In this sphere calculator, you can enter the value of a sphere's radius, diameter, surface area, or volume and it will calculate the other three for you.

Also, the calculator will tell you not just the answers, but also you can calculate them.

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Radius $\hspace{0.2em} (r) = \hspace{0.2em}$

In this sphere calculator, you can enter the value of a sphere's radius, diameter, surface area, or volume and it will calculate the other three for you.

Also, the calculator will tell you not just the answers, but also you can calculate them.

Each of the inputs provided must be a non-negative real number. In other words, the input must be 0 or greater. Here are a few examples.

- Whole numbers or decimals → $\hspace{0.2em} 2 \hspace{0.2em}$, $\hspace{0.2em} 4.25 \hspace{0.2em}$, $\hspace{0.2em} 0 \hspace{0.2em}$, $\hspace{0.2em} 0.33 \hspace{0.2em}$
- Fractions → $\hspace{0.2em} 2/3 \hspace{0.2em}$, $\hspace{0.2em} 1/5 \hspace{0.2em}$
- Mixed numbers → $\hspace{0.2em} 5 \hspace{0.4em} 1/4 \hspace{0.2em}$

If you would like to see an example of the calculator's working, just click the "example" button.

As mentioned earlier, the calculator won't just tell you the answer but also the steps you can follow to do the calculation yourself. The "show/hide solution" button would be available to you after the calculator has processed your input.

We would love to see you share our calculators with your family, friends, or anyone else who might find it useful.

By checking the "include calculation" checkbox, you can share your calculation as well.

Here's a quick overview of what a circle is and a few different concepts related to the shape.

A sphere is a three-dimensional shape formed by the set of all points in space that are within a fixed distance (radius, usually denoted by $\hspace{0.2em} r \hspace{0.2em}$) from a certain fixed point (center, $\hspace{0.2em} O \hspace{0.2em}$).

Now, if you draw a line segment from one point on a sphere's surface to another such that it also passes through the sphere's center, you get a diameter of the sphere. And the length of the diameter is twice the length of the radius.

$d = 2r$

Alright, let's look at the formulas for the volume and surface area of a sphere.

For a sphere of radius $\hspace{0.2em} r \hspace{0.2em}$, the volume would be

$V = \frac{4}{3} \pi r^3$

And its surface area would be

$V = 4 \pi r^2$

If a sphere is divided into two identical halves, we get each half is a hemisphere.

Not surprisingly, the volume of a hemisphere is the half of that of a sphere with the same radius. So,

$V = \frac{2}{3} \pi r^3$

When it comes to the surface area, things get slightly tricky. Here's why.

If you compare a sphere and a hemisphere of the same radius, you'll see that the hemisphere has a surface in addition to half the surface of a sphere.

So, to get a hemisphere's surface area, we must add the area of this flat surface to half of the corresponding sphere's surface area. That means

$\begin{align*} S \hspace{0.3em} &= \hspace{0.3em} \frac{1}{2} \cdot 4 \pi r^2 + \pi r^2 \\[1.3em] &= \hspace{0.3em} 3 \pi r^2 \end{align*}$

Example

Calculate the volume and surface area of a sphere with a radius of $\hspace{0.2em} 4 \text{ cm} \hspace{0.2em}$.

Solution

Let's find the volume first. The formula for a sphere's volume is

$V \hspace{0.25em} = \hspace{0.25em} \frac{4}{3} \pi r^3$

Substituting the value of $\hspace{0.2em} r \hspace{0.2em}$, we get

$\begin{align*} V \hspace{0.25em} &= \hspace{0.25em} \frac{4}{3} \pi \cdot 4^3 \\[1.75em] &= \hspace{0.25em} 268.08 \end{align*}$

And now we go for the surface area.

$\begin{align*} S \hspace{0.25em} &= \hspace{0.25em} 4 \pi r^2 \\[1em] &= \hspace{0.25em} 4 \pi \cdot 4^2 \\[1em] &= \hspace{0.25em} 201.06 \end{align*}$

So the volume and surface area of the sphere are $\hspace{0.2em} 268.08 \text{ cm}^3 \hspace{0.2em}$ and $\hspace{0.2em} 201.06 \text{ cm}^2 \hspace{0.2em}$ respectively.

Example

The volume of a sphere with a radius of $\hspace{0.2em} 288 \pi \text{ in}^3 \hspace{0.2em}$. Calculate its radius.

Solution

Plugging the value of $\hspace{0.2em} V \hspace{0.2em}$ into the volume formula and solving for $\hspace{0.2em} r \hspace{0.2em}$, we have

$\begin{align*} V \hspace{0.25em} &= \hspace{0.25em} \frac{4}{3} \pi r^3 \\[1.75em] 288 \pi \hspace{0.25em} &= \hspace{0.25em} \frac{4}{3} \pi r^3 \\[1.75em] r \hspace{0.25em} &= \hspace{0.25em} 6 \end{align*}$

The sphere has a radius of $\hspace{0.2em} 6 \text{ in} \hspace{0.2em}$.

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