Let's solve each problem step by step: --- ### **3) Shell Method for Volume** **Given:** - Region bounded by \( y = x^2 \), \( x = 1 \), and \( y = 0 \). - Revolved about the line \( y = -8 \). #### **Step 1: Visualize the Region** - The region is above the \( x \)-axis (\( y = 0 \)), below \( y = x^2 \), and to the left of \( x = 1 \). - Lower bound for \( x \) is 0, upper bound is 1. #### **Step 2: Setup Shell Method** - Shells are **vertical**, so integrate with respect to \( x \). - **Radius** of a typical shell: Distance from \( y = x^2 \) to \( y = -8 \): \( \text{Radius} = x^2 - (-8) = x^2 + 8 \) - **Height** of a typical shell: \( x \) runs from 0 to 1, so height is just from \( y = 0 \) to \( y = x^2 \), but since we're using vertical slices, the height is \( x \). - **Circumference**: \( 2\pi \times \text{radius} \). - **Height**: \( x^2 \) (from \( y = x^2 \) to \( y = 0 \)). #### **Step 3: Write the Integral** \[ V = 2\pi \int_{x=0}^{x=1} (\text{radius}) \cdot (\text{height})\,dx \] \[ V = 2\pi \int_{0}^{1} (x^2 + 8) \cdot x \, dx \] \[ V = 2\pi \int_{0}^{1} (x^3 + 8x) \, dx \] #### **Step 4: Integrate** \[ \int_{0}^{1} x^3 \, dx = \left. \frac{x^4}{4} \right|_0^1 = \frac{1}{4} \] \[ \int_{0}^{1} 8x \, dx = 8 \left. \frac{x^2}{2} \right|_0^1 = 8 \cdot \frac{1}{2} = 4 \] So, \[ V = 2\pi \left( \frac{1}{4} + 4 \right) = 2\pi \left( \frac{17}{4} \right) = \frac{17\pi}{2} \] #### **Final Answer (3):** \[ \boxed{V = \frac{17\pi}{2}} \] --- ### **4) Both Washer and Shell Methods** **Given:** - Region in the first quadrant bounded by \( y = x^2 \), \( y = 25 \), and \( x = 0 \). - Revolved about \( x = 5 \). #### **Step 1: Bounds** - \( y = x^2 \implies x = \sqrt{y} \) - \( y \) ranges from \( 0 \) to \( 25 \) - \( x \) ranges from \( 0 \) to \( 5 \) (since \( y = 25 \implies x = 5 \)) --- #### **(a) Washer Method (Horizontal Slices)** - Slices perpendicular to the axis of rotation, so integrate with respect to \( y \). - At height \( y \), \( x = \sqrt{y} \). - **Outer radius**: Distance from \( x = 5 \) to \( x = 0 \): \( 5 - 0 = 5 \) - **Inner radius**: Distance from \( x = 5 \) to \( x = \sqrt{y} \): \( 5 - \sqrt{y} \) \[ V = \pi \int_{y=0}^{y=25} \left[ (5)^2 - (5 - \sqrt{y})^2 \right] dy \] \[ = \pi \int_{0}^{25} \left[ 25 - (25 - 10\sqrt{y} + y) \right] dy \] \[ = \pi \int_{0}^{25} (10\sqrt{y} - y) dy \] Integrate: \[ \int_{0}^{25} 10\sqrt{y} \, dy = 10 \int_{0}^{25} y^{1/2} dy = 10 \left. \frac{2}{3} y^{3/2} \right|_0^{25} = 10 \cdot \frac{2}{3} \cdot 125 = \frac{2500}{3} \] \[ \int_{0}^{25} y \, dy = \left. \frac{y^2}{2} \right|_0^{25} = \frac{625}{2} \] So, \[ V = \pi \left( \frac{2500}{3} - \frac{625}{2} \right) = \pi \left( \frac{5000 - 1875}{6} \right) = \pi \left( \frac{3125}{6} \right) \] #### **Washer Method Answer:** \[ \boxed{V = \frac{3125\pi}{6}} \] --- #### **(b) Shell Method (Vertical Slices)** - Slices **parallel** to axis of rotation, integrate with respect to \( x \). - **Radius** (distance from shell at \( x \) to \( x = 5 \)): \( 5 - x \) - **Height**: From \( y = x^2 \) to \( y = 25 \): \( 25 - x^2 \) - \( x \) from 0 to 5. \[ V = 2\pi \int_{x=0}^{x=5} (\text{radius}) \cdot (\text{height}) dx \] \[ = 2\pi \int_{0}^{5} (5 - x)(25 - x^2) dx \] Expand: \[ = 2\pi \int_{0}^{5} \left[ 125 - 5x^2 - 25x + x^3 \right] dx \] \[ = 2\pi \int_{0}^{5} (125 - 25x - 5x^2 + x^3) dx \] Integrate each term: - \(\int_{0}^{5} 125 dx = 125x|_0^5 = 625\) - \(\int_{0}^{5} 25x dx = 25 \cdot \frac{x^2}{2}|_0^5 = 25 \cdot \frac{25}{2} = 312.5\) - \(\int_{0}^{5} 5x^2 dx = 5 \cdot \frac{x^3}{3}|_0^5 = 5 \cdot \frac{125}{3} = \frac{625}{3}\) - \(\int_{0}^{5} x^3 dx = \frac{x^4}{4}|_0^5 = \frac{625}{4}\) Sum: \[ = 2\pi \left[ 625 - 312.5 - \frac{625}{3} + \frac{625}{4} \right] \] Convert to common denominator (12): - \(625 = \frac{7500}{12}\) - \(312.5 = \frac{3750}{12}\) - \(\frac{625}{3} = \frac{2500}{12}\) - \(\frac{625}{4} = \frac{1875}{12}\) So sum: \[ 625 - 312.5 - \frac{625}{3} + \frac{625}{4} = \frac{7500 - 3750 - 2500 + 1875}{12} = \frac{3125}{12} \] Multiply by 2π: \[ V = 2\pi \cdot \frac{3125}{12} = \frac{3125\pi}{6} \] --- #### **Shell Method Answer:** \[ \boxed{V = \frac{3125\pi}{6}} \] --- ## **Final Answers** ### **3)** \[ \boxed{V = \frac{17\pi}{2}} \] ### **4)** \[ \boxed{V = \frac{3125\pi}{6}} \] (using either method)