Finding The Radius Of A Sphere Given Its Volume A Step-by-Step Guide

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In the realm of geometry, the sphere stands as a fundamental three-dimensional shape, characterized by its perfect symmetry and constant curvature. Its volume, a measure of the space it occupies, is intrinsically linked to its radius, the distance from the center to any point on its surface. The relationship between these two properties is elegantly expressed by the formula $V(r) = \frac{4}{3} \pi r^3$, where $V$ represents the volume and $r$ denotes the radius. This equation provides a direct means to calculate the volume of a sphere given its radius. However, in many practical scenarios, we encounter the inverse problem: determining the radius of a sphere when its volume is known. This requires us to express the radius, $r$, as a function of the volume, $V$, a task that involves algebraic manipulation and a clear understanding of the underlying mathematical principles.

This article delves into the process of expressing the radius of a sphere as a function of its volume. We will explore the steps involved in rearranging the volume formula to isolate $r$, thereby obtaining an equation that directly calculates the radius from the volume. Furthermore, we will apply this derived formula to a specific example, calculating the radius of a sphere with a volume of 200 cubic units. This practical application will solidify our understanding of the relationship between volume and radius and demonstrate the utility of the derived formula in real-world scenarios. Understanding this relationship is crucial in various fields, including engineering, physics, and even art, where spherical shapes are prevalent.

To express the radius, $r$, of a sphere as a function of its volume, $V$, we begin with the well-known formula for the volume of a sphere:

V=43Ï€r3V = \frac{4}{3} \pi r^3

Our goal is to isolate $r$ on one side of the equation. To achieve this, we will perform a series of algebraic manipulations. First, we multiply both sides of the equation by $\frac{3}{4}$ to eliminate the fraction on the right side:

34V=34â‹…43Ï€r3\frac{3}{4}V = \frac{3}{4} \cdot \frac{4}{3} \pi r^3

This simplifies to:

34V=Ï€r3\frac{3}{4}V = \pi r^3

Next, we divide both sides by $\pi$ to isolate the $r^3$ term:

34Ï€V=r3\frac{3}{4\pi}V = r^3

Finally, to solve for $r$, we take the cube root of both sides of the equation:

r=34Ï€V3r = \sqrt[3]{\frac{3}{4\pi}V}

This equation expresses the radius, $r$, as a function of the volume, $V$. It allows us to directly calculate the radius of a sphere given its volume. This derived formula is a powerful tool in various applications, as it provides a straightforward method to determine a sphere's radius from its volumetric measurement. The cube root operation is crucial here, as it undoes the cubing effect on the radius in the original volume formula. The constant $\frac{3}{4\pi}$ acts as a scaling factor, adjusting the volume to correctly reflect the radius. Understanding this derivation not only provides the formula but also enhances our comprehension of the relationship between a sphere's volume and its radius.

Now that we have derived the formula for the radius of a sphere as a function of its volume, let's apply it to a specific example. We are given a sphere with a volume of 200 cubic units, and our task is to find its radius. Using the formula we derived:

r=34Ï€V3r = \sqrt[3]{\frac{3}{4\pi}V}

We substitute $V = 200$ into the equation:

r=34Ï€(200)3r = \sqrt[3]{\frac{3}{4\pi}(200)}

First, we simplify the expression inside the cube root:

r=6004Ï€3r = \sqrt[3]{\frac{600}{4\pi}}

r=150Ï€3r = \sqrt[3]{\frac{150}{\pi}}

Now, we approximate the value of $\pi$ as 3.14159 and calculate the value inside the cube root:

r=1503.141593r = \sqrt[3]{\frac{150}{3.14159}}

r=47.7463r = \sqrt[3]{47.746}

Finally, we take the cube root of 47.746 to find the radius:

r≈3.626r \approx 3.626

Therefore, the radius of a sphere with a volume of 200 cubic units is approximately 3.626 units. This calculation demonstrates the practical application of the derived formula. By substituting the given volume into the formula, we were able to directly compute the corresponding radius. This process highlights the importance of algebraic manipulation in solving real-world problems. The ability to calculate the radius from the volume is essential in various fields, including engineering design, where precise dimensions are crucial.

In this exploration, we have successfully expressed the radius, $r$, of a sphere as a function of its volume, $V$. Starting with the fundamental formula for the volume of a sphere, $V = \frac{4}{3} \pi r^3$, we systematically manipulated the equation to isolate $r$, arriving at the formula:

r=34Ï€V3r = \sqrt[3]{\frac{3}{4\pi}V}

This derived formula provides a direct and efficient method to calculate the radius of a sphere when its volume is known. We then applied this formula to a practical example, determining the radius of a sphere with a volume of 200 cubic units. Through this calculation, we found the radius to be approximately 3.626 units, demonstrating the real-world applicability of the derived formula.

The ability to express one variable as a function of another is a fundamental skill in mathematics and its applications. In this case, understanding the relationship between the volume and radius of a sphere is crucial in various fields, including geometry, physics, and engineering. For instance, in engineering, this relationship is vital for designing spherical containers or calculating the size of spherical components in machinery. In physics, it is essential for understanding concepts such as buoyancy and fluid displacement. The process of deriving and applying this formula reinforces the importance of algebraic manipulation and problem-solving skills. Furthermore, it highlights the interconnectedness of mathematical concepts and their relevance to practical scenarios. This understanding not only enhances our mathematical proficiency but also our ability to analyze and solve real-world problems involving spherical objects.