Let's break down the problem step by step. ### **Given:** The shape of the cooling tower is described by \[ y = 20 \sqrt{1 + \frac{x^2}{72{,}900}} \] for \(-210 \leq x \leq 270\). We are to find the **volume** of the solid formed by rotating this curve about the x-axis. --- ## **Step 1: Set Up the Volume Integral** The volume of a solid of revolution about the x-axis is given by: \[ V = \pi \int_{a}^{b} [y(x)]^2 dx \] Plug in the given function and bounds: \[ V = \pi \int_{-210}^{270} \left[20 \sqrt{1 + \frac{x^2}{72{,}900}}\right]^2 dx \] --- ## **Step 2: Simplify the Integrand** \[ \left(20 \sqrt{1 + \frac{x^2}{72{,}900}}\right)^2 = 400 \left(1 + \frac{x^2}{72{,}900}\right) \] So, \[ V = \pi \int_{-210}^{270} 400 \left(1 + \frac{x^2}{72{,}900}\right) dx \] \[ = 400\pi \int_{-210}^{270} \left(1 + \frac{x^2}{72{,}900}\right) dx \] \[ = 400\pi \left( \int_{-210}^{270} 1\, dx + \int_{-210}^{270} \frac{x^2}{72{,}900}\, dx \right) \] --- ## **Step 3: Calculate Each Integral** ### **First Integral:** \[ \int_{-210}^{270} 1\, dx = (270) - (-210) = 480 \] ### **Second Integral:** \[ \int_{-210}^{270} x^2\, dx = \left[ \frac{x^3}{3} \right]_{-210}^{270} \] Calculate \(270^3\) and \((-210)^3\): - \(270^3 = 19,683,000\) - \((-210)^3 = -9,261,000\) So, \[ \left[ \frac{x^3}{3} \right]_{-210}^{270} = \frac{19,683,000 - (-9,261,000)}{3} = \frac{28,944,000}{3} = 9,648,000 \] Now divide by \(72,900\): \[ \int_{-210}^{270} \frac{x^2}{72,900}\, dx = \frac{9,648,000}{72,900} = 132.32 \] --- ## **Step 4: Add the Results** \[ \text{Total integral} = 480 + 132.32 = 612.32 \] --- ## **Step 5: Multiply by \(400\pi\)** \[ V = 400\pi \times 612.32 = 244,928\pi \] --- ## **Final Answer:** \[ \boxed{244,928\pi} \] This is the volume, in cubic feet, of the tower formed by rotating the region bounded by the graph about the x-axis.