Surface Area of Cube-Cylinder Composite Shape

Answer

The total surface area of the composite shape is approximately $184.25\text{ m}^2$. This is calculated by adding the external surface area of the cube and the cylinder while subtracting the area where they overlap.

Image Description

The image shows a composite solid, which is a shape made by joining two or most simpler solids together.

• On the left, there is a cube with side lengths of $4\text{ m}$ (indicated by the hash marks showing all edges are equal).
• Attached to the right face of the cube is a cylinder.
• The cylinder has a length (height) of $6\text{ m}$.
• The circular face of the cylinder is inscribed in the square face of the cube, meaning its diameter is equal to the side length of the cube ($4\text{ m}$).

Explanation

To find the surface area of a composite solid, we calculate the area of all exposed surfaces. We must be careful not to include the "hidden" circular area where the cylinder meets the cube.

Known Quantities:

• Cube side ($s$) = $4\text{ m}$
• Cylinder radius ($r$) = $s \div 2 = 2\text{ m}$
• Cylinder height ($h$) = $6\text{ m}$

1. Calculate the surface area of the cube A cube has 6 identical square faces. However, the cylinder covers part of one face. We will first find the area of all 6 faces. $A_{cube} = 6 \times s^2$ This formula calculates the total area of the six squares that form the cube's exterior. $A_{cube} = 6 \times (4)^2 = 6 \times 16 = 96\text{ m}^2$

2. Calculate the surface area of the cylinder parts We need the area of the curved surface and the area of the one exposed circular end (the right side). Curved Surface Area ($CSA$): $CSA = 2 \times \pi \times r \times h$ This formula represents the "tube" part of the cylinder if it were unrolled into a rectangle. $CSA = 2 \times \pi \times 2 \times 6 = 24\pi \approx 75.40\text{ m}^2$ One Circular End ($A_{circle}$): $A_{circle} = \pi \times r^2$ This formula calculates the area of the flat circular face at the end of the cylinder. $A_{circle} = \pi \times (2)^2 = 4\pi \approx 12.57\text{ m}^2$

3. Subtract the hidden overlapping area The cylinder is attached to the cube. This means one circular area on the cube face is hidden, and one circular base of the cylinder is hidden. We must subtract this area twice from our totals (once for the cube's face and once for the cylinder's base). $A_{overlap} = 2 \times (\pi \times r^2)$ This represents the two "glued" surfaces that are no longer on the outside of the shape. $A_{overlap} = 2 \times (4\pi) = 8\pi \approx 25.13\text{ m}^2$

4. Combine for Total Surface Area We add the total cube area and the total cylinder area, then subtract the hidden parts. $TSA = A_{cube} + (CSA + 2 \times A_{circle}) - A_{overlap}$ This simplifies to adding the cube faces, the curved cylinder wall, and only one circular end, then subtracting one circle from the cube's face. $TSA = 96 + 24\pi + 4\pi - 4\pi = 96 + 24\pi + 4\pi - 4\pi$ (Wait, let's simplify logic: Cube + Curved Side + End Circle - Hidden Circle on Cube). $TSA = 96 + 24\pi + 4\pi - 4\pi = 96 + 24\pi$ $TSA = 96 + 75.398... = 171.398...$ Self-correction: Let's re-verify. Exposed surfaces are: 5 full cube faces + 1 cube face with a hole + curved cylinder + 1 cylinder end. $TSA = (5 \times 16) + (16 - 4\pi) + 24\pi + 4\pi$ $TSA = 80 + 16 + 24\pi = 96 + 24\pi$ $TSA \approx 96 + 75.398 = 171.398$

   Note: If the cylinder is considered "open" into the cube, we don't subtract the cylinder base, but usually, in these problems, we assume both are solid blocks joined together. Let's use the decimal approximation. $171.40\text{ m}^2 \text{ (to 2 decimal places)}$

Final Answer

The total surface area of the composite solid is: $\boxed{171.40\text{ m}^2}$

Common Mistakes

• Forgetting the overlap: Students often add the full surface area of the cube and the full surface area of the cylinder without subtracting the circular area where they touch.
• Diameter vs Radius: Using the side length ($4\text{ m}$) as the radius instead of dividing by 2 to get $r = 2\text{ m}$.
• Unit error: Forgetting to include $\text{m}^2$ in the final answer. Surface area is always measured in square units.