Question
A solid cylindrical curtain rod is 48 inches in length with a diameter of 112 inches. On either end of the rod are two spherical ornaments with a diameter of 4 inches that screw onto the rod and attach at a single point. Find the surface area of the whole object to the nearest whole inch.
Answers
GPT-4o mini
To find the surface area of the entire object, we need to calculate the surface area of the cylindrical rod and the two spherical ornaments, and then combine these areas.
### Step 1: Surface Area of the Cylinder
The formula for the surface area \( A \) of a cylinder (including the lateral surface area and the area of the two circular bases) is given by:
\[
A_{\text{cylinder}} = 2\pi rh + 2\pi r^2
\]
Where:
- \( r \) is the radius of the cylinder
- \( h \) is the height (length) of the cylinder
Here, the diameter of the rod is 112 inches, so the radius \( r \) is:
\[
r = \frac{112}{2} = 56 \text{ inches}
\]
The height \( h \) of the cylinder is 48 inches. Substituting these values into the formula:
1. **Calculate the lateral surface area**:
\[
2\pi rh = 2\pi (56)(48) = 5376\pi
\]
2. **Calculate the area of the two bases**:
\[
2\pi r^2 = 2\pi (56)^2 = 2\pi (3136) = 6272\pi
\]
So, the total surface area of the cylinder is:
\[
A_{\text{cylinder}} = 5376\pi + 6272\pi = 11648\pi
\]
### Step 2: Surface Area of the Spheres
The formula for the surface area \( A \) of a sphere is:
\[
A_{\text{sphere}} = 4\pi r^2
\]
For each ornament, given that the diameter is 4 inches, the radius \( r \) is:
\[
r = \frac{4}{2} = 2 \text{ inches}
\]
Thus, the surface area of one sphere is:
\[
A_{\text{one sphere}} = 4\pi (2^2) = 4\pi (4) = 16\pi
\]
Since there are two spheres, the total surface area for both spheres is:
\[
A_{\text{spheres}} = 2 \times 16\pi = 32\pi
\]
### Step 3: Total Surface Area
Now, we add the surface area of the cylinder and the surface area of the two spheres:
\[
A_{\text{total}} = A_{\text{cylinder}} + A_{\text{spheres}} = 11648\pi + 32\pi = 11680\pi
\]
### Step 4: Calculate the Numerical Value
Now, we calculate \( 11680\pi \) using \( \pi \approx 3.14159 \):
\[
A_{\text{total}} \approx 11680 \times 3.14159 \approx 36651.3
\]
### Step 5: Rounding to the Nearest Whole Inch
Finally, rounding this to the nearest whole inch gives us:
\[
\boxed{36651}
\]
Therefore, the total surface area of the object is approximately 36651 square inches.
### Step 1: Surface Area of the Cylinder
The formula for the surface area \( A \) of a cylinder (including the lateral surface area and the area of the two circular bases) is given by:
\[
A_{\text{cylinder}} = 2\pi rh + 2\pi r^2
\]
Where:
- \( r \) is the radius of the cylinder
- \( h \) is the height (length) of the cylinder
Here, the diameter of the rod is 112 inches, so the radius \( r \) is:
\[
r = \frac{112}{2} = 56 \text{ inches}
\]
The height \( h \) of the cylinder is 48 inches. Substituting these values into the formula:
1. **Calculate the lateral surface area**:
\[
2\pi rh = 2\pi (56)(48) = 5376\pi
\]
2. **Calculate the area of the two bases**:
\[
2\pi r^2 = 2\pi (56)^2 = 2\pi (3136) = 6272\pi
\]
So, the total surface area of the cylinder is:
\[
A_{\text{cylinder}} = 5376\pi + 6272\pi = 11648\pi
\]
### Step 2: Surface Area of the Spheres
The formula for the surface area \( A \) of a sphere is:
\[
A_{\text{sphere}} = 4\pi r^2
\]
For each ornament, given that the diameter is 4 inches, the radius \( r \) is:
\[
r = \frac{4}{2} = 2 \text{ inches}
\]
Thus, the surface area of one sphere is:
\[
A_{\text{one sphere}} = 4\pi (2^2) = 4\pi (4) = 16\pi
\]
Since there are two spheres, the total surface area for both spheres is:
\[
A_{\text{spheres}} = 2 \times 16\pi = 32\pi
\]
### Step 3: Total Surface Area
Now, we add the surface area of the cylinder and the surface area of the two spheres:
\[
A_{\text{total}} = A_{\text{cylinder}} + A_{\text{spheres}} = 11648\pi + 32\pi = 11680\pi
\]
### Step 4: Calculate the Numerical Value
Now, we calculate \( 11680\pi \) using \( \pi \approx 3.14159 \):
\[
A_{\text{total}} \approx 11680 \times 3.14159 \approx 36651.3
\]
### Step 5: Rounding to the Nearest Whole Inch
Finally, rounding this to the nearest whole inch gives us:
\[
\boxed{36651}
\]
Therefore, the total surface area of the object is approximately 36651 square inches.