## Step 5: Evaluating the Integral We have: \[ \int_{-3}^{3} (144 - 16y^2) \, dy = \left[ -16 \left( \frac{3^3}{3} \right) - \left( \frac{-3^2}{3} \right) + 144y \right]_{-3}^{3} \] **Explanation:** This equation represents the definite integral of the function \(144 - 16y^2\) with respect to \(y\), over the interval from \(-3\) to \(3\). 1. **Function:** The function inside the integral, \(144 - 16y^2\), is a downward-opening quadratic, which forms a parabola along the \(y\)-axis. 2. **Evaluation:** The right side shows the evaluation of the integral by applying the fundamental theorem of calculus. - The expression \(-16 \left( \frac{3^3}{3} \right)\) simplifies the integration process for the term \(-16y^2\). - Similarly, \(- \left( \frac{-3^2}{3} \right)\) simplifies the integration process for the square term. - The \(144y\) term accounts for the linear portion within the integrand. 3. **Limits of Integration:** Evaluated at the limits \(-3\) and \(3\), this expression will yield the area under the curve between these bounds. This step is crucial in calculating the exact area of the region defined by the given function and interval. **Finding the Area of a Region Enclosed by Given Curves** To determine the area of the region enclosed by the curves, follow these steps: 1. **Identify and Define the Curves:** - The curves provided are: \[ x = 72 - 8y^2 \] \[ x = 8y^2 - 72 \] 2. **Decide on the Integration Method:** - Choose whether to integrate with respect to \(x\) or \(y\). 3. **Graph the Region:** - Sketch the region that the curves enclose. **Step 1: Sketch the Region** A diagram illustrates a coordinate plane with a vertical axis labeled \(y\). Two curves intersect, enclosing a space. There are two distinct arms extending from the point where the curves intersect on the \(y\)-axis, labeled with a value of 3, indicating significant points on the vertical axis.