Surface Area and Volume of Spheres
Every question with a fully worked solution.
- Surface Area and Volume of Spheres - Video - Volume and Surface Area of a Sphere Formula, Examples, Word Problems, Geometry Watch
Theory
A sphere with radius \(r\) has surface area \(4\pi r^2\) and volume \(\dfrac{4}{3}\pi r^3\). This Year 10A maths page covers hemispheres, working backwards to the radius with a square root or cube root, composite solids, 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).
- Composite solids: add the parts you can see. A cylinder with a hemisphere on each end has no flat circles: the two hemispheres together make one whole sphere, so its surface is \(2\pi rh + 4\pi r^2\).
- Sphere inside a cylinder: a sphere that just fits has the same radius as the cylinder, and the cylinder's height is the diameter, \(2r\).
- 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.