Questions: Compute the volume of the given frustum. Present your answer as a decimal, rounded to three decimal places. (Picture is not to scale.) V=490.016 × m^3

Solution Steps

Let $h$ be the height of the smaller cone removed from the top of the larger cone. The radii of the bases of the frustum are $r_1=3$ m and $r_2=9$ m, and the height of the frustum is $h'=4$ m.

The ratio of corresponding lengths of similar triangles is constant. Thus, we have $$ \frac{h}{h+4} = \frac{3}{9} = \frac{1}{3} $$ So, $3h = h+4$, which gives $2h=4$ and $h=2$ m. The height of the full cone is $H=h+h'=2+4=6$ m.

The volume of the larger cone is $$V_{large} = \frac{1}{3} \pi r_2^2 H = \frac{1}{3} \pi (9^2)(6) = 162\pi$$

The volume of the smaller cone is $$V_{small} = \frac{1}{3} \pi r_1^2 h = \frac{1}{3} \pi (3^2)(2) = 6\pi$$

The volume of the frustum is the difference between the volumes of the larger and smaller cones. $$V = V_{large} - V_{small} = 162\pi - 6\pi = 156\pi \approx 490.088 \text{ m}^3$$ Rounding to three decimal places, we get $490.088$ m$^3$.

Final Answer: The final answer is $\boxed{490.088}$