1. **State the problem:** Calculate the surface area of the solid formed by revolving the curve $f(x) = \frac{1}{5}x^3$ around the x-axis from $x=3$ to $x=5$. 2. **Formula for surface area of revolution about the x-axis:** $$ S = 2\pi \int_a^b f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx $$ 3. **Find the derivative:** $$ f(x) = \frac{1}{5}x^3 \implies f'(x) = \frac{3}{5}x^2 $$ 4. **Substitute into the formula:** $$ S = 2\pi \int_3^5 \frac{1}{5}x^3 \sqrt{1 + \left(\frac{3}{5}x^2\right)^2} \, dx = 2\pi \int_3^5 \frac{1}{5}x^3 \sqrt{1 + \frac{9}{25}x^4} \, dx $$ 5. **Simplify inside the square root:** $$ \sqrt{1 + \frac{9}{25}x^4} = \sqrt{\frac{25}{25} + \frac{9}{25}x^4} = \frac{\sqrt{25 + 9x^4}}{5} $$ 6. **Rewrite the integral:** $$ S = 2\pi \int_3^5 \frac{1}{5}x^3 \cdot \frac{\sqrt{25 + 9x^4}}{5} \, dx = 2\pi \int_3^5 \frac{x^3 \sqrt{25 + 9x^4}}{25} \, dx $$ 7. **Factor constants out:** $$ S = \frac{2\pi}{25} \int_3^5 x^3 \sqrt{25 + 9x^4} \, dx $$ 8. **Use substitution:** Let $$ u = 25 + 9x^4 \implies du = 36x^3 dx $$ 9. **Rewrite the integral in terms of $u$:** $$ x^3 dx = \frac{du}{36} $$ 10. **Change limits:** When $x=3$, $$ u = 25 + 9 \times 3^4 = 25 + 9 \times 81 = 25 + 729 = 754 $$ When $x=5$, $$ u = 25 + 9 \times 5^4 = 25 + 9 \times 625 = 25 + 5625 = 5650 $$ 11. **Integral becomes:** $$ S = \frac{2\pi}{25} \int_{754}^{5650} \sqrt{u} \cdot \frac{du}{36} = \frac{2\pi}{25 \times 36} \int_{754}^{5650} u^{1/2} du $$ 12. **Simplify constants:** $$ \frac{2\pi}{900} = \frac{\pi}{450} $$ 13. **Integrate:** $$ \int u^{1/2} du = \frac{2}{3} u^{3/2} $$ 14. **Evaluate definite integral:** $$ S = \frac{\pi}{450} \times \frac{2}{3} \left[u^{3/2}\right]_{754}^{5650} = \frac{2\pi}{1350} \left(5650^{3/2} - 754^{3/2}\right) $$ 15. **Calculate powers:** $$ 5650^{3/2} = 5650 \times \sqrt{5650} \approx 5650 \times 75.166 = 424,683.9 $$ $$ 754^{3/2} = 754 \times \sqrt{754} \approx 754 \times 27.466 = 20,711.6 $$ 16. **Subtract:** $$ 424,683.9 - 20,711.6 = 403,972.3 $$ 17. **Multiply by constant:** $$ S \approx \frac{2\pi}{1350} \times 403,972.3 = \frac{2 \times 3.1416}{1350} \times 403,972.3 \approx 0.004654 \times 403,972.3 = 1,879.5 $$ 18. **Final answer rounded to 2 decimals:** $$ \boxed{1879.50} $$ This is the surface area of the solid of revolution.