Let's solve the problem step by step: --- ### 1. **Understanding the Problem** - The base of the solid is the region between \( y = -3x^2 + 12 \) and the x-axis, from \( x = 0 \) to \( x = 2 \). - Cross sections perpendicular to the x-axis are **squares**. --- ### 2. **Graph the Region** Let's plot \( y = -3x^2 + 12 \) from \( x = 0 \) to \( x = 2 \): - At \( x = 0 \), \( y = 12 \) - At \( x = 2 \), \( y = -3(2^2) + 12 = -12 + 12 = 0 \) So, the region is above the x-axis, under the curve from \( x = 0 \) to \( x = 2 \). #### Sketch (please draw this on your paper): ``` y ^ | * | * | * | * | * | * |-----------------------> x 0 1 2 ``` - The curve starts at (0,12) and touches (2,0) on the x-axis. - Shade the area under the curve from \( x = 0 \) to \( x = 2 \). --- ### 3. **Find the Volume** The **side length** of each square at position \( x \) is the value of the function: \[ s = -3x^2 + 12 \] The **area** of the square cross section: \[ A(x) = s^2 = (-3x^2 + 12)^2 \] Volume of the solid: \[ V = \int_{0}^{2} A(x) \, dx = \int_{0}^{2} (-3x^2 + 12)^2 dx \] --- ### 4. **Expand the Integrand** Expand \( (-3x^2 + 12)^2 \): \[ (-3x^2 + 12)^2 = ( -3x^2 )^2 + 2(-3x^2)(12) + (12)^2 = 9x^4 - 72x^2 + 144 \] So, \[ V = \int_{0}^{2} [9x^4 - 72x^2 + 144] dx \] --- ### 5. **Integrate** Integrate term by term: \[ \int 9x^4 dx = 9 \cdot \frac{x^5}{5} \] \[ \int -72x^2 dx = -72 \cdot \frac{x^3}{3} = -24x^3 \] \[ \int 144 dx = 144x \] Put together: \[ V = \left[ \frac{9}{5}x^5 - 24x^3 + 144x \right]_{0}^{2} \] --- ### 6. **Evaluate at the Bounds** Plug in \( x = 2 \): \[ \frac{9}{5}(2)^5 - 24(2)^3 + 144(2) = \frac{9}{5} \cdot 32 - 24 \cdot 8 + 288 = \frac{288}{5} - 192 + 288 \] Now at \( x = 0 \), all terms are zero. So, \[ V = \left( \frac{288}{5} - 192 + 288 \right) - 0 \] --- ### 7. **Simplify** \[ V = \frac{288}{5} + 288 - 192 \] \[ = \frac{288}{5} + (288 - 192) = \frac{288}{5} + 96 \] Convert 96 to fifths: \[ 96 = \frac{480}{5} \] \[ V = \frac{288 + 480}{5} = \frac{768}{5} \] --- ### 8. **Final Boxed Answer** \[ \boxed{V = \frac{768}{5}} \] Or as a decimal: \( V = 153.6 \) cubic units. --- ### 9. **Summary Table** | Step | Work Shown | |------|------------| | Formula | \( V = \int_{0}^{2} (-3x^2 + 12)^2 dx \) | | Expand | \( = \int_{0}^{2} (9x^4 - 72x^2 + 144) dx \) | | Integrate | \( = \left[ \frac{9}{5}x^5 - 24x^3 + 144x \right]_{0}^{2} \) | | Evaluate | \( = \frac{288}{5} - 192 + 288 \) | | Simplify | \( = \frac{288}{5} + 96 = \frac{768}{5} \) | --- ### 10. **Graph** Here is a sketch of the base region (draw this on your paper): ``` 14 +-----------------------------+ | * y | * | * 12 + | * | * ---- +-----------------------------+ 0 1 2 x ``` - Curve: \( y = -3x^2 + 12 \) - Shaded below curve, above x-axis, from \( x = 0 \) to \( x = 2 \) --- Let me know if you need the drawing in a different style!