Question
A bowling-ball maker starts with an 8.5-inch-diameter resin sphere and drills 3 cylindrical finger holes in it. Each hole is 1 inch in diameter and 3.5 inches deep. Which is the best estimate of the volume of resin in the finished ball?
Answers
GPT-4o mini
To find the volume of resin in the finished bowling ball, we need to calculate the volume of the original sphere and then subtract the volume of the finger holes drilled into it.
### Step 1: Calculate the volume of the sphere
The formula for the volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r \) is the radius of the sphere. The diameter of the resin sphere is 8.5 inches, thus the radius \( r \) is:
\[
r = \frac{8.5}{2} = 4.25 \text{ inches}
\]
Now, substituting \( r \) into the volume formula:
\[
V = \frac{4}{3} \pi (4.25)^3
\]
Calculating \( (4.25)^3 \):
\[
4.25^3 = 4.25 \times 4.25 \times 4.25 = 76.765625
\]
Now we can calculate the volume of the sphere:
\[
V = \frac{4}{3} \pi (76.765625)
\]
\[
V \approx \frac{4}{3} \times 3.14 \times 76.765625 \approx \frac{4}{3} \times 241.829 \approx 322.39 \text{ cubic inches}
\]
### Step 2: Calculate the volume of one finger hole
The formula for the volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
Where \( r \) is the radius and \( h \) is the height (or depth). Each finger hole has a diameter of 1 inch, so the radius is:
\[
r = \frac{1}{2} = 0.5 \text{ inch}
\]
The depth of each hole is given as 3.5 inches. Now, substituting \( r \) and \( h \) into the cylinder volume formula:
\[
V = \pi (0.5)^2 (3.5) = \pi (0.25)(3.5) = \pi (0.875)
\]
\[
V \approx 3.14 \times 0.875 \approx 2.74375 \text{ cubic inches}
\]
### Step 3: Calculate the total volume of the finger holes
Since there are 3 finger holes:
\[
\text{Total volume of finger holes} = 3 \times 2.74375 \approx 8.23125 \text{ cubic inches}
\]
### Step 4: Calculate the volume of resin remaining in the ball
Now, subtract the total volume of the finger holes from the volume of the sphere:
\[
\text{Volume of resin} = \text{Volume of sphere} - \text{Total volume of finger holes}
\]
\[
\text{Volume of resin} \approx 322.39 - 8.23125 \approx 314.15875 \text{ cubic inches}
\]
### Conclusion
The best estimate of the volume of resin in the finished ball is approximately:
\[
\boxed{314 \text{ cubic inches}}
\]
### Step 1: Calculate the volume of the sphere
The formula for the volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r \) is the radius of the sphere. The diameter of the resin sphere is 8.5 inches, thus the radius \( r \) is:
\[
r = \frac{8.5}{2} = 4.25 \text{ inches}
\]
Now, substituting \( r \) into the volume formula:
\[
V = \frac{4}{3} \pi (4.25)^3
\]
Calculating \( (4.25)^3 \):
\[
4.25^3 = 4.25 \times 4.25 \times 4.25 = 76.765625
\]
Now we can calculate the volume of the sphere:
\[
V = \frac{4}{3} \pi (76.765625)
\]
\[
V \approx \frac{4}{3} \times 3.14 \times 76.765625 \approx \frac{4}{3} \times 241.829 \approx 322.39 \text{ cubic inches}
\]
### Step 2: Calculate the volume of one finger hole
The formula for the volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
Where \( r \) is the radius and \( h \) is the height (or depth). Each finger hole has a diameter of 1 inch, so the radius is:
\[
r = \frac{1}{2} = 0.5 \text{ inch}
\]
The depth of each hole is given as 3.5 inches. Now, substituting \( r \) and \( h \) into the cylinder volume formula:
\[
V = \pi (0.5)^2 (3.5) = \pi (0.25)(3.5) = \pi (0.875)
\]
\[
V \approx 3.14 \times 0.875 \approx 2.74375 \text{ cubic inches}
\]
### Step 3: Calculate the total volume of the finger holes
Since there are 3 finger holes:
\[
\text{Total volume of finger holes} = 3 \times 2.74375 \approx 8.23125 \text{ cubic inches}
\]
### Step 4: Calculate the volume of resin remaining in the ball
Now, subtract the total volume of the finger holes from the volume of the sphere:
\[
\text{Volume of resin} = \text{Volume of sphere} - \text{Total volume of finger holes}
\]
\[
\text{Volume of resin} \approx 322.39 - 8.23125 \approx 314.15875 \text{ cubic inches}
\]
### Conclusion
The best estimate of the volume of resin in the finished ball is approximately:
\[
\boxed{314 \text{ cubic inches}}
\]