Question

Which of the following expressions is equivalent to the diameter of the sphere portrayed above, with a radius of \(\mathrm{r}\) and volume \(\mathrm{V}\) ? (A) \(2 \sqrt[3]{\frac{3 V}{4 \pi}}\) (B) \(\pi r 3\) (C) \(4 \sqrt{\frac{2 r^{3}}{3}}\) (D) \(\frac{4 V^{3}}{3 r^{2}}\)

Short answer

The equivalent expression is option (A) \(2 \sqrt[3]{\frac{3 V}{4 \pi}}\).

Step-by-step solution

Understand the problem

We need to find the expression equivalent to the diameter of a sphere, given its radius \(r\) and volume \(V\). The diameter of a sphere can be expressed in terms of the radius as \(d = 2r\). We are also given that the volume \(V\) of the sphere is calculated using the formula \(V = \frac{4}{3} \pi r^3\).

Express the diameter in terms of the volume

Since we have the volume formula \(V = \frac{4}{3} \pi r^3\), we need to express \(r\) in terms of \(V\) first. Starting with the equation, multiply by \(3/4\) to both sides: \(V = \frac{4}{3} \pi r^3\), we obtain \(\frac{3V}{4\pi} = r^3\). Taking the cube root of both sides, we get \(r = \sqrt[3]{\frac{3V}{4\pi}}\). Next, substitute \(r\) in the diameter formula \(d = 2r\) gives \(d = 2\sqrt[3]{\frac{3V}{4\pi}}\).

Match with the given options

Now that we have found \(d = 2\sqrt[3]{\frac{3V}{4\pi}}\), compare it with the options given:- (A) \(2\sqrt[3]{\frac{3V}{4\pi}}\)- (B) \(\pi r 3\)- (C) \(4\sqrt{\frac{2r^{3}}{3}}\)- (D) \(\frac{4V^{3}}{3r^{2}}\)The option that matches our derivation is option (A).

Key concepts

Sphere

A sphere is a perfectly round three-dimensional shape, similar to the surface of a ball. All points on the surface of a sphere are equidistant from its center. This unique property makes spheres one of the most symmetrical shapes in geometry.
To imagine a sphere, think of ordinary objects like a basketball or a soap bubble. The center of the sphere is called the 'center', while the distance from this point to any point on the surface is called the 'radius.'
In mathematics, spheres are important because they help us understand symmetry, geometry, and properties of three-dimensional shapes. Whether studying physics, astronomy, or engineering, spheres are a foundational concept that appears frequently.

Volume of a Sphere

The volume of a sphere is the amount of space it occupies in three-dimensional space. It is a measure of how much material you would need to fill the sphere entirely. The formula to calculate the volume of a sphere given its radius \( r \) is:\[ V = \frac{4}{3} \pi r^3 \]This formula tells us that you need to cube the radius, multiply it by \( \pi \), and then by \( \frac{4}{3} \).

• The \( r^3 \) represents the exponential increase in volume as the sphere gets larger.
• The constant \( \pi \) brings in the circle's geometry since a sphere's cross-section is a circle.
• The fraction \( \frac{4}{3} \) adjusts the formula to account for the three-dimensional aspect of the shape.

Understanding these components helps clarify why spheres have such unique volume calculations compared to other 3D shapes.

Diameter

The diameter of a sphere is twice the length of its radius. It is the longest straight line that can be drawn through the sphere, touching two opposite sides of its surface.
If you know the radius \( r \) of a sphere, you can easily find the diameter \( d \) using the formula:\[ d = 2r \]This simple relationship shows that the diameter is exactly twice as large as the radius.
Knowing the diameter is crucial because it provides a complete span of the sphere, allowing us to understand the full size of the sphere in one straight measurement. Whether in construction, design, or physics, knowing a sphere's diameter can be as important as knowing its volume or radius.