Problem 54 - Calculus Volume 1

Question

[T] The volume \(V\) of a sphere depends on the length of its radius as \(V=(4 / 3) \pi r^{3}\) . Because Earth is not a perfect sphere, we can use the mean radius when measuring from the center to its surface. The mean radius is the average distance from the physical center to the surface, based on a large number of samples. Find the volume of Earth with mean radius \(6.371 \times 10^{6} \mathrm{m} .\)

Step-by-Step Solution

Answer

The volume of Earth is approximately \(1.083 \times 10^{21}\) cubic meters.

Step 1: Understand the formula

The formula to calculate the volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). We need to plug in the radius of Earth into this formula to find its volume.

Step 2: Identify the given radius

The problem provides the mean radius of Earth as \( 6.371 \times 10^6 \text{ m} \). We will use this value for \( r \) in our formula.

Step 3: Substitute the radius into the formula

Replace \( r \) in the formula with \( 6.371 \times 10^6 \text{ m} \):\[ V = \frac{4}{3} \pi (6.371 \times 10^6)^3 \]

Step 4: Calculate \( r^3 \)

First, compute the cube of the radius: \( (6.371 \times 10^6)^3 = 2.58 \times 10^{20} \text{ m}^3 \).This values comes from calculating \((6.371^3) \times (10^6)^3 = 258637000087.9 \times 10^{18} = 2.58 \times 10^{20}\).

Step 5: Plug into volume formula

Substitute \( r^3 = 2.58 \times 10^{20} \) back into the volume formula:\[ V = \frac{4}{3} \pi \times 2.58 \times 10^{20} \]

Step 6: Calculate the resulting volume

Calculate the volume: \[ V = \frac{4}{3} \times 3.14159 \times 2.58 \times 10^{20} \approx 1.083 \times 3.14159 \times 2.58 \times 10^{20} \]\[ V \approx 1.083 \times 8.10 \times 10^{20} \approx 10.88 \times 10^{20} \]\[ V \approx 1.083 \times 10^{21} \text{ cubic meters} \]

Key Concepts

Earth's Radius

The concept of Earth's radius is crucial when calculating the volume of our planet, despite its non-perfect spherical shape. Earth, being an oblate spheroid, doesn't have a single fixed radius at all points. Therefore, scientists use a mean radius, which is the average distance from Earth's center to its surface. This mean radius accounts for the varying distances due to Earth's irregularities, such as mountains and valleys.
For calculations like volume, this mean radius provides a standardized value. In this exercise, the mean radius is given as \(6.371 \times 10^6 \text{ meters}\). This value is essential to apply in the formula to compute volumes and other related measurements with greater precision. Understanding why we use mean radius helps us recognize the efforts to account for Earth's true shape in mathematical models.

Volume Calculation

Calculating the volume involves using a mathematical approach where geometry and measurement play key roles. The volume of a sphere tells us how much space it occupies, and it relies heavily on the radius. In the context of our problem, we aim to find Earth's volume based on its mean radius.

The computation steps include:

• Understanding and identifying the correct radius value, which here is \(6.371 \times 10^6 \text{ m}\).
• Applying this radius in the right formula to find the volume.
• Ensuring the accuracy of our calculations, especially when raising the radius value to a power such as 3, to maintain the precision of space measurements.

This step-by-step calculation ensures we can understand the space Earth occupies in mathematical terms.

Sphere Formula

The sphere formula is essential in geometry and is especially useful when determining volumes. The formula for the volume \(V\) of a sphere is \[ V = \frac{4}{3} \pi r^3 \]. Here, \(r\) represents the sphere's radius. This formula stems from integrating the basic principles of geometry to determine the space enclosed by a sphere.

This formula builds from:

• The notion of a sphere as a 3D object in which every surface point is equidistant from the center.
• The mathematical constant \(\pi\), crucial in any circle-related computation, adding precision and universality to the formula.
• The cube of the radius \(r^3\), representing how three-dimensional space expands with changes in radius size.

By plugging Earth's mean radius into this formula, we can approximate Earth's volume more accurately, appreciating its scale and enormity.

Mathematical Formula Application

Mathematically applying the sphere formula involves precision and understanding each step. In our Earth volume example, we've used \( V = \frac{4}{3} \pi r^3 \) to perform the calculation. Here is how:

• Substituting the mean radius \(6.371 \times 10^6 \text{ m}\) into \( r \), ensuring the formula is set to handle large values.
• Calculating \(r^3\), which reflects the exponential growth of space with radius expansion, yielding \(2.58 \times 10^{20} \text{ m}^3 \).
• Finally applying \(\pi\) and other constants to derive the numerical volume, giving us \(1.083 \times 10^{21} \text{ cubic meters}\), indicating the immense space Earth occupies.

This methodical application exemplifies how math formulas are tools to interpret real-world dimensions, demonstrating the grandeur of our planet in calculable terms.