To solve this problem, we’ll start by recalling the formulas for the volumes of a cylinder, sphere, and cone:
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The volume \(V_c\) of a cylinder is given by: \[ V_c = \pi r^2 h \]
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The volume \(V_s\) of a sphere is given by: \[ V_s = \frac{4}{3} \pi r^3 \]
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The volume \(V_{cone}\) of a cone is given by: \[ V_{cone} = \frac{1}{3} \pi r^2 h \]
Now, according to the problem:
- The volume of the cylinder is \(36 , \text{cm}^3\).
- The sphere has the same radius \(r\) as the cylinder.
- The height of the cylinder \(h\) is equal to the diameter of the sphere, which can be expressed as \(2r\).
Since we know the volume of the cylinder: \[ V_c = \pi r^2 h = 36 \]
Substituting \(h\) with \(2r\): \[ V_c = \pi r^2 (2r) = 2\pi r^3 = 36 \]
To find \(r^3\): \[ 2\pi r^3 = 36 \] \[ r^3 = \frac{36}{2\pi} = \frac{18}{\pi} \]
Now we will use this value of \(r^3\) to find the volume of the sphere: \[ V_s = \frac{4}{3} \pi r^3 \] \[ V_s = \frac{4}{3} \pi \left(\frac{18}{\pi}\right) \]
Now, simplify the expression for \(V_s\): \[ V_s = \frac{4 \cdot 18}{3} = \frac{72}{3} = 24 , \text{cm}^3 \]
Therefore, the volume of the sphere is: \[ \boxed{24 , \text{cm}^3} \]