Let's break down your **Calculus Final Project** into clear steps and provide a sample solution you can follow. We'll use a **parabola and a line** to create a bounded region, and then find its area and volume when rotated. --- ## **Part 1: Bounded Region and Area** ### **A. Create a Bounded Region** Let's pick: - **Nonlinear curve:** \( y = x^2 \) - **Line:** \( y = 4 \) The region is bounded above by the line \( y = 4 \) and below by the parabola \( y = x^2 \). ### **B. Graph the Region** 1. **Find intersection points:** \[ x^2 = 4 \implies x = -2, 2 \] 2. **Shade the region between \( x = -2 \) and \( x = 2 \), from \( y = x^2 \) to \( y = 4 \).** - Draw both curves on the coordinate plane. - Mark the intersection points at (-2, 4) and (2, 4). - Shade the region between the parabola and the line. **Diagram:** ``` | | ---- y = 4 (line) | / \ | / \ |---/------------------\---- x-axis | -2 2 | |____|____|____|____|____| ``` ### **C. Area Calculation** **Set up the definite integral:** \[ A = \int_{-2}^{2} [4 - x^2] \, dx \] **Compute the area:** 1. Integrate: \[ A = \int_{-2}^{2} 4\, dx - \int_{-2}^{2} x^2\, dx \] \[ = 4[x]_{-2}^{2} - \left[\frac{x^3}{3}\right]_{-2}^{2} \] 2. Evaluate: - \( 4[x]_{-2}^{2} = 4(2 - (-2)) = 4 \times 4 = 16 \) - \( \left[\frac{x^3}{3}\right]_{-2}^{2} = \left(\frac{8}{3} - \left(-\frac{8}{3}\right)\right) = \frac{16}{3} \) 3. So, \[ A = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \] \[ A \approx 10.667 \ \text{units}^2 \] --- ## **Part 2: Volume by Rotation** ### **A. Rotation Application** Let’s rotate the region around the **x-axis**. #### **i. Volume Calculation (Washer Method)** \[ V = \pi \int_{-2}^{2} \left[(4)^2 - (x^2)^2\right] dx \] \[ = \pi \int_{-2}^{2} (16 - x^4) dx \] Integrate: \[ V = \pi \left[16x - \frac{x^5}{5}\right]_{-2}^{2} \] Evaluate: - For \( 16x \): \( 16(2 - (-2)) = 16 \times 4 = 64 \) - For \( \frac{x^5}{5} \): \( \frac{32}{5} - \left(-\frac{32}{5}\right) = \frac{64}{5} \) So, \[ V = \pi \left(64 - \frac{64}{5}\right) = \pi \left(\frac{320}{5} - \frac{64}{5}\right) = \pi \left(\frac{256}{5}\right) \] \[ V \approx \pi \times 51.2 \approx 160.221 \ \text{units}^3 \] #### **ii. Record Your Work** Label the diagram, show the area and volume calculations, and write the final answers with three decimal places. --- ### **B. Create a Physical Volume** - You could use clay, foam, or cardboard to model the solid formed above. - The shape will look like a “cap” or “bulge” since you are rotating the region between the parabola and the line. --- ## **Summary Table** | Step | Description | Value (units) | |------|-------------|---------------| | Area | \(\int_{-2}^{2} (4 - x^2) dx\) | \(\frac{32}{3} \approx 10.667\) | | Volume | \(\pi \int_{-2}^{2} [16 - x^4] dx\) | \( \approx 160.221 \) | --- ### **Tips for Your Poster** - Use color to highlight the bounded region and label all axes. - Show all steps clearly and label the intersection points. - For the physical model, try to match the dimensions used in your graph. --- If you want to use different curves or rotate around a different axis, let me know! I can adjust the explanation and calculations for you.

Certainly! Here's a **detailed step-by-step solution** for your project, including all calculations, explanations, and final answers. --- # **Part 1: Bounded Region and Area** ## **A. Choose the Curves** - **Nonlinear curve:** \( y = x^2 \) - **Line:** \( y = 4 \) **Region:** The bounded area between \( y = x^2 \) (parabola) and \( y = 4 \) (horizontal line), from the intersections at \( x = -2 \) to \( x = 2 \). --- ## **B. Graph the Region** - Find the intersection points: \[ x^2 = 4 \implies x = \pm 2 \] - The region is between \( x = -2 \) and \( x = 2 \), bounded above by \( y=4 \) and below by \( y=x^2 \). *Visual description:* Draw the parabola opening upward with vertex at (0,0), and the horizontal line at \( y=4 \). Shade the area between these curves from \( x=-2 \) to \( x=2 \). --- ## **C. Set up the Definite Integral for Area** The **area \(A\)** between the curves is: \[ A = \int_{x=-2}^{2} [\text{top} - \text{bottom}] \, dx = \int_{-2}^{2} [4 - x^2] \, dx \] --- ## **D. Calculate the Area** ### Step 1: Write the integral: \[ A = \int_{-2}^{2} (4 - x^2) \, dx \] ### Step 2: Integrate term-by-term: \[ A = \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} x^2 \, dx \] - Integral of 4: \[ \int 4 \, dx = 4x \] - Integral of \( x^2 \): \[ \int x^2 \, dx = \frac{x^3}{3} \] ### Step 3: Evaluate the definite integrals: \[ A = \left[ 4x \right]_{-2}^{2} - \left[ \frac{x^3}{3} \right]_{-2}^{2} \] Calculate each: - \( 4x \) from \(-2\) to \(2\): \[ 4(2) - 4(-2) = 8 - (-8) = 8 + 8 = 16 \] - \( \frac{x^3}{3} \) from \(-2\) to \(2\): \[ \frac{(2)^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \] ### **Final Area:** \[ A = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \] Expressed as a decimal: \[ A \approx 10.6667 \text{ square units} \] --- # **Part 2: Volume of the Solid of Revolution** ## **A. Choose the Rotation Axis and Method** **Rotation around the x-axis** is a common choice here. **Method:** Washer method (since the region is between two curves, revolving around the x-axis). --- ## **B. Set up the Volume Integral** The volume \(V\) is: \[ V = \pi \int_{-2}^{2} \left[ (\text{outer radius})^2 - (\text{inner radius})^2 \right] dx \] - Outer radius: distance from x-axis to the top curve \( y=4 \): **radius = 4** - Inner radius: distance from x-axis to the bottom curve \( y=x^2 \): **radius = x^2** Thus: \[ V = \pi \int_{-2}^{2} \left[ (4)^2 - (x^2)^2 \right] dx \] Simplify: \[ V = \pi \int_{-2}^{2} \left[ 16 - x^4 \right] dx \] --- ## **C. Calculate the Volume** ### Step 1: Write the integral: \[ V = \pi \int_{-2}^{2} (16 - x^4) \, dx \] ### Step 2: Integrate term-by-term: \[ V = \pi \left[ \int_{-2}^{2} 16 \, dx - \int_{-2}^{2} x^4 \, dx \right] \] - Integral of 16: \[ \int 16 \, dx = 16x \] - Integral of \( x^4 \): \[ \int x^4 \, dx = \frac{x^5}{5} \] ### Step 3: Evaluate integrals: \[ V = \pi \left[ 16x \big|_{-2}^{2} - \frac{x^5}{5} \big|_{-2}^{2} \right] \] - Evaluate \( 16x \): \[ 16(2) - 16(-2) = 32 - (-32) = 32 + 32 = 64 \] - Evaluate \( \frac{x^5}{5} \): \[ \frac{(2)^5}{5} - \frac{(-2)^5}{5} = \frac{32}{5} - \frac{-32}{5} = \frac{32}{5} + \frac{32}{5} = \frac{64}{5} \] ### **Final volume:** \[ V = \pi \left( 64 - \frac{64}{5} \right) \] Express as a single fraction: \[ V = \pi \left( \frac{320}{5} - \frac{64}{5} \right) = \pi \times \frac{256}{5} \] **Numerical approximation:** \[ V \approx 3.1416 \times 51.2 \approx 160.220 \text{ cubic units} \] --- # **Final Answers Summary** | **Part** | **Calculation** | **Result** (rounded to 3 decimal places) | |------------|------------------|------------------------------------------| | **Area of region** | \(\int_{-2}^{2} (4 - x^2) dx\) | \(\boxed{10.667}\) units\(^2\) | | **Volume of solid** | \(V = \pi \int_{-2}^{2} (16 - x^4) dx\) | \(\boxed{160.220}\) units\(^3\) | --- ## **Additional Suggestions** - **Label all parts clearly** on your poster. - **Show all steps** of the integration. - **Include diagrams** with labeled curves and shaded regions. - **Create a physical model** that visually demonstrates the volume—such as stacking slices or using modeling clay. --- If you'd like to explore rotation around a different line or with different curves, I can help adjust the calculations accordingly!