What is the volume of a cone (in cubic units) with radius $3$ units and slant height $3\sqrt{2}$ units?

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**Solution:**

1. Given:
   - Radius $r = 3$ units
   - Slant height $l = 3\sqrt{2}$ units

2. We need to find the height $h$ of the cone using the Pythagorean theorem:
   $$h = \sqrt{l^2 - r^2}$$

3. Calculate $h$:
   $$h = \sqrt{(3\sqrt{2})^2 - 3^2} = \sqrt{9 	imes 2 - 9} = \sqrt{18 - 9} = \sqrt{9} = 3$$

4. Volume $V$ of a cone is given by:
   $$V = \frac{1}{3} \pi r^2 h$$

5. Substitute the values:
   $$V = \frac{1}{3} \pi 	imes 3^2 	imes 3 = \frac{1}{3} \pi 	imes 9 	imes 3 = \frac{27\pi}{3} = 9\pi$$

**Answer:** $9\pi$

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