Problem 31 - Basic Technical Mathematics with Calculus

Question

Find the coordinates of the centroids of the given figures. Each region is covered by a thin, flat plate. Find the location of the centroid of a hemisphere of radius \(a\). Using this result, locate the centroid of the northern hemisphere of Earth for which the radius is \(6370 \mathrm{km}\).

Step-by-Step Solution

Answer

The centroid of a hemisphere of radius \\( a \\\) is at \\( z = \frac{3a}{8} \\\). Thus, the northern hemisphere of Earth has its centroid at \\( (0, 0, 2388.75) \, \text{km} \\\).

Step 1: Understand the Problem

We are tasked with finding the centroid of a hemisphere with a known radius. After calculating this, we will apply it to the Earth treated as a hemisphere to find its centroid.

Step 2: Recall the Centroid Formula for a Hemisphere

The centroid of a solid hemisphere of radius \( a \) lies on the axis of symmetry at \( rac{3a}{8} \) above the base (or the plane of the equator for a full sphere). This is derived from the integration of the hemispherical volume.

Step 3: Calculate the Centroid for a Hemisphere of Radius \\( a \\\)

Using the formula derived: \[ z_{centroid} = \frac{3a}{8} \]where \( a \) is the radius of the hemisphere.

Step 4: Apply the Formula to Earth

Given the radius of the Earth \( a = 6370 \, \text{km} \), substitute this value into the centroid formula:\[ z_{centroid} = \frac{3 \times 6370}{8} \ = 2388.75 \, \text{km} \]

Step 5: Determine the Final Coordinates

Since we are dealing with a hemisphere, the coordinates of the centroid will be on the axis passing through its center. Thus, for the Earth, the centroid is located at \( (0, 0, 2388.75) \, \text{km} \).

Key Concepts

Integration

Integration is a mathematical tool that allows us to calculate quantities like area, volume, and centroids, especially for irregular shapes. In the problem of finding the centroid of a hemisphere, integration helps us express and solve for parts of the volume that contribute to the centroid's position.

By integrating over a volume, we can find cumulative measurements such as mass or centroid. In this exercise, the integration process involves summing infinitesimally small slices of the hemisphere and calculating their contribution to find the centroid. This continuous sum over the entire volume gives a precise location of the centroid.

Integration simplifies the calculation process for geometrically complex shapes, making it essential for these types of problems.

Axis of Symmetry

The axis of symmetry in a geometry context is an imaginary line where a shape can be folded or reflected, producing two identical halves. For a hemisphere, the axis of symmetry is a straight line running from the base of the hemisphere (which is a circle) to its topmost point.

This line is crucial for finding the centroid, as it helps in determining the balanced center of mass. It simplifies the problem since you only need to focus on the variation along this axis to calculate the centroid.

• In the hemisphere problem, the axis of symmetry is vertical (often referred to as the z-axis in coordinate systems).
• The centroid lies along this axis, making complex volume calculations manageable.

Hemispherical Volume

Understanding the hemispherical volume is essential in finding the centroid. A hemisphere is exactly half of a complete sphere, and its volume can be calculated using the formula for the volume of a sphere divided by two.

The formula for the volume of a full sphere is \[ V = \frac{4}{3}\pi a^3 \] where \( a \) is the radius.

Thus, the volume of a hemisphere becomes \[ V_{hemisphere} = \frac{1}{2} \cdot \frac{4}{3}\pi a^3 = \frac{2}{3}\pi a^3 \]

This volume formula is useful in achieving a proper understanding of the mass distribution along the hemisphere, hence directly influencing the centroid's position.

Centroid Formula

The centroid of a hemisphere is calculated through a specific formula derived from integration, particularly focusing on its volumetric symmetry.

The centroid's coordinates indicate where the average mass of the shape is located. For a hemisphere of radius \( a \), the centroid is found at \[ z_{centroid} = \frac{3a}{8} \]

Here's why this formula makes sense:

• This formula results from integrating the volume of the hemisphere to find the balance point along the axis of symmetry.
• The height of the centroid above the base is less than that of a full sphere since a hemisphere lacks mass at one end.

By plugging different radius values into this formula: for example, using the Earth's radius (6370 km), we find that the Earth's northern hemisphere centroid lies 2388.75 km from the surface on the axis of symmetry.