Let's solve partb) step step as requested--- ## **Step 1: Show \( S \) is a Smooth Surface in Space** Given: \[ f(x, y, z) = x^2 + y^2 + z^2 - 4x^2 + y^2 + 3 \] However, this expression looks like it might have a small typo. Assuming the intention is: \[ f(x, y, z) = x^2 + y^2 + z^2 - 4x^2 + y^2 + 3 \] Combine like terms: \[ = (x^2 - 4x^2) + (y^2 + y^2) + z^2 + 3 = -3x^2 + 2y^2 + z^2 + 3 \] Let \( S \) be the solution set to \( f(x, y, z) = \): \[ -3x^2 + 2y^2 + z^2 + 3 = \] To show \( S \) is a smooth surface, check that \(\nabla f \neq \) at all points on \( S \): \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (-6x, 4y, 2z) \] If \(\nabla f = \), then \(x = \), \(y = \), \(z = \). Substitute into the surface equation: \[ -3()^2 + 2()^2 + ()^2 + 3 = 3 \neq \] So, the gradient is never zero on the surface. By the Implicit Function Theorem, \( S \) is a smooth surface in \(\mathbb{R}^3\). **Explanation:** We showed the gradient is never zero on the surface, so \( S \) is a smooth 2D manifold (surface) in \(\mathbb{R}^3\). --- ## **Step 2: Identify the Surface and Convert to Cylindrical Coordinates** Recall the equation: \[ -3x^2 + 2y^2 + z^2 + 3 = \] Write in standard quadratic form: \[ -3x^2 + 2y^2 + z^2 = -3 \implies 3x^2 - 2y^2 - z^2 = 3 \] Divide both sides by 3: \[ x^2 - \frac{2}{3}y^2 - \frac{1}{3}z^2 = 1 \] This is the equation of a hyperboloid of one sheet. **Convert to cylindrical coordinates (\(x = r\cos\theta,\, y = r\sin\theta\)):** \[ x^2 - \frac{2}{3}y^2 - \frac{1}{3}z^2 = 1 \] \[ (r\cos\theta)^2 - \frac{2}{3}(r\sin\theta)^2 - \frac{1}{3}z^2 = 1 \] \[ r^2 \cos^2\theta - \frac{2}{3} r^2 \sin^2\theta - \frac{1}{3}z^2 = 1 \] \[ r^2 (\cos^2\theta - \frac{2}{3}\sin^2\theta) - \frac{1}{3}z^2 = 1 \] **Explanation:** The surface is a hyperboloid of one sheet. Its equation in cylindrical coordinates is: \[ r^2 \left(\cos^2\theta - \frac{2}{3}\sin^2\theta \right) - \frac{1}{3}z^2 = 1 \] --- ## **Final Answer** **(1) \( S \) is a smooth surface in space, as the gradient of \( f \) is never zero on \( S \).** **(2) The surface is a hyperboloid of one sheet. In cylindrical coordinates:** \[ r^2 \left(\cos^2\theta - \frac{2}{3}\sin^2\theta \right) - \frac{1}{3}z^2 = 1 \]

# Part (b) Solution: Smooth Surface Analysis ## **Step 1: Show \( S \) is a Smooth Surface in Space** Given the function: \[ f(x, y, z) = x^2 + y^2 + z^2 - 4x^2 + y^2 + 3 \] First, simplify: \[ f(x, y, z) = -3x^2 + 2y^2 + z^2 + 3 \] Let \( S \) be the solution set where: \[ f(x, y, z) = 0 \implies -3x^2 + 2y^2 + z^2 + 3 = 0 \] To show \( S \) is a smooth surface, we need to check the gradient \( \nabla f \): \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (-6x, 4y, 2z) \] For \( S \) to be smooth, \( \nabla f \) must not be zero at any point on \( S \). This happens when: \[ -6x = 0 \quad \text{and} \quad 4y = 0 \quad \text{and} \quad 2z = 0 \] This implies \( x = 0 \), \( y = 0 \), \( z = 0 \). Substituting into the surface equation: \[ -3(0)^2 + 2(0)^2 + (0)^2 + 3 = 3 \neq 0 \] Thus, \( \nabla f \neq 0 \) for all points on \( S \). **Explanation:** The gradient \( \nabla f \) is non-zero on \( S \), confirming that \( S \) is a smooth 2-dimensional manifold (surface) in \(\mathbb{R}^3\). --- ## **Step 2: Identify the Surface and Convert to Cylindrical Coordinates** Rewrite the equation derived earlier: \[ -3x^2 + 2y^2 + z^2 + 3 = 0 \implies 3x^2 - 2y^2 - z^2 = 3 \] Dividing by 3 gives: \[ x^2 - \frac{2}{3}y^2 - \frac{1}{3}z^2 = 1 \] This describes a hyperboloid of one sheet. ### **Convert to Cylindrical Coordinates** Using cylindrical coordinates where \( x = r\cos\theta \) and \( y = r\sin\theta \): \[ (r\cos\theta)^2 - \frac{2}{3}(r\sin\theta)^2 - \frac{1}{3}z^2 = 1 \] Expanding this yields: \[ r^2\cos^2\theta - \frac{2}{3}r^2\sin^2\theta - \frac{1}{3}z^2 = 1 \] Rearranging results in: \[ r^2\left(\cos^2\theta - \frac{2}{3}\sin^2\theta\right) - \frac{1}{3}z^2 = 1 \] **Explanation:** The surface is a hyperboloid of one sheet, represented in cylindrical coordinates by: \[ r^2\left(\cos^2\theta - \frac{2}{3}\sin^2\theta\right) - \frac{1}{3}z^2 = 1 \] --- ## **Final Answer** 1. **\( S \) is a smooth surface in space, confirmed by the non-zero gradient of \( f \) at all points on \( S \).** 2. **The surface is a hyperboloid of one sheet, expressed in cylindrical coordinates as:** \[ r^2\left(\cos^2\theta - \frac{2}{3}\sin^2\theta\right) - \frac{1}{3}z^2 = 1 \]