1. **State the problem:** A spherical balloon's radius decreases at a constant rate of 15 cm/min. We need to find the rate at which the volume of air is removed when the radius is 9 cm. 2. **Formula and rules:** The volume $V$ of a sphere with radius $r$ is given by: $$V = \frac{4}{3} \pi r^3$$ We want to find the rate of change of volume $\frac{dV}{dt}$ when $r=9$ cm, given $\frac{dr}{dt} = -15$ cm/min (negative because radius is decreasing). 3. **Differentiate the volume formula with respect to time $t$:** $$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt}$$ 4. **Substitute the known values:** $$r = 9, \quad \frac{dr}{dt} = -15$$ 5. **Calculate the rate of volume change:** $$\frac{dV}{dt} = 4 \pi (9)^2 (-15) = 4 \pi \times 81 \times (-15) = -4860 \pi$$ 6. **Interpretation:** The volume is decreasing at a rate of $4860 \pi$ cubic centimeters per minute when the radius is 9 cm. **Final answer:** $$\frac{dV}{dt} = -4860 \pi \text{ cm}^3/\text{min}$$