ab + 4b^2 = 8a^2 + 8b^2.$$ So, $$f(a,b) = 8(a^2 + b^2).$$ 3. **Roots:** Set $f(a,b) = 0$. Since $8(a^2 + b^2) = 0$, we get $$a = 0, \quad b = 0.$$ 4. **Given relationship for roots:** $$b = -i a, \quad b = i a,$$ which are complex values consistent with roots if $a = 0$. The integer root is at $(a,b) = (0,0)$. 5. **Derivative with respect to $a$:** $$\frac{\partial}{\partial a} f(a,b) = \frac{\partial}{\partial a} 8(a^2 + b^2) = 16 a.$$ 6. **Indefinite integral with respect to $a$:** $$\int f(a,b) \ da = \int 8(a^2 + b^2) \ da = 8\left( \frac{a^3}{3} + a b^2 \right) + C.$$ 7. **Definite integral over the disk of radius $R$:** Convert to polar coordinates $(r, \theta)$ where $a = r \cos \theta$, $b = r \sin \theta$: $$\iint_{a^2 + b^2 < R^2} f(a,b) \ da db = \iint_{r = 0}^R \int_{\theta=0}^{2\pi} 8 r^2 r \, d\theta dr = 8 \int_0^{2\pi} d\theta \int_0^{R} r^3 dr.$$ Calculate integrals: $$8(2\pi) \left. \frac{r^4}{4} \right|_0^{R} = 16 \pi \frac{R^4}{4} = 4 \pi R^4.$$ **Final answer:** - Simplified function: $$f(a,b) = 8(a^2 + b^2).$$ - Roots: $$a=0, b=0.$$ - Derivative: $$\frac{\partial f}{\partial a} = 16a.$$ - Indefinite integral: $$8 \left( \frac{a^3}{3} + a b^2 \right) + C.$$ - Definite integral over disk of radius $R$: $$4 \pi R^4.$$