Answer
The volume of the solid and the right circular cone with base radius $3$ and height $12$ is the same and equal to $36 \pi$.
Work Step by Step
Area $=\pi r^2=(\dfrac{\pi}{16})x^2$
We integrate the integral to calculate the volume of the solid as follows:
$V_s= \int_{0}^{12} (\dfrac{\pi}{16})x^2 dx =\dfrac{ \pi}{48})x^3]_{0}^{12}=36 \pi$
Now, the volume of the circular cone is as follows: $V_c=\dfrac{1}{3} \pi r^2 h=\dfrac{1}{3} (9\pi) (12)=36 \pi$
Thus, the volume of the solid and the right circular cone with base radius $3$ and height $12$.
