1. **State the problem:** Find the surface area of the solid formed by rotating the curve $y = x^2 + 3$ for $1 < x < 5$ about the y-axis. 2. **Formula for surface area of revolution about y-axis:** $$ S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ 3. **Calculate the derivative:** $$ \frac{dy}{dx} = 2x $$ 4. **Substitute into the formula:** $$ S = \int_1^5 2\pi x \sqrt{1 + (2x)^2} \, dx = \int_1^5 2\pi x \sqrt{1 + 4x^2} \, dx $$ 5. **Simplify the integral:** Let $$ u = 1 + 4x^2 \Rightarrow du = 8x \, dx \Rightarrow x \, dx = \frac{du}{8} $$ Rewrite the integral: $$ S = 2\pi \int_1^5 x \sqrt{u} \, dx = 2\pi \int_{u(1)}^{u(5)} \sqrt{u} \frac{du}{8} = \frac{\pi}{4} \int_5^{101} u^{1/2} \, du $$ 6. **Integrate:** $$ \int u^{1/2} \, du = \frac{2}{3} u^{3/2} $$ 7. **Evaluate definite integral:** $$ S = \frac{\pi}{4} \times \frac{2}{3} \left[ u^{3/2} \right]_5^{101} = \frac{\pi}{6} \left(101^{3/2} - 5^{3/2} \right) $$ Calculate values: $$ 101^{3/2} = 101 \times \sqrt{101} \approx 101 \times 10.05 = 1015.05 $$ $$ 5^{3/2} = 5 \times \sqrt{5} \approx 5 \times 2.236 = 11.18 $$ 8. **Final surface area:** $$ S \approx \frac{\pi}{6} (1015.05 - 11.18) = \frac{\pi}{6} \times 1003.87 \approx 3.1416 \times 167.31 = 525.56 $$ **Answer:** The surface area is approximately **525.56** square units.