# Solution to Problem 4 **Using Pappus-Guldinus Theorems to Find the Surface Area and Volume** ## Given Data - Outer radius of revolution: \( r = 50 \) mm - Dimensions on the section: - Height = 25 mm - Width = 25 mm (for each section) - Inner gap = 40 mm ## Step 1: Understanding the Shape The figure is a solid of revolution generated by rotating a closed area around an axis (not passing through the area), specifically a rectangular section with an internal "cut," creating a ring segment. ## Step 2: Area and Centroid of the Generating Section ### **Area of the Generating Shape** Let’s calculate the area \( A_{\text{gen}} \) of the shaded region: - Large rectangle: \( 75\, \text{mm} \) (height) \(\times\) \( 25\, \text{mm} \) (width) - Subtract the inner rectangle: \( 25\, \text{mm} \) (height) \(\times\) \( 40\, \text{mm} \) (width) So: \[ A_{\text{gen}} = (75 \times 25) - (25 \times 40) = 1875 - 100 = 875\, \text{mm}^2 \] ### **Centroid of the Area** Since the section is symmetric, the centroid is at a distance \( R = 50\, \text{mm} \) from the axis of revolution. ## Step 3: **Pappus-Guldinus Theorems** ### **1. Surface Area** The surface area generated by rotating the curve is: \[ A = (\text{perimeter of the generating shape}) \times (\text{distance travelled by the centroid}) \] But for the solid, we use the area: \[ A_{\text{surface}} = (\text{perimeter of generating section}) \times (2\pi R) \] #### But usually, for the volume: ### **2. Volume** \[ V = (\text{area of generating section}) \times (2\pi R) \] \[ V = 875\, \text{mm}^2 \times 2\pi \times 50\, \text{mm} \] \[ V = 875 \times 2 \times 3.1416 \times 50 \] \[ V = 875 \times 314.16 \] \[ V = 274,530\, \text{mm}^3 \] ### **3. Surface Area** \[ A = (\text{perimeter of generating section}) \times (2\pi R) \] The perimeter \( P \) of the section: - Outer rectangle: \(2 \times (75 + 25) = 200\, \text{mm}\) - Inner rectangle: \(2 \times (25 + 40) = 130\, \text{mm}\) - Since it's a "hole," add the outer perimeter and inner perimeter. Total perimeter = \( 200 + 130 = 330\, \text{mm} \) Now: \[ A_{\text{surface}} = 330\, \text{mm} \times 2\pi \times 50\, \text{mm} \] \[ A_{\text{surface}} = 330 \times 314.16 \] \[ A_{\text{surface}} = 103,672.8\, \text{mm}^2 \] --- ## **Final Answers** \[ \boxed{ \begin{aligned} \text{Surface Area (A)} &= 103,673\, \text{mm}^2 \\ \text{Volume (V)} &= 274,530\, \text{mm}^3 \\ \end{aligned} } \] --- ### **Summary Table** | Quantity | Value | |------------------|-------------| | Surface Area (A) | 103,673 mm² | | Volume (V) | 274,530 mm³ |