A cylindrical vase has a diameter of 6 inches. At the bottom of the vase, there are 9 marbles, each of diameter 3 inches. The vase is filled with water up to a height of 12 inches.

Which of the following could be used to calculate the volume of water in the vase?

 π(3in)2(12in) − 1.5(four over threeπ(9in)3)
 π(12in)2(3in) − 1.5(four over threeπ(9in)3)
 π(3in)2(12in) − 9(four over threeπ(1.5in)3)
 π(12in)2(3in) − 9(four over threeπ(1.5in)3)

We are tasked with calculating the volume of water in a cylindrical vase that contains marbles. To do this, we need to find the volume of the water-filled space in the vase, excluding the volume of the marbles.

### Step 1: Volume of the vase (without marbles)
The vase is cylindrical, so the volume \( V_{\text{vase}} \) can be calculated using the formula for the volume of a cylinder:

\[
V_{\text{vase}} = \pi r^2 h
\]

- The **diameter** of the vase is 6 inches, so the **radius** \( r \) is half of that: \( r = \frac{6}{2} = 3 \) inches.
- The **height** of the water is given as \( h = 12 \) inches.

Thus, the volume of the vase filled with water is:

\[
V_{\text{vase}} = \pi (3 \, \text{in})^2 (12 \, \text{in}) = \pi (9 \, \text{in}^2)(12 \, \text{in}) = 108\pi \, \text{in}^3
\]

### Step 2: Volume of the marbles
Each marble is a sphere, and its volume \( V_{\text{marble}} \) can be calculated using the formula for the volume of a sphere:

\[
V_{\text{marble}} = \frac{4}{3} \pi r^3
\]

- The **diameter** of each marble is 3 inches, so the **radius** \( r \) of each marble is \( \frac{3}{2} = 1.5 \) inches.

Thus, the volume of one marble is:

\[
V_{\text{marble}} = \frac{4}{3} \pi (1.5 \, \text{in})^3 = \frac{4}{3} \pi (3.375 \, \text{in}^3) = 4.5 \pi \, \text{in}^3
\]

Since there are 9 marbles, the total volume of all the marbles is:

\[
V_{\text{marbles}} = 9 \times 4.5 \pi \, \text{in}^3 = 40.5\pi \, \text{in}^3
\]

### Step 3: Volume of the water
The volume of the water in the vase is the volume of the vase minus the volume of the marbles:

\[
V_{\text{water}} = V_{\text{vase}} - V_{\text{marbles}} = 108\pi \, \text{in}^3 - 40.5\pi \, \text{in}^3 = 67.5\pi \, \text{in}^3
\]

### Step 4: Identify the correct expression
Among the given options, the correct one is:

\[
\pi (3 \, \text{in})^2 (12 \, \text{in}) - 9 \left( \frac{4}{3} \pi (1.5 \, \text{in})^3 \right)
\]

This corresponds to the option:

\[
\boxed{\pi (3 \, \text{in})^2 (12 \, \text{in}) - 9 \left( \frac{4}{3} \pi (1.5 \, \text{in})^3 \right)}
\]

This correctly represents the volume of the vase minus the volume of the 9 marbles.

---

Do you want more details, or have any questions?

Here are 5 related questions to consider:
1. How does the volume of a sphere change if the radius is doubled?
2. What is the formula for the volume of a cone, and how does it compare to that of a cylinder?
3. How would the volume calculation change if the marbles were stacked to occupy the full height of the vase?
4. What happens to the water volume if the number of marbles is reduced to 5?
5. How would you calculate the surface area of the cylindrical vase?

**Tip:** Remember, the volume of a sphere grows rapidly as its radius increases, since the radius is cubed in the formula.

A cylindrical vase has a diameter of 6 inches. At the bottom of the vase, there are 9 marbles, each of diameter 3 inches. The vase is filled with water up to a height of 12 inches. Which of the following could be used to calculate the volume of water in the vase? π(3in)^2(12in) − 1.5(four over threeπ(9in)^3) π(12in)^2(3in) − 1.5(four over threeπ(9in)^3) π(3in)^2(12in) − 9(four over threeπ(1.5in)^3) π(12in)^2(3in) − 9(four over threeπ(1.5in)^3)

Learn how to calculate the volume of water in a cylindrical vase with a diameter of 6 inches, filled with 9 marbles of 3 inches in diameter. Understand the use of volume formulas for cylinders and spheres, with a step-by-step guide. Suitable for students in Grades 9-12.

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