Consider the shape made by rotating the area between the curves \( y=x^{3} \) and \( y=x^{2} \) about the \( x \)-axis for \( 1 \leq x \leq 3 \) calculate: a) Volume b) Surface area

To find the volume and surface area of the solid formed by rotating the area between the curves \( y = x^3 \) and \( y = x^2 \) about the \( x \)-axis for \( 1 \leq x \leq 3 \), we will use the methods of disks/washers for volume and the surface area formula for revolution. ### a) Volume The volume \( V \) of the solid of revolution can be calculated using the washer method. The volume is given by the integral: \[ V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) \, dx \] where \( R(x) \) is the outer radius and \( r(x) \) is the inner radius. In this case: - The outer curve is \( y = x^2 \) (since it is above \( y = x^3 \) in the interval). - The inner curve is \( y = x^3 \). Thus, we have: \[ R(x) = x^2 \quad \text{and} \quad r(x) = x^3 \] The volume integral becomes: \[ V = \pi \int_{1}^{3} \left( (x^2)^2 - (x^3)^2 \right) \, dx \] Calculating the squares: \[ V = \pi \int_{1}^{3} \left( x^4 - x^6 \right) \, dx \] Now, we compute the integral: \[ V = \pi \left[ \frac{x^5}{5} - \frac{x^7}{7} \right]_{1}^{3} \] Calculating at the bounds: 1. At \( x = 3 \): \[ \frac{3^5}{5} - \frac{3^7}{7} = \frac{243}{5} - \frac{2187}{7} \] Finding a common denominator (35): \[ \frac{243 \cdot 7}{35} - \frac{2187 \cdot 5}{35} = \frac{1701 - 10935}{35} = \frac{-9234}{35} \] 2. At \( x = 1 \): \[ \frac{1^5}{5} - \frac{1^7}{7} = \frac{1}{5} - \frac{1}{7} \] Finding a common denominator (35): \[ \frac{7}{35} - \frac{5}{35} = \frac{2}{35} \] Now, substituting back into the volume formula: \[ V = \pi \left( \frac{-9234}{35} - \frac{2}{35} \right) = \pi \left( \frac{-9236}{35} \right) \] Thus, the volume is: \[ V = \frac{-9236\pi}{35} \] ### b) Surface Area The surface area \( S \) of the solid of revolution is given by the formula: \[ S = 2\pi \int_{a}^{b} y \sqrt{(f'(x))^2 + 1} \, dx \] where \( y = f(x) \) is the function being rotated. We need to consider both curves, so we will calculate the surface area for each and then combine them. 1. For \( y = x^2 \): \[ f(x) = x^2 \quad \Rightarrow \quad f'(x) = 2x \] The surface area integral becomes: \[ S_1 = 2\pi \int_{1}^{3} x^2 \sqrt{(2x)^2 + 1} \, dx = 2\pi \int_{1}^{3} x^2 \sqrt{4x^2 + 1} \, dx \] 2. For \( y = x^3 \): \[ g(x) = x^3 \quad \Rightarrow \quad g'(x) = 3x^2 \] The surface area integral becomes: \[ S_2 = 2\pi \int_{1}^{3} x^3 \sqrt{(3x^2)^2 + 1} \, dx = 2\pi \int_{1}^{3} x^3 \sqrt{9x^4 + 1} \, dx \] Now we need to compute both integrals \( S_1 \) and \( S_2 \). ### Final Calculation The integrals \( S_1