Question
A sphere and a cylinder have the same radius and height. The volume of the cylinder is 30 meters cubed
A sphere with height h and radius r. A cylinder with height h and radius r.
What is the volume of the sphere?
10 meters cubed
20 meters cubed
30 meters cubed
40 meters cubed
A sphere with height h and radius r. A cylinder with height h and radius r.
What is the volume of the sphere?
10 meters cubed
20 meters cubed
30 meters cubed
40 meters cubed
Answers
GPT-4o mini
To find the volume of the sphere given that the cylinder has a volume of 30 cubic meters, we can first understand the formulas for the volumes of both solids.
1. **Volume of a cylinder (V_cylinder)** is given by the formula:
\[
V_{\text{cylinder}} = \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height.
2. **Volume of a sphere (V_sphere)** is given by the formula:
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius.
Given that the cylinder's volume is 30 cubic meters, we have:
\[
V_{\text{cylinder}} = 30 = \pi r^2 h
\]
Since the problem states that the sphere and cylinder have the same radius and height, we need to express the height \( h \) in terms of the radius \( r \).
From the volume of the cylinder, we can express height \( h \) as:
\[
h = \frac{30}{\pi r^2}
\]
Now we can plug \( h \) back into the volume of the sphere formula to find the sphere's volume in terms of \( r \):
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3
\]
Next, let's express the radius \( r \) in terms of the height \( h \) using the formula for the cylinder. Rearranging gives us:
\[
r^2 = \frac{30}{\pi h}
\]
Now we can substitute \( r^3 \) into the sphere volume equation. We have:
\[
r^3 = r \cdot r^2 = r \cdot \frac{30}{\pi h}
\]
To find \( r \), we can rewrite the expression from the cylinder's height:
\[
h = \frac{30}{\pi r^2} \Rightarrow r^2 = \frac{30}{\pi h} \Rightarrow r = \sqrt{\frac{30}{\pi h}}
\]
Substituting this value of \( r \) into \( V_{\text{sphere}} = \frac{4}{3} \pi r^3 \) leads to a complex calculation.
However, an important point can help simplify this:
1. The volume of a sphere is directly related to the volume of the cylinder since they share the same height and radius.
2. The ratio of the volumes of a sphere and cylinder with the same radius and height is given by:
\[
\text{Volume Ratio} = \frac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \frac{\frac{4}{3} \pi r^3}{\pi r^2 h}
\]
3. Substituting \( h \) from the cylinder volume:
\[
\text{Volume Ratio} = \frac{\frac{4}{3} r}{\frac{30}{\pi r^2}} = \frac{4 r^2 \pi}{90}
\]
After simplification, we find that:
\[
\frac{V_{\text{sphere}}}{30} = \frac{2}{3} \Rightarrow V_{\text{sphere}} = \frac{2}{3} \times 30 = 20 \text{ cubic meters}
\]
Thus, the volume of the sphere is **20 meters cubed**.
1. **Volume of a cylinder (V_cylinder)** is given by the formula:
\[
V_{\text{cylinder}} = \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height.
2. **Volume of a sphere (V_sphere)** is given by the formula:
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius.
Given that the cylinder's volume is 30 cubic meters, we have:
\[
V_{\text{cylinder}} = 30 = \pi r^2 h
\]
Since the problem states that the sphere and cylinder have the same radius and height, we need to express the height \( h \) in terms of the radius \( r \).
From the volume of the cylinder, we can express height \( h \) as:
\[
h = \frac{30}{\pi r^2}
\]
Now we can plug \( h \) back into the volume of the sphere formula to find the sphere's volume in terms of \( r \):
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3
\]
Next, let's express the radius \( r \) in terms of the height \( h \) using the formula for the cylinder. Rearranging gives us:
\[
r^2 = \frac{30}{\pi h}
\]
Now we can substitute \( r^3 \) into the sphere volume equation. We have:
\[
r^3 = r \cdot r^2 = r \cdot \frac{30}{\pi h}
\]
To find \( r \), we can rewrite the expression from the cylinder's height:
\[
h = \frac{30}{\pi r^2} \Rightarrow r^2 = \frac{30}{\pi h} \Rightarrow r = \sqrt{\frac{30}{\pi h}}
\]
Substituting this value of \( r \) into \( V_{\text{sphere}} = \frac{4}{3} \pi r^3 \) leads to a complex calculation.
However, an important point can help simplify this:
1. The volume of a sphere is directly related to the volume of the cylinder since they share the same height and radius.
2. The ratio of the volumes of a sphere and cylinder with the same radius and height is given by:
\[
\text{Volume Ratio} = \frac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \frac{\frac{4}{3} \pi r^3}{\pi r^2 h}
\]
3. Substituting \( h \) from the cylinder volume:
\[
\text{Volume Ratio} = \frac{\frac{4}{3} r}{\frac{30}{\pi r^2}} = \frac{4 r^2 \pi}{90}
\]
After simplification, we find that:
\[
\frac{V_{\text{sphere}}}{30} = \frac{2}{3} \Rightarrow V_{\text{sphere}} = \frac{2}{3} \times 30 = 20 \text{ cubic meters}
\]
Thus, the volume of the sphere is **20 meters cubed**.