1. **State the problem:** A cone with base radius $r=6$ cm and height $H=36$ cm is cut parallel to its base, forming a smaller cone on top and a frustum below. The frustum volume is $572$ cm$^3$. We need to find the volume of a sphere with radius equal to the smaller cone's radius after cutting, using $\pi=\frac{22}{7}$. 2. **Find the volume of the original cone:** $$V=\frac{1}{3}\pi r^2 H=\frac{1}{3} \times \frac{22}{7} \times 6^2 \times 36=\frac{1}{3} \times \frac{22}{7} \times 36 \times 36=\frac{1}{3} \times \frac{22}{7} \times 1296=\frac{22 \times 1296}{21}=\frac{28512}{21}=1358.666...$$ 3. **Let the smaller cone's height be $h$ and radius $r_s$.** Since the cut is parallel, the smaller cone is similar to the original cone, so $$\frac{r_s}{6}=\frac{h}{36} \implies r_s=\frac{h}{36} \times 6=\frac{h}{6}$$ 4. **Volume of smaller cone:** $$V_s=\frac{1}{3} \pi r_s^2 h=\frac{1}{3} \times \frac{22}{7} \times \left(\frac{h}{6}\right)^2 \times h=\frac{1}{3} \times \frac{22}{7} \times \frac{h^2}{36} \times h=\frac{22}{7} \times \frac{h^3}{108}$$ 5. **Volume of frustum:** $$V_f=V - V_s=572$$ Substitute $V=1358.666...$ and $V_s$: $$1358.666... - \frac{22}{7} \times \frac{h^3}{108} = 572$$ 6. **Solve for $h^3$:** $$\frac{22}{7} \times \frac{h^3}{108} = 1358.666... - 572 = 786.666...$$ $$\frac{22}{7} \times \frac{h^3}{108} = 786.666...$$ Multiply both sides by $\frac{7 \times 108}{22}$: $$h^3 = 786.666... \times \frac{7 \times 108}{22} = 786.666... \times 34.3636... = 27030$$ 7. **Calculate $h$:** $$h = \sqrt[3]{27030} \approx 30$$ 8. **Find smaller cone radius $r_s$:** $$r_s = \frac{h}{6} = \frac{30}{6} = 5$$ 9. **Find volume of sphere with radius $r_s=5$:** $$V_{sphere} = \frac{4}{3} \pi r_s^3 = \frac{4}{3} \times \frac{22}{7} \times 5^3 = \frac{4}{3} \times \frac{22}{7} \times 125 = \frac{4}{3} \times \frac{2750}{7} = \frac{11000}{21} \approx 523.81$$ 10. **Final answer:** The volume of the sphere is approximately $524$ cm$^3$. **Answer: 524 cm^3**