Let's solve the given problem **step by step**: --- ### **Problem Restatement** **Sketch the solid consisting of all points with spherical coordinates \((\rho, \theta, \phi)\) such that:** - \(0 \leq \theta \leq \pi/2\) - \(0 \leq \phi \leq \pi/6\) - \(0 \leq \rho \leq 2\cos(\phi)\) --- ### **Step 1: Understand Spherical Coordinates** Recall the relationship: - \(\rho\): Distance from the origin. - \(\theta\): Angle in the \(xy\)-plane from the positive \(x\)-axis (azimuthal angle). - \(\phi\): Angle from the positive \(z\)-axis (polar angle). The conversion to Cartesian: \[ x = \rho \sin\phi \cos\theta\\ y = \rho \sin\phi \sin\theta\\ z = \rho \cos\phi \] --- ### **Step 2: Analyze the Bounds** #### **a. \(\theta\) Range (\(0 \leq \theta \leq \pi/2\))** - This sweeps out the first quadrant of the \(xy\)-plane. #### **b. \(\phi\) Range (\(0 \leq \phi \leq \pi/6\))** - This is a small "cap" near the positive \(z\)-axis. - Since \(\phi = 0\) is the positive \(z\)-axis, and \(\phi = \pi/6\) is a cone opening downward from the \(z\)-axis with half-angle \(\pi/6\). #### **c. \(\rho\) Range (\(0 \leq \rho \leq 2\cos(\phi)\))** - For each direction (\(\theta, \phi\)), \(\rho\) runs from the origin out to the surface \(\rho = 2\cos(\phi)\). - This surface is not a sphere, but a "cap" defined by \(\rho\) as a function of \(\phi\). --- ### **Step 3: Describe the Surface \(\rho = 2\cos(\phi)\)** For any fixed \(\phi\), the maximum radius is \(2\cos(\phi)\). - At \(\phi = 0\): \(\rho = 2\). - At \(\phi = \pi/6\): \(\rho = 2\cos(\pi/6) = 2 \cdot \sqrt{3}/2 = \sqrt{3}\). So, the "cap" height decreases as we move away from the \(z\)-axis. --- ### **Step 4: Convert the Boundary to Cartesian for Visualization** For any point on the boundary: \[ x = \rho \sin\phi \cos\theta \\ y = \rho \sin\phi \sin\theta \\ z = \rho \cos\phi \] Let \(\rho = 2\cos\phi\): \[ x = 2\cos\phi \cdot \sin\phi \cdot \cos\theta = \sin(2\phi)\cos\theta \\ y = 2\cos\phi \cdot \sin\phi \cdot \sin\theta = \sin(2\phi)\sin\theta \\ z = 2\cos^2\phi \] --- ### **Step 5: Final Description** - The solid is a "quarter cap" (since \(\theta\) only goes from \(0\) to \(\pi/2\)), above the \(xy\)-plane, in the region where both \(x \geq 0\) and \(y \geq 0\). - It is bounded by the cone \(\phi = \pi/6\), and the surface \(\rho = 2\cos\phi\). - The base of the cap is a quarter disk at \(z = 2\cos^2(\pi/6) = 2 \cdot (3/4) = 3/2\). --- ### **Step 6: Sketch (Description)** 1. Draw the \(x, y, z\) axes. 2. Consider only the first quadrant in the \(xy\)-plane (\(x \geq 0, y \geq 0\)). 3. Draw a cone with vertex at the origin, opening upward, with half-angle \(\pi/6\). 4. Draw a cap surface starting at radius 2 on the \(z\)-axis (\(\phi=0\)), curving downward, getting closer to the origin as you approach the cone (\(\phi = \pi/6\)), where the radius is \(\sqrt{3}\). 5. The solid is the region inside this cap, above the \(xy\)-plane in the first quadrant, and below the surface \(\rho = 2\cos\phi\) for \(0 \leq \phi \leq \pi/6\), \(0 \leq \theta \leq \pi/2\). --- ### **Final Answer (Summary)** > **The solid is the region inside the spherical cap defined by \(\rho = 2\cos\phi\), for \(0 \leq \phi \leq \pi/6\), and within the first quadrant of the \(xy\)-plane (\(0 \leq \theta \leq \pi/2\)). It forms a quarter "cap" above the \(xy\)-plane, bounded by the cone \(\phi = \pi/6\) and the surface \(\rho = 2\cos\phi\).** --- If you want a sketch, you can visualize a quarter of a shallow dome sitting above the \(xy\)-plane and bounded by a cone with a small opening angle.