Let's solve both **the preliminary example** and **Example 1** step by step: --- ## **Preliminary Example Solution**: We are given: - The trunk is **300 ft** tall. - Every **60 ft**, a diameter was measured: **8 ft, 8 ft, 9 ft, 10 ft, 12 ft, 14 ft** (from top to bottom). We can model the trunk as 5 frustums (truncated cones), each 60 ft tall. ### **Step 1: List the Radii** Convert diameters to **radii**: | Section | Diameter (ft) | Radius (ft) | |---------|---------------|-------------| | Top | 8 | 4 | | | 8 | 4 | | | 9 | 4.5 | | | 10 | 5 | | | 12 | 6 | | Bottom | 14 | 7 | ### **Step 2: Volume of a Frustum** The volume of a frustum of a cone is: \[ V = \frac{1}{3}\pi h (R_1^2 + R_1 R_2 + R_2^2) \] where \( h \) is the height, \( R_1 \) and \( R_2 \) are the radii of the ends. ### **Step 3: Calculate Each Frustum's Volume** Each frustum is **60 ft** tall. #### **a) Between 4 ft and 4 ft:** \[ V_1 = \frac{1}{3}\pi (60) (4^2 + 4 \times 4 + 4^2) = \frac{1}{3}\pi (60) (16 + 16 + 16) = \frac{1}{3}\pi (60) (48) = 20\pi (48) = 960\pi \text{ ft}^3 \] #### **b) Between 4 ft and 4.5 ft:** \[ V_2 = \frac{1}{3}\pi (60) (4^2 + 4 \times 4.5 + 4.5^2) = \frac{1}{3}\pi (60) (16 + 18 + 20.25) = \frac{1}{3}\pi (60) (54.25) = 20\pi (54.25) = 1085\pi \text{ ft}^3 \] #### **c) Between 4.5 ft and 5 ft:** \[ V_3 = \frac{1}{3}\pi (60) (4.5^2 + 4.5 \times 5 + 5^2) = \frac{1}{3}\pi (60) (20.25 + 22.5 + 25) = \frac{1}{3}\pi (60) (67.75) = 20\pi (67.75) = 1355\pi \text{ ft}^3 \] #### **d) Between 5 ft and 6 ft:** \[ V_4 = \frac{1}{3}\pi (60) (5^2 + 5 \times 6 + 6^2) = \frac{1}{3}\pi (60) (25 + 30 + 36) = \frac{1}{3}\pi (60) (91) = 20\pi (91) = 1820\pi \text{ ft}^3 \] #### **e) Between 6 ft and 7 ft:** \[ V_5 = \frac{1}{3}\pi (60) (6^2 + 6 \times 7 + 7^2) = \frac{1}{3}\pi (60) (36 + 42 + 49) = \frac{1}{3}\pi (60) (127) = 20\pi (127) = 2540\pi \text{ ft}^3 \] ### **Step 4: Add Up All Volumes** \[ V_\text{total} = (960 + 1085 + 1355 + 1820 + 2540)\pi = 7760\pi \text{ ft}^3 \] ### **Step 5: Plug in \(\pi \approx 3.1416\)** \[ V_\text{total} \approx 7760 \times 3.1416 = 24,372 \text{ ft}^3 \] --- ## **Final Answer (Preliminary Example):** \[ \boxed{24,372\ \text{ft}^3} \] (Rounded to the nearest ft³.) --- ## **Example 1 Solution** **Find the volume of the solid generated by rotating \(y = \sin x\) about the x-axis, \(x \in [0, \pi]\).** ### **Step 1: Formula for Volume of Revolution** \[ V = \pi \int_{a}^{b} [f(x)]^2 dx \] Here, \(f(x) = \sin x\), \(a = 0\), \(b = \pi\): \[ V = \pi \int_{0}^{\pi} [\sin x]^2 dx \] ### **Step 2: Simplify the Integral** Recall: \[ \sin^2 x = \frac{1 - \cos(2x)}{2} \] So: \[ V = \pi \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} dx = \frac{\pi}{2} \int_{0}^{\pi} (1 - \cos(2x)) dx \] ### **Step 3: Integrate** \[ \int (1 - \cos(2x)) dx = x - \frac{1}{2}\sin(2x) \] So: \[ V = \frac{\pi}{2} \left[ x - \frac{1}{2}\sin(2x) \right]_{0}^{\pi} \] ### **Step 4: Evaluate the Limits** - At \(x = \pi\): \(x = \pi\), \(\sin(2\pi) = 0\) - At \(x = 0\): \(x = 0\), \(\sin(0) = 0\) \[ V = \frac{\pi}{2} \left[ (\pi - 0) - \frac{1}{2}(0 - 0) \right] = \frac{\pi}{2} (\pi) = \frac{\pi^2}{2} \] --- ## **Final Answer (Example 1):** \[ \boxed{\frac{\pi^2}{2}} \] Or, as a decimal (using \(\pi \approx 3.1416\)): \[ \frac{(3.1416)^2}{2} \approx \frac{9.8696}{2} \approx 4.9348 \] --- ### **Summary of Final Answers:** - **Tree Trunk Volume Estimate:** \(\boxed{24,372\ \text{ft}^3}\) - **Solid of Revolution Volume:** \(\boxed{\frac{\pi^2}{2}}\) (or \(\approx 4.9348\))