1. **State the problem:** We need to write the integral to find the area of the region bounded by the curves $u(y) = 2\sqrt{y} + 2$ and $v(y) = \frac{2(y + 2)}{3}$, integrating with respect to $y$. 2. **Identify the formula:** When integrating with respect to $y$, the area between two curves $x = u(y)$ and $x = v(y)$ from $y=a$ to $y=b$ is given by: $$\text{Area} = \int_a^b |u(y) - v(y)| \, dy$$ 3. **Find the intersection points (limits of integration):** Solve $u(y) = v(y)$: $$2\sqrt{y} + 2 = \frac{2(y + 2)}{3}$$ Multiply both sides by 3 to clear denominator: $$3(2\sqrt{y} + 2) = 2(y + 2)$$ $$6\sqrt{y} + 6 = 2y + 4$$ Rearranged: $$6\sqrt{y} + 6 - 2y - 4 = 0$$ $$6\sqrt{y} - 2y + 2 = 0$$ Let $t = \sqrt{y}$, so $y = t^2$: $$6t - 2t^2 + 2 = 0$$ Divide entire equation by 2: $$3t - t^2 + 1 = 0$$ Rewrite: $$-t^2 + 3t + 1 = 0$$ Multiply both sides by $-1$: $$t^2 - 3t - 1 = 0$$ Use quadratic formula: $$t = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$$ Since $t = \sqrt{y} \geq 0$, take positive roots: $$t_1 = \frac{3 - \sqrt{13}}{2} \approx 0.697, \quad t_2 = \frac{3 + \sqrt{13}}{2} \approx 3.303$$ Corresponding $y$ values: $$y_1 = t_1^2 \approx 0.485, \quad y_2 = t_2^2 \approx 10.91$$ 4. **Set up the integral:** The area is $$\int_{y_1}^{y_2} \bigl| u(y) - v(y) \bigr| \, dy = \int_{0.485}^{8} \bigl( u(y) - v(y) \bigr) \, dy$$ Note: The problem states the horizontal grid line at $y=8$, so we use $8$ as upper limit instead of $10.91$. 5. **Write the integral explicitly:** $$\int_{0.485}^{8} \left( 2\sqrt{y} + 2 - \frac{2(y + 2)}{3} \right) dy$$ This integral represents the area bounded by the curves when integrating with respect to $y$. **Final answer:** $$\boxed{\int_{0.485}^{8} \left( 2\sqrt{y} + 2 - \frac{2(y + 2)}{3} \right) dy}$$