# Problem 4: Using Pappus-Guldinus Theorems **Statement (translated):** Using the Pappus-Guldinus theorems, find the surface area and the volume of the piece. --- Let's break down the solution step by step. ## 1. **Understanding the Solid** The figure is a ring (torus segment) formed by rotating a rectangular section around an axis. ### Dimensions: - The center of rotation is at a distance \( r = 50 \) mm. - The rectangular section has a width of 25 mm and a height of 25 mm. - The inner “hole” is 40 mm wide. ## 2. **Pappus-Guldinus Theorems** 1. **Surface Area:** The surface area generated by revolving a curve (of length \( L \)) about an external axis is \[ A = L \cdot (2\pi R) \] where \( R \) is the distance from the centroid of the curve to the axis of rotation. 2. **Volume:** The volume generated by revolving a plane area (\( A \)) about an external axis is \[ V = A \cdot (2\pi R) \] where \( R \) is the distance from the centroid of the area to the axis of rotation. --- ## 3. **Calculating the Area and Volume** ### a) **Area of Cross Section** The cross-section is a rectangle: - Height = 25 mm - Width = 25 mm So, the area: \[ A_{section} = 25\, \text{mm} \times 25\, \text{mm} = 625\, \text{mm}^2 \] ### b) **Centroid Distance** The centroid of the rectangle is at its center. The center of the rectangle is 25 mm away from the inner edge, and the inner edge is at \( r = 50 \) mm from the axis. So, the distance from the axis to the centroid is: \[ R = r + 25\, \text{mm}/2 = 50\, \text{mm} + 12.5\, \text{mm} = 62.5\, \text{mm} \] ### c) **Volume** Using Pappus’ theorem: \[ V = A_{section} \cdot (2\pi R) = 625\, \text{mm}^2 \cdot 2\pi \cdot 62.5\, \text{mm} \] \[ V = 625 \times 2 \times \pi \times 62.5 \] \[ V = 625 \times 125 \times \pi \] \[ V = 78,125 \times \pi \] \[ V \approx 245,436.93\, \text{mm}^3 \] --- ### d) **Perimeter of the Curve (for Surface Area)** The curve to be rotated is the **outer and inner boundaries**. The outer boundary is a rectangle’s perimeter minus the back, i.e., the sum of the two verticals and the top: \[ L = 2 \times 25\, \text{mm} + 40\, \text{mm} = 90\, \text{mm} \] But for the **outer surface area**, we need the perimeter of the rectangle being revolved (excluding the base attached to the axis): \[ L = 2 \times (\text{height}) + (\text{width}) = 2 \times 25\, \text{mm} + 40\, \text{mm} = 90\, \text{mm} \] ### e) **Surface Area** Using Pappus’ theorem: \[ A = L \cdot (2\pi R) = 90\, \text{mm} \cdot 2\pi \cdot 62.5\, \text{mm} \] \[ A = 90 \times 125 \times \pi \] \[ A = 11,250 \times \pi \] \[ A \approx 35,342.92\, \text{mm}^2 \] --- ## **Final Answers** ### **Surface Area** \[ \boxed{A \approx 35,343\ \text{mm}^2} \] ### **Volume** \[ \boxed{V \approx 245,437\ \text{mm}^3} \] --- **Note:** Please double-check whether you need to include more surfaces (inner/outer) or just the lateral surface, depending on your course's conventions. The above considers the main rotation surface. If you need more detail, let me know!