Question
A cylindrical rod has a diameter of 2 cm and a length of 40 cm. Two hemispherical ends are added to the rod to make a solid shape. What is the total surface area of the resulting shape?
Answers
GPT-4o mini
To find the total surface area of the solid shape formed by the cylindrical rod with two hemispherical ends, we can break down the problem into two parts: the surface area of the cylinder and the surface area of the two hemispheres.
1. **Surface area of the cylindrical rod**:
The formula for the lateral surface area of a cylinder is given by:
\[
A_{\text{cylinder}} = 2\pi r h
\]
where \( r \) is the radius of the cylinder and \( h \) is the height (or length) of the cylinder.
The diameter of the cylindrical rod is 2 cm, thus the radius \( r \) is:
\[
r = \frac{2 \, \text{cm}}{2} = 1 \, \text{cm}
\]
The height \( h \) of the cylinder is 40 cm. Plugging in the values:
\[
A_{\text{cylinder}} = 2\pi (1 \, \text{cm})(40 \, \text{cm}) = 80\pi \, \text{cm}^2
\]
2. **Surface area of the hemispherical ends**:
The total surface area of one hemisphere is given by the formula:
\[
A_{\text{hemisphere}} = 2\pi r^2
\]
Since there are two hemispheres:
\[
A_{\text{two hemispheres}} = 2 \times A_{\text{hemisphere}} = 2 \times 2\pi r^2 = 4\pi r^2
\]
Plugging in the radius \( r = 1 \, \text{cm} \):
\[
A_{\text{two hemispheres}} = 4\pi (1 \, \text{cm})^2 = 4\pi \, \text{cm}^2
\]
3. **Total surface area of the solid shape**:
Now, we can combine the surface areas of the cylinder and the hemispheres. The total surface area \( A_{\text{total}} \) is given by:
\[
A_{\text{total}} = A_{\text{cylinder}} + A_{\text{two hemispheres}}
\]
Thus:
\[
A_{\text{total}} = 80\pi \, \text{cm}^2 + 4\pi \, \text{cm}^2 = 84\pi \, \text{cm}^2
\]
4. **Final result**:
To provide a numerical approximation, we can calculate \( 84\pi \):
\[
A_{\text{total}} \approx 84 \times 3.14159 \approx 263.76 \, \text{cm}^2
\]
Therefore, the total surface area of the resulting shape is \( 84\pi \, \text{cm}^2 \), or approximately \( 263.76 \, \text{cm}^2 \).
1. **Surface area of the cylindrical rod**:
The formula for the lateral surface area of a cylinder is given by:
\[
A_{\text{cylinder}} = 2\pi r h
\]
where \( r \) is the radius of the cylinder and \( h \) is the height (or length) of the cylinder.
The diameter of the cylindrical rod is 2 cm, thus the radius \( r \) is:
\[
r = \frac{2 \, \text{cm}}{2} = 1 \, \text{cm}
\]
The height \( h \) of the cylinder is 40 cm. Plugging in the values:
\[
A_{\text{cylinder}} = 2\pi (1 \, \text{cm})(40 \, \text{cm}) = 80\pi \, \text{cm}^2
\]
2. **Surface area of the hemispherical ends**:
The total surface area of one hemisphere is given by the formula:
\[
A_{\text{hemisphere}} = 2\pi r^2
\]
Since there are two hemispheres:
\[
A_{\text{two hemispheres}} = 2 \times A_{\text{hemisphere}} = 2 \times 2\pi r^2 = 4\pi r^2
\]
Plugging in the radius \( r = 1 \, \text{cm} \):
\[
A_{\text{two hemispheres}} = 4\pi (1 \, \text{cm})^2 = 4\pi \, \text{cm}^2
\]
3. **Total surface area of the solid shape**:
Now, we can combine the surface areas of the cylinder and the hemispheres. The total surface area \( A_{\text{total}} \) is given by:
\[
A_{\text{total}} = A_{\text{cylinder}} + A_{\text{two hemispheres}}
\]
Thus:
\[
A_{\text{total}} = 80\pi \, \text{cm}^2 + 4\pi \, \text{cm}^2 = 84\pi \, \text{cm}^2
\]
4. **Final result**:
To provide a numerical approximation, we can calculate \( 84\pi \):
\[
A_{\text{total}} \approx 84 \times 3.14159 \approx 263.76 \, \text{cm}^2
\]
Therefore, the total surface area of the resulting shape is \( 84\pi \, \text{cm}^2 \), or approximately \( 263.76 \, \text{cm}^2 \).