Surface Area and Volume of Spheres
A sphere of radius \(r\) has surface area \(4\pi r^2\) and volume \(\frac{4}{3}\pi r^3\). A hemisphere is half a sphere, so a solid hemisphere has a curved surface plus a flat circular face. In NSW Year 10 Advanced maths you use these formulas forwards and backwards, giving exact answers in terms of \(\pi\) or rounded ones (NESA outcomes MA5‑ARE‑P‑01 and MA5‑VOL‑P‑01).
You will find surface areas and volumes of spheres and hemispheres, find a radius from a surface area or volume, and find the volume of a basketball from its circumference. You will estimate the surface area of the Earth, find how much of a box a ball fills, paint a dome, fill cups from a punch bowl, and find the mass of a hollow rubber ball. Harder questions show how a sphere compares with the cylinder that just holds it. It is part of Year 10 Advanced Measurement.
Every question with a fully worked solution.
- Surface Area and Volume of Spheres - Video - Volume and Surface Area of a Sphere Formula Watch
Theory
A sphere with radius \(r\) has surface area \(4\pi r^2\) and volume \(\dfrac{4}{3}\pi r^3\). This Year 10 Advanced maths page covers hemispheres, working backwards to the radius with a square root or cube root, hollow balls, and water rising when a ball is dropped in.
A sphere is the set of points that are all the same distance \(r\) (the radius) from a centre. A hemisphere is exactly half a sphere, cut through the centre.
The surface area of a sphere is \(4\pi r^2\): exactly four times the area of a circle with the same radius. The volume is \(\dfrac{4}{3}\pi r^3\).
A hemisphere's curved surface is half the sphere's, \(2\pi r^2\). A solid hemisphere also has a flat circular face, \(\pi r^2\). An open bowl or a dome has the curved part only.
The distance around the widest part of a sphere (its great circle) is \(2\pi r\), so a circumference gives the radius.
| Solid | Surface area | Volume |
|---|---|---|
| Sphere | \(4\pi r^2\) | \(\dfrac{4}{3}\pi r^3\) |
| Hemisphere (curved part only) | \(2\pi r^2\) | \(\dfrac{2}{3}\pi r^3\) |
| Solid hemisphere | \(3\pi r^2\) |
How to find the surface area or volume
- Find the radius (halve a diameter; \(r = \dfrac{C}{2\pi}\) from a circumference).
- Choose the formula: \(4\pi r^2\) for area, \(\dfrac{4}{3}\pi r^3\) for volume; halve them for a hemisphere and add \(\pi r^2\) for a flat face.
- Substitute, keeping \(\pi\) exact if an exact answer is wanted.
- Round once at the end, with square or cubic units.
Other questions
- Very large or small answers: give them in scientific notation to the significant figures asked.
- A ball that just fits in a box or can: the diameter equals the edge of the box (or the can's width).
- Hollow ball: volume of material \(= \dfrac{4}{3}\pi(R^3 - r^3)\), then mass = volume \(\times\) density.
- Displacement: a sunken ball raises the water by its own volume. Rise = ball volume \(\div\) base area of the container. For several balls, use an inequality and round up.
- Containers and paint: capacity in mL (\(1\text{ cm}^3 = 1\) mL), cups filled round down; tins of paint round up.
- Comparing formulas: to find when two expressions are equal, set them equal and divide by the common factor (such as \(4\pi r^2\)), since \(r \gt 0\).
(i) its surface area
(ii) its volume, each exactly and correct to two decimal places.
The radius is \(5\) cm.
(i)
| \(\text{SA}\) | \(=\) | \(4\pi r^2 = 4\pi \times 5^2\) |
| \(\) | \(=\) | \(\mathbf{100\pi}\text{ cm}^2\) |
| \(\) | \(\approx\) | \(\mathbf{314.16}\text{ cm}^2\) |
(ii)
| \(V\) | \(=\) | \(\dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi \times 125\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{500\pi}{3}}\text{ cm}^3\) |
| \(\) | \(\approx\) | \(\mathbf{523.60}\text{ cm}^3\) |
Square the radius for area, cube it for volume.
(i) its radius, correct to two decimal places
(ii) its surface area, correct to one decimal place.
(i) \(1\text{ L} = 1000\text{ cm}^3\). Make \(r^3\) the subject, then take the cube root:
| \(\dfrac{4}{3}\pi r^3\) | \(=\) | \(1000\) |
| \(r^3\) | \(=\) | \(\dfrac{3000}{4\pi} = \dfrac{750}{\pi}\) |
| \(\) | \(=\) | \(238.7324\ldots\) |
| \(r\) | \(=\) | \(\sqrt[3]{238.73\ldots} = 6.2035\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{6.20}\text{ cm}\) |
(ii) Use the unrounded radius:
| \(\text{SA}\) | \(=\) | \(4\pi \times (6.2035\ldots)^2\) |
| \(\) | \(=\) | \(483.59\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{483.6}\text{ cm}^2\) |
(i) its total surface area
(ii) its volume, each correct to two decimal places.
The radius is \(3.5\) cm.
(i) Curved surface plus the flat circle:
| \(\text{SA}\) | \(=\) | \(2\pi r^2 + \pi r^2 = 3\pi r^2\) |
| \(\) | \(=\) | \(3\pi \times 3.5^2 = 36.75\pi\) |
| \(\) | \(\approx\) | \(\mathbf{115.45}\text{ cm}^2\) |
(ii) Half of a sphere:
| \(V\) | \(=\) | \(\dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi \times 3.5^3\) |
| \(\) | \(=\) | \(89.7971\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{89.80}\text{ cm}^3\) |
The ball pushes aside its own volume of water. That water forms a layer over the whole base:
| \(V_{\text{ball}}\) | \(=\) | \(\dfrac{4}{3}\pi \times 4.5^3 = 121.5\pi\) |
| \(40 \times 30 \times d\) | \(=\) | \(121.5\pi\) |
| \(d\) | \(=\) | \(\dfrac{121.5\pi}{1200}\) |
| \(\) | \(=\) | \(0.3180\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{0.32}\text{ cm}\) |
Common pitfalls
Frequently asked questions
What is the formula for the surface area of a sphere?
\(\text{SA} = 4\pi r^2\), where \(r\) is the radius. It is exactly four times the area of a circle with the same radius.
What is the formula for the volume of a sphere?
\(V = \dfrac{4}{3}\pi r^3\). A hemisphere has half of this, \(\dfrac{2}{3}\pi r^3\).
How do you find the radius of a sphere from its volume?
Multiply the volume by \(3\), divide by \(4\pi\), then take the cube root: \(r = \sqrt[3]{\dfrac{3V}{4\pi}}\).
What is the surface area of a hemisphere?
The curved part is \(2\pi r^2\). A solid hemisphere also has a flat circle, \(\pi r^2\), giving \(3\pi r^2\) in total.
What happens to the volume of a sphere if the radius doubles?
The volume is multiplied by 2 cubed, which is 8. The surface area is multiplied by 2 squared, which is 4.