1. The problem asks to integrate the function $x = 4 + \sqrt{y+4}$ with respect to $y$ from $y=3$ to $y=6$. 2. Set up the definite integral: $$\int_3^6 \left(4 + \sqrt{y+4}\right) dy$$ 3. Split the integral into two parts: $$\int_3^6 4 \, dy + \int_3^6 \sqrt{y+4} \, dy$$ 4. First integral: $$\int_3^6 4 \, dy = 4(y) \Big|_3^6 = 4(6) - 4(3) = 24 - 12 = 12$$ 5. Second integral: Let $u = y+4$, then $du = dy$. Bounds change from $y=3$ to $u=7$ and from $y=6$ to $u=10$. We have: $$\int_7^{10} \sqrt{u} \, du = \int_7^{10} u^{1/2} \, du$$ 6. Integrate: $$\int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C$$ Evaluate from 7 to 10: $$\frac{2}{3} (10)^{3/2} - \frac{2}{3} (7)^{3/2}$$ 7. Calculate the values: $10^{3/2} = 10^{1} \times 10^{1/2} = 10 \times \sqrt{10} \approx 10 \times 3.1623 = 31.623$ $7^{3/2} = 7^{1} \times 7^{1/2} = 7 \times \sqrt{7} \approx 7 \times 2.6458 = 18.5206$ So, $$\frac{2}{3} (31.623 - 18.5206) = \frac{2}{3} (13.1024) = 8.7349$$ 8. Add both results: $$12 + 8.7349 = 20.7349$$ 9. Final answer: $$\boxed{20.7349}$$