SOLUTION: A diameter of a sphere of radius R coincides with an element of a right circular cylinder of diameter R. For the solid common to the sphere and the cylinder, find the area of a sec

Question 1181748: A diameter of a sphere of radius R coincides with an element of a right circular cylinder of diameter R. For the solid common to the sphere and the cylinder, find the area of a section made by (a) a plane containing the axis of the cylinder and the diameter of the sphere which coincides with the element of the cylinder, (b) the plane perpendicular to the axis of the cylinder at its midpoint, (c) a plane containing the axis of the cylinder and perpendicular to the plane of (a).
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to find the area of the sections:
**1. Visualize the Problem:** Imagine a sphere and a cylinder intersecting. The cylinder's diameter is equal to the sphere's radius, and one of the cylinder's sides (an element) is aligned with a diameter of the sphere. The common solid is the region where both the sphere and cylinder exist.
**2. Define Coordinate System:** It's helpful to use a coordinate system. Let the axis of the cylinder and the diameter of the sphere be along the z-axis. The center of the sphere is at the origin (0,0,0).
**a) Plane Containing the Axis and the Diameter:**
* This plane is essentially a cross-section through the center of both the sphere and the cylinder.
* The intersection with the sphere is a circle of radius R.
* The intersection with the cylinder is a rectangle with height 2R (the diameter of the sphere) and width R (the diameter of the cylinder).
* The section of the common solid is the *smaller* of these two shapes, which is the rectangle.
* Area of the rectangle = height * width = 2R * R = 2R²
**b) Plane Perpendicular to the Axis at Midpoint:**
* This plane is horizontal (parallel to the x-y plane) and passes through the center of the sphere (z = 0).
* The intersection with the sphere is a circle of radius R.
* The intersection with the cylinder is a circle of radius R/2.
* The section of the common solid is the smaller circle, which is the intersection with the cylinder.
* Area of the circle = π * (radius)² = π * (R/2)² = (πR²)/4
**c) Plane Containing the Axis and Perpendicular to the Plane in (a):**
* This plane is perpendicular to the plane in (a). It's effectively a vertical plane passing through the center of the cylinder.
* The intersection with the sphere is a circle of radius R.
* The intersection with the cylinder is a rectangle with height 2R and width R.
* The section of the common solid is again the smaller of the two, which will be two rectangles, each with height 2R and width R/2.
* Total area of the two rectangles = 2 * (height * width) = 2 * (2R * R/2) = 2R²
**Summary of Areas:**
* (a) Plane containing the axis and the diameter: 2R²
* (b) Plane perpendicular to the axis at midpoint: (πR²)/4
* (c) Plane containing the axis and perpendicular to the plane in (a): 2R²

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