15 Aug, 2024
· Mathematics

How to find the volume of a sphere with a radius of 3 m

π3.14\pi \approx 3.14

Short Answer
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Long Explanation

Explanation

Understanding the Formula

The volume VV of a sphere can be found using the formula:

V=43πr3V = \frac{4}{3} \pi r^3

Where:

  • π\pi is approximately 3.14
  • rr is the radius of the sphere

Substituting the Given Radius

Given that the radius rr of the sphere is 3 meters:

r=3mr = 3 \, \text{m}

Calculating the Volume

Now, substitute the values into the formula:

V=43π(3)3V = \frac{4}{3} \pi (3)^3

First, calculate 333^3:

33=3×3×3=273^3 = 3 \times 3 \times 3 = 27

Next, multiply by π\pi (using 3.14 as an approximation):

π×27=3.14×27=84.78\pi \times 27 = 3.14 \times 27 = 84.78

Now, multiply by 43\frac{4}{3}:

V=43×84.78V = \frac{4}{3} \times 84.78

This simplifies to:

V=4×84.783=339.123=113.04V = \frac{4 \times 84.78}{3} = \frac{339.12}{3} = 113.04

Final Volume

Therefore, the volume of the sphere is:

113.04cubic meters\boxed{113.04 \, \text{cubic meters}}
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Emily Rosen

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Emily Rosen is a recent graduate with a Master's in Mathematics from the University of Otago. She has been tutoring math students and working as a part-time contract writer for the past three years. She is passionate about helping students overcome their fear of math through easily digestible content.

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Concept

Volume Of A Sphere Formula

Understanding the Volume of a Sphere Formula

The volume of a sphere refers to the amount of space that is enclosed within it. To calculate the volume, you can use a specific formula based on the radius of the sphere.

Formula Derivation

A sphere's volume formula is derived from integral calculus, but for practical purposes, it is often just given in its simplified form:

V=43πr3V = \frac{4}{3} \pi r^3

Explanation

  • VV represents the volume of the sphere.
  • π\pi (pi) is a mathematical constant, approximately equal to 3.14159.
  • rr is the radius of the sphere, which is the distance from the center of the sphere to any point on its surface.

Key Points

  • The volume scales with the cube of the radius. This means that even small changes in the radius can lead to significant changes in the volume.
  • The constant 43\frac{4}{3} in the formula has been derived from the nature of a sphere and involves integrating the surface area over the radius.

Illustration

To further illustrate this, consider a sphere with a radius of rr units. The volume would be computed as follows:

V=43πr3V = \frac{4}{3} \pi r^3

For example, if r=5r = 5 units:

V=43π(53)=43π(125)=5003π523.6 units3V = \frac{4}{3} \pi (5^3) = \frac{4}{3} \pi (125) = \frac{500}{3} \pi \approx 523.6 \text{ units}^3

Practical Application

Understanding this formula is crucial in various fields such as physics, engineering, and architecture where the measurement of space inside a spherical object is necessary.

By mastering this formula, you can solve many practical problems involving spherical volumes efficiently.

Concept

Substitution Of Given Values

Explanation

Substitution of given values is a fundamental concept in algebra and calculus where specific values are substituted into variables within mathematical expressions or equations. This technique is essential for solving equations, evaluating functions, and simplifying expressions.

When substituting, you replace each variable with its corresponding given value, then perform the necessary arithmetic operations to simplify the expression and find the result.

Example 1: Simple Algebraic Expression

Given the expression 3x+53x + 5 and x=2x = 2:

3x+5=32+5=6+5=113x + 5 = 3 \cdot 2 + 5 = 6 + 5 = 11

Example 2: Quadratic Equation

Consider the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 and substitute a=1a = 1, b=3b = -3, and c=2c = 2:

x23x+2=0x^2 - 3x + 2 = 0

Example 3: Function Evaluation

For the function f(x,y)=2x2yf(x, y) = 2x^2 - y, if x=3x = 3 and y=1y = 1:

f(3,1)=2(3)21=291=181=17f(3, 1) = 2(3)^2 - 1 = 2 \cdot 9 - 1 = 18 - 1 = 17

Example 4: Multi-variable Equation

Given the equation z=x2+y2z = x^2 + y^2 and substituting x=4x = 4 and y=3y = 3:

z=x2+y2=42+32=16+9=25\begin{aligned} z &= x^2 + y^2 \\ &= 4^2 + 3^2 \\ &= 16 + 9 \\ &= 25 \end{aligned}

Importance

  • Simplifies complex expressions: By substituting specific values, we can reduce complex algebraic expressions to simple numerical results.
  • Solves equations: Helps in finding the numerical solutions of equations by substituting values of variables.
  • Evaluates functions: Allows the calculation of function outputs for given inputs.
  • Applications: Widely used in physics, engineering, economics, and various other fields for practical problem-solving.

Steps to Substitute Given Values

  1. Identify the variables: Determine which variables in the expression or equation need to be substituted.
  2. Substitute values: Replace each variable with the given value.
  3. Simplify: Perform arithmetic operations to simplify the result.
  4. Check: Verify the calculations to ensure accuracy, especially in more complex expressions.