To solve the problem, we will start by recalling the formulas for the volumes of a cylinder, a sphere, and a cone:
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Volume of a cylinder: \[ V_{cylinder} = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height.
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Volume of a sphere: \[ V_{sphere} = \frac{4}{3} \pi r^3 \]
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Volume of a cone: \[ V_{cone} = \frac{1}{3} \pi r^2 h \]
From the problem statement, we know that the volume of the cylinder is \( 36 , \text{cm}^3 \). We will set up the equations using the given information:
Let the radius of the cylinder (and sphere) be \( r \) and the height of the cylinder be \( h \). According to the problem, the height \( h \) of the cylinder is equal to the sphere's diameter. Since the diameter of a sphere is \( 2r \), we have: \[ h = 2r \]
Now we can substitute \( h \) in the volume formula of the cylinder: \[ V_{cylinder} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3 \]
Set this equal to the given volume of the cylinder: \[ 2\pi r^3 = 36 \]
Now, we can solve for \( r^3 \): \[ r^3 = \frac{36}{2\pi} = \frac{18}{\pi} \]
Next, we will substitute \( r^3 \) into the volume formula for the sphere: \[ V_{sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{18}{\pi}\right) \]
Simplifying the sphere's volume: \[ V_{sphere} = \frac{4}{3} \cdot 18 = \frac{72}{3} = 24 , \text{cm}^3 \]
Thus, the volume of the sphere is: \[ \boxed{24 , \text{cm}^3} \]