1. **State the problem:** Find the surface area of the solid formed by rotating the curve $y=2x^2$ for $0 \leq x \leq 1$ about the y-axis. 2. **Formula for surface area when rotating about the y-axis:** $$ S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ 3. **Calculate the derivative:** $$ \frac{dy}{dx} = \frac{d}{dx}(2x^2) = 4x $$ 4. **Substitute into the formula:** $$ S = \int_0^1 2\pi x \sqrt{1 + (4x)^2} \, dx = \int_0^1 2\pi x \sqrt{1 + 16x^2} \, dx $$ 5. **Use substitution:** Let $$ u = 1 + 16x^2 \implies du = 32x \, dx \implies x \, dx = \frac{du}{32} $$ 6. **Rewrite the integral:** $$ S = 2\pi \int_0^1 x \sqrt{1 + 16x^2} \, dx = 2\pi \int_{u=1}^{u=17} \sqrt{u} \frac{du}{32} = \frac{\pi}{16} \int_1^{17} u^{1/2} \, du $$ 7. **Integrate:** $$ \int u^{1/2} \, du = \frac{2}{3} u^{3/2} $$ 8. **Evaluate definite integral:** $$ S = \frac{\pi}{16} \times \frac{2}{3} \left[ u^{3/2} \right]_1^{17} = \frac{\pi}{24} (17^{3/2} - 1) $$ 9. **Calculate $17^{3/2}$:** $$ 17^{3/2} = (\sqrt{17})^3 = (4.1231)^3 \approx 70.992 $$ 10. **Final surface area:** $$ S \approx \frac{\pi}{24} (70.992 - 1) = \frac{\pi}{24} \times 69.992 \approx 9.16 $$ **Answer:** The surface area is approximately **9.16** square units.