1. **State the problem:** Find the surface area of the solid generated by revolving the curve $y=\sqrt{7+3x^2}$ from $x=0$ to $x=1$ about the y-axis. 2. **Formula for surface area about y-axis:** $$ S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ 3. **Find $\frac{dy}{dx}$:** Given $y=\sqrt{7+3x^2} = (7+3x^2)^{1/2}$, $$ \frac{dy}{dx} = \frac{1}{2}(7+3x^2)^{-1/2} \cdot 6x = \frac{3x}{\sqrt{7+3x^2}} $$ 4. **Compute $1 + \left(\frac{dy}{dx}\right)^2$:** $$ 1 + \left(\frac{3x}{\sqrt{7+3x^2}}\right)^2 = 1 + \frac{9x^2}{7+3x^2} = \frac{7+3x^2 + 9x^2}{7+3x^2} = \frac{7 + 12x^2}{7 + 3x^2} $$ 5. **Substitute into the integral:** $$ S = \int_0^1 2\pi x \sqrt{\frac{7 + 12x^2}{7 + 3x^2}} \, dx = 2\pi \int_0^1 x \frac{\sqrt{7 + 12x^2}}{\sqrt{7 + 3x^2}} \, dx $$ 6. **Simplify and evaluate the integral:** Let $u = 7 + 3x^2$, then $du = 6x \, dx$, so $x \, dx = \frac{du}{6}$. Rewrite the integral: $$ S = 2\pi \int_{u=7}^{10} \frac{\sqrt{7 + 12x^2}}{\sqrt{u}} \cdot \frac{du}{6} $$ Note $7 + 12x^2 = 7 + 12 \cdot \frac{u-7}{3} = 7 + 4(u-7) = 4u - 21$. So, $$ S = \frac{\pi}{3} \int_7^{10} \frac{\sqrt{4u - 21}}{\sqrt{u}} \, du = \frac{\pi}{3} \int_7^{10} \sqrt{\frac{4u - 21}{u}} \, du = \frac{\pi}{3} \int_7^{10} \sqrt{4 - \frac{21}{u}} \, du $$ 7. **Rewrite the integral:** $$ S = \frac{\pi}{3} \int_7^{10} \sqrt{4 - \frac{21}{u}} \, du = \frac{\pi}{3} \int_7^{10} \sqrt{\frac{4u - 21}{u}} \, du $$ 8. **Split the square root:** $$ \sqrt{\frac{4u - 21}{u}} = \frac{\sqrt{4u - 21}}{\sqrt{u}} $$ 9. **Integral becomes:** $$ S = \frac{\pi}{3} \int_7^{10} \frac{\sqrt{4u - 21}}{\sqrt{u}} \, du $$ 10. **Use substitution $t = \sqrt{u}$, so $u = t^2$, $du = 2t dt$:** Limits: when $u=7$, $t=\sqrt{7}$; when $u=10$, $t=\sqrt{10}$. Integral: $$ S = \frac{\pi}{3} \int_{\sqrt{7}}^{\sqrt{10}} \frac{\sqrt{4t^2 - 21}}{t} \cdot 2t \, dt = \frac{2\pi}{3} \int_{\sqrt{7}}^{\sqrt{10}} \sqrt{4t^2 - 21} \, dt $$ 11. **Evaluate $\int \sqrt{4t^2 - 21} \, dt$:** Use formula: $$ \int \sqrt{a^2 t^2 - b^2} \, dt = \frac{t}{2} \sqrt{a^2 t^2 - b^2} - \frac{b^2}{2a} \ln \left| a t + \sqrt{a^2 t^2 - b^2} \right| + C $$ Here, $a=2$, $b=\sqrt{21}$. 12. **Apply limits:** $$ \int_{\sqrt{7}}^{\sqrt{10}} \sqrt{4t^2 - 21} \, dt = \left[ \frac{t}{2} \sqrt{4t^2 - 21} - \frac{21}{4} \ln \left| 2t + \sqrt{4t^2 - 21} \right| \right]_{\sqrt{7}}^{\sqrt{10}} $$ 13. **Calculate at $t=\sqrt{10}$:** $$ \frac{\sqrt{10}}{2} \sqrt{40 - 21} - \frac{21}{4} \ln \left( 2\sqrt{10} + \sqrt{40 - 21} \right) = \frac{\sqrt{10}}{2} \sqrt{19} - \frac{21}{4} \ln \left( 2\sqrt{10} + \sqrt{19} \right) $$ 14. **Calculate at $t=\sqrt{7}$:** $$ \frac{\sqrt{7}}{2} \sqrt{28 - 21} - \frac{21}{4} \ln \left( 2\sqrt{7} + \sqrt{28 - 21} \right) = \frac{\sqrt{7}}{2} \sqrt{7} - \frac{21}{4} \ln \left( 2\sqrt{7} + \sqrt{7} \right) = \frac{7}{2} - \frac{21}{4} \ln \left( \sqrt{7}(2 + 1) \right) $$ 15. **Simplify the difference:** $$ \int_{\sqrt{7}}^{\sqrt{10}} \sqrt{4t^2 - 21} \, dt = \frac{\sqrt{10}}{2} \sqrt{19} - \frac{21}{4} \ln \left( 2\sqrt{10} + \sqrt{19} \right) - \left( \frac{7}{2} - \frac{21}{4} \ln \left( 3\sqrt{7} \right) \right) $$ 16. **Final surface area:** $$ S = \frac{2\pi}{3} \left[ \frac{\sqrt{10}}{2} \sqrt{19} - \frac{7}{2} - \frac{21}{4} \ln \left( \frac{2\sqrt{10} + \sqrt{19}}{3\sqrt{7}} \right) \right] $$ This is the exact surface area of the solid of revolution about the y-axis.