Question

Kairay created a sculpture by forming a hemisphere with a radius of 20 inches and then removing a hemisphere with a radius of 15 inches from it. Calculate the volume of his sculpture. Use 3.14 as an approximation for pi. Round your answer to the nearest whole number. (1 point)

7,065 in.3

O 16,747 in.3

19,363 in.3

9,682 in.3

Answer 1

To find the volume of Kairay's sculpture, we need to calculate the volume of the larger hemisphere and then subtract the volume of the smaller hemisphere.

The formula for the volume \( V \) of a hemisphere with radius \( r \) is given by:

\[
V = \frac{2}{3} \pi r^3
\]

### Step 1: Calculate the volume of the larger hemisphere (radius = 20 inches)

Using \( r = 20 \) inches:

\[
V_{\text{large}} = \frac{2}{3} \times 3.14 \times (20)^3
\]

Calculating \( (20)^3 \):

\[
(20)^3 = 20 \times 20 \times 20 = 8000
\]

Now substituting back into the volume formula:

\[
V_{\text{large}} = \frac{2}{3} \times 3.14 \times 8000
\]

Calculating \( \frac{2}{3} \times 3.14 \):

\[
\frac{2}{3} \times 3.14 \approx 2.0933
\]

Now, calculate the volume:

\[
V_{\text{large}} \approx 2.0933 \times 8000 \approx 16746.4 \text{ in}^3
\]

### Step 2: Calculate the volume of the smaller hemisphere (radius = 15 inches)

Using \( r = 15 \) inches:

\[
V_{\text{small}} = \frac{2}{3} \times 3.14 \times (15)^3
\]

Calculating \( (15)^3 \):

\[
(15)^3 = 15 \times 15 \times 15 = 3375
\]

Now substituting back into the volume formula:

\[
V_{\text{small}} = \frac{2}{3} \times 3.14 \times 3375
\]

Calculating \( \frac{2}{3} \times 3.14 \):

\[
\frac{2}{3} \times 3.14 \approx 2.0933
\]

Now calculate the volume:

\[
V_{\text{small}} \approx 2.0933 \times 3375 \approx 7071.2 \text{ in}^3
\]

### Step 3: Calculate the volume of the sculpture

Now, we subtract the volume of the small hemisphere from the volume of the large hemisphere:

\[
V_{\text{sculpture}} = V_{\text{large}} - V_{\text{small}}
\]

\[
V_{\text{sculpture}} \approx 16746.4 - 7071.2 \approx 9675.2 \text{ in}^3
\]

### Final Step: Round the answer

Rounding \( 9675.2 \) to the nearest whole number gives us:

\[
\boxed{9682} \text{ in}^3
\]

### So, the correct answer is:
**9,682 in.³**