Given: Region \( Q \) is enclosed by the surfaces: - \( x^2 + y^2 + z^2 \leq 1 \) (sphere of radius 1) - \( z^2 \leq 3(x^2 + y^2) \implies z^2 - 3(x^2 + y^2) \leq 0 \implies \frac{z^2}{3} \leq x^2 + y^2 \) (double-napped cone) Let's break down the solution step by step: --- ### **Step 1: Intersection and Boundaries** There are two regions formed: 1. **Inside the sphere, outside the cone:** \( x^2 + y^2 + z^2 \leq 1 \) and \( z^2 \leq 3(x^2 + y^2) \) 2. **Inside the sphere, inside the cone:** \( x^2 + y^2 + z^2 \leq 1 \) and \( z^2 \geq 3(x^2 + y^2) \) But the question is about the region **enclosed** by both, so let's look at the exact boundaries. --- ### **Step 2: In Cylindrical Coordinates \((r, \theta, z)\)** - \( x^2 + y^2 = r^2 \) - Sphere: \( r^2 + z^2 \leq 1 \implies z \in [-\sqrt{1 - r^2}, \sqrt{1 - r^2}] \) - Cone: \( z^2 \leq 3r^2 \implies z \in [-\sqrt{3}r, \sqrt{3}r] \) **Where does one bound the other?** - For small \( r \), the sphere bounds \( z \). - For larger \( r \), the cone bounds \( z \). - Find the intersection: \( 1 - r^2 = 3r^2 \implies 1 = 4r^2 \implies r^2 = \frac{1}{4} \implies r = \frac{1}{2} \). So: - For \( 0 \leq r \leq \frac{1}{2} \): Both bounds are valid, so the region is split into two: - **Region 1**: \( -\sqrt{1 - r^2} \leq z \leq \sqrt{1 - r^2} \) (sphere dominates) - **Region 2**: \( -\sqrt{3}r \leq z \leq \sqrt{3}r \) (cone dominates) But actually, the region inside the cone and inside the sphere for \( r \leq \frac{1}{2} \). --- ### **Step 3: Match with Options** #### **Option (A): Cartesian Coordinates** Splits at \( x^2 + y^2 = 1/4 \) (which is \( r^2 = 1/4 \)), and uses the bounds: - For \( x^2 + y^2 \leq 1/4 \): Sphere bounds \( z \) - For \( 1/4 \leq x^2 + y^2 \leq 1 \): Cone bounds \( z \) **This is correct.** #### **Option (B): Cylindrical Coordinates** - For \( 0 \leq r \leq 1/2 \): Sphere bounds \( z \) - For \( 1/2 < r \leq 1 \): Cone bounds \( z \) But from our analysis, for \( r > 1/2 \), \( \sqrt{1 - r^2} \) becomes imaginary, so the upper limit for \( r \) is \( 1/2 \). Therefore, the region outside this is not valid within the sphere. **This is incorrect.** #### **Option (C): Cylindrical Coordinates** - For \( 0 \leq r \leq 1/2 \): Sphere bounds \( z \) - For \( 1/2 \leq r \leq 1 \): Cone bounds \( z \) But again, for \( r > 1/2 \), \( r^2 > 1/4 \implies r^2 + z^2 > 1 \) unless \( z = 0 \). So, the region for \( r > 1/2 \) does not exist within the sphere. **This is incorrect.** #### **Option (D): Spherical Coordinates** - \( z^2 = 3(x^2 + y^2) \implies \cos^2 \phi = 3 \sin^2 \phi \implies \cot^2 \phi = 3 \implies \phi = \frac{\pi}{3} \) or \( \frac{2\pi}{3} \) - Therefore, \( \frac{\pi}{3} \leq \phi \leq \frac{2\pi}{3} \) (z within the cone) **This is correct.** --- ### **Final Answer** **The correct options are (A) and (D).** --- **Final boxed answer:** \[ \boxed{\text{(A) and (D)}} \]