Sphere submerged in the cone

A right circular cone with a top width of 24 cm and an altitude of 8 cm is filled with water. A spherical steel ball with a radius of 3.0cm is submerged in the cone. Find the volume of water below the sphere.

Correct answer:

V =  -22.3505 cm³

Step-by-step explanation:

$$\begin{gathered} D=24\text{cm} \\ h=8\text{cm} \\ r=3.0\text{cm} \\ R=D/2=24/2=12\text{cm} \\ \alpha =\arctan (R/h)=\arctan (12/8)\doteq 0.9828\text{rad} \\ \tan \alpha =r/s \\ s=r/\tan \alpha =3/\tan 0.9828=2\text{cm} \\ \cos \alpha =h_1:s \\ \sin \alpha =r_1:s \\ {h}_1=s\cdot \cos \alpha =2\cdot \cos 0.9828\doteq 1.1094\text{cm} \\ {r}_1=s\cdot \sin \alpha =2\cdot \sin 0.9828\doteq 1.6641\text{cm} \\ \sin \alpha =r:h_3 \\ {h}_3=r\cdot \sin \alpha =3\cdot \sin 0.9828\doteq 2.4962\text{cm} \\ {h}_2=h_1-(h_3-r)=1.1094-(2.4962-3)\doteq 1.6132\text{cm} \\ {V}_1=\frac{1}{3}\cdot \pi \cdot r_1^2\cdot h_1=\frac{1}{3}\cdot 3.1416\cdot 1.6641^2\cdot 1.1094\doteq 3.2172{\text{cm}}^3 \\ {V}_2=\pi \cdot (r_1^2+2\cdot r_1\cdot h_2)=3.1416\cdot (1.6641^2+2\cdot 1.6641\cdot 1.6132)\doteq 25.5677{\text{cm}}^3 \\ V=V_1-V_2=3.2172-25.5677=-22.3505{\text{cm}}^3 \end{gathered}$$

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