Let's solve each problem step by step: --- ## **Problem 1: Disk/Washer Method** **Given:** - Bounded by \( y = x^2 \) and \( y = x^3 \) - Rotate about the x-axis. **Step 1: Find intersection points.** Set \( x^2 = x^3 \) \[ x^2 = x^3 \implies x^3 - x^2 = 0 \implies x^2(x-1) = 0 \implies x = 0, 1 \] **Step 2: Set up the integral.** Since \( x^2 \geq x^3 \) for \( x \in [0,1] \), the outer radius is \( x^2 \), inner radius is \( x^3 \): \[ V = \pi \int_{0}^{1} \left[ (x^2)^2 - (x^3)^2 \right] dx = \pi \int_{0}^{1} [x^4 - x^6] dx \] **Step 3: Integrate.** \[ = \pi \left[ \frac{x^5}{5} - \frac{x^7}{7} \right]_{0}^{1} = \pi \left( \frac{1}{5} - \frac{1}{7} \right) = \pi \left( \frac{7-5}{35} \right) = \pi \left( \frac{2}{35} \right) \] **Final Answer:** \[ \boxed{\frac{2\pi}{35}} \] --- ## **Problem 2: Cylinders/Shell Method** **Given:** - \( y = x^2 - 6x + 9 \) and \( y = -x^2 + 6x - 1 \) - Rotated about \( x = 8 \) **Step 1: Find intersection points.** Set \( x^2 - 6x + 9 = -x^2 + 6x - 1 \): \[ x^2 - 6x + 9 = -x^2 + 6x - 1 \\ x^2 + x^2 - 6x - 6x + 9 + 1 = 0 \\ 2x^2 - 12x + 10 = 0 \\ x^2 - 6x + 5 = 0 \\ (x-5)(x-1) = 0 \implies x = 1, 5 \] **Step 2: Shell radius and height.** - Radius from \( x \) to \( x = 8 \) is \( r = 8 - x \) - Height: \( H = [-x^2 + 6x - 1] - [x^2 - 6x + 9] \) \[ H = -x^2 + 6x - 1 - x^2 + 6x - 9 = -2x^2 + 12x - 10 \] **Step 3: Set up the shell integral.** \[ V = 2\pi \int_{x=1}^{5} (8-x) \left[ -2x^2 + 12x - 10 \right] dx \] Expand: \[ (8-x)(-2x^2 + 12x - 10) = 8(-2x^2 + 12x - 10) - x(-2x^2 + 12x - 10) = -16x^2 + 96x - 80 + 2x^3 - 12x^2 + 10x = 2x^3 - 28x^2 + 106x - 80 \] So, \[ V = 2\pi \int_{1}^{5} [2x^3 - 28x^2 + 106x - 80] dx \] **Step 4: Integrate.** \[ \int [2x^3 - 28x^2 + 106x - 80] dx = \frac{2x^4}{4} - \frac{28x^3}{3} + 53x^2 - 80x = \frac{x^4}{2} - \frac{28}{3}x^3 + 53x^2 - 80x \] Evaluate from 1 to 5: For \( x=5 \): \[ \frac{5^4}{2} - \frac{28}{3} \cdot 125 + 53 \cdot 25 - 80 \cdot 5 = \frac{625}{2} - \frac{3500}{3} + 1325 - 400 \] For \( x=1 \): \[ \frac{1}{2} - \frac{28}{3} + 53 - 80 \] Subtract (upper - lower): \[ \left(\frac{625}{2} - \frac{3500}{3} + 1325 - 400\right) - \left(\frac{1}{2} - \frac{28}{3} + 53 - 80\right) \] \[ = \left(\frac{625-1}{2}\right) - \left(\frac{3500-28}{3}\right) + (1325-53) - (400-80) = \frac{624}{2} - \frac{3472}{3} + 1272 - 320 = 312 - \frac{3472}{3} + 1272 - 320 = (312 + 1272 - 320) - \frac{3472}{3} = (1564 - 320) - \frac{3472}{3} = 1244 - \frac{3472}{3} \] \[ V = 2\pi \left(1244 - \frac{3472}{3}\right) = 2\pi \left(\frac{3732 - 3472}{3}\right) = 2\pi \left(\frac{732}{3}\right) = 2\pi \times 244 = 488\pi \] **Final Answer:** \[ \boxed{488\pi} \] --- ## **Problem 3: Arc Length** Given \( f(x) = 3 + \cosh(x) \), from \( x=0 \) to \( x=2 \) Arc length formula: \[ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx \] \( f'(x) = \sinh(x) \) \[ L = \int_{0}^{2} \sqrt{1 + \sinh^2(x)} dx \] But \( 1 + \sinh^2(x) = \cosh^2(x) \) So, \[ L = \int_{0}^{2} \cosh(x) dx = \sinh(x) \Big|_{0}^{2} = \sinh(2) - \sinh(0) = \sinh(2) - 0 = \sinh(2) \] **Final Answer:** \[ \boxed{\sinh(2)} \] --- ## **Problem 4: Parametric Curves** Given \( x = t^2, y = t^3 - t \) ### a) **Concavity** First, find \( \frac{dy}{dx} \): \( \frac{dx}{dt} = 2t \), \( \frac{dy}{dt} = 3t^2 - 1 \) \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 1}{2t} \] Second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt} \] \[ \frac{d}{dt}\left(\frac{3t^2-1}{2t}\right) = \frac{6t \cdot 2t - (3t^2-1)\cdot 2}{(2t)^2} = \frac{12t^2 - [6t^2-2]}{4t^2} = \frac{12t^2 - 6t^2 + 2}{4t^2} = \frac{6t^2 + 2}{4t^2} \] \[ \frac{d^2y}{dx^2} = \frac{6t^2+2}{4t^2} \div 2t = \frac{6t^2+2}{4t^2} \cdot \frac{1}{2t} = \frac{6t^2+2}{8t^3} \] Set \( \frac{6t^2+2}{8t^3} = 0 \Rightarrow 6t^2+2 = 0 \Rightarrow t^2 = -\frac{1}{3} \) But this has **no real solutions** for \( t \). - For \( t > 0 \), numerator \( > 0 \), denominator \( > 0 \): positive (curve is **concave up**). - For \( t < 0 \), numerator \( > 0 \), denominator \( < 0 \): negative (**concave down**). **So,** - **Concave up** for \( t > 0 \) - **Concave down** for \( t < 0 \) --- ### b) **Sketch for \( t = -1 \) to \( t = 2 \) with arrow** - At \( t = -1: x = 1, y = -1 + 1 = 0 \) → (1, 0) - At \( t = 0: x = 0, y = 0 \) → (0, 0) - At \( t = 1: x = 1, y = 1 - 1 = 0 \) → (1, 0) - At \( t = 2: x = 4, y = 8 - 2 = 6 \) → (4, 6) Plot these points and draw a smooth curve from (1, 0) through (0, 0), back to (1, 0), up to (4, 6). Draw an **arrow** in the direction of increasing \( t \). --- ### c) **Convert to Cartesian** Given: \[ x = t^2 \implies t = \pm \sqrt{x} \] \[ y = t^3 - t = t(t^2 - 1) \] Substitute: \[ y = t^3 - t = (\pm\sqrt{x})^3 - (\pm\sqrt{x}) = \pm x^{3/2} \mp x^{1/2} \] So: \[ y = t^3 - t = t(t^2 - 1) = t(t-1)(t+1) \] So, in terms of \( x \): \[ y = (\sqrt{x})^3 - \sqrt{x} = x^{3/2} - x^{1/2} \] or \[ y = -x^{3/2} + x^{1/2} \] So, the full Cartesian equation is: \[ \boxed{y = x^{3/2} - x^{1/2} \text{ or } y = -x^{3/2} + x^{1/2}} \] --- ## **Summary of Final Answers** 1. **Volume**: \( \boxed{\frac{2\pi}{35}} \) 2. **Volume**: \( \boxed{488\pi} \) 3. **Exact length**: \( \boxed{\sinh(2)} \) 4. a) Concave up for \( t > 0 \), concave down for \( t < 0 \) b) See explanation for sketch c) \( \boxed{y = x^{3/2} - x^{1/2} \text{ or } y = -x^{3/2} + x^{1/2}} \)