Cone, sphere, and hemisphere

This subtopic adds the three iconic round solids to your toolkit: the right circular cone, the sphere, and the hemisphere. The formulas are simple but worth memorising because they appear in nearly every board paper.

Cone

A right circular cone is generated by rotating a right triangle around one of its legs. Its key dimensions are the radius $r$ (of the circular base), the height $h$ (perpendicular from the apex to the centre of the base), and the slant height $\ell$ (the distance from the apex to any point on the boundary of the base). These three quantities are related by

$\ell^2 = r^2 + h^2.$

If you cut a cone along a straight line from the apex to the base and unroll it flat, you get a sector of a circle of radius $\ell$ and arc length $2\pi r$ (the circumference of the cone's base). The area of this sector is its curved surface area (CSA):

$\text{CSA}_{\text{cone}} = \pi r \ell.$

(Derivation: the sector area is $(1/2)(\ell)(\text{arc length}) = (1/2)(\ell)(2\pi r) = \pi r \ell$ .)

The total surface area is the CSA plus the area of the base:

$\text{TSA}_{\text{cone}} = \pi r \ell + \pi r^2 = \pi r (\ell + r).$

The volume of a cone is one-third the volume of the cylinder of the same radius and height:

$V_{\text{cone}} = \frac{1}{3} \pi r^2 h.$

(This fact requires calculus to prove rigorously, but you can demonstrate it experimentally by filling a cone three times into a cylinder of the same base and height , the water exactly fills the cylinder.)

Sphere

A sphere is the set of all points equidistant from a centre , a perfectly round ball. With radius $r$ :

Surface area: $S = 4\pi r^2$ . (Equal to the curved surface area of the smallest cylinder that contains it.)
Volume: $V = (4/3) \pi r^3$ .

These formulas were discovered by Archimedes around $250$ BCE; he was so proud of them that he asked for a sphere inscribed in a cylinder to be carved on his tombstone.

Hemisphere

A hemisphere is half a sphere , cut along a great circle (a circle passing through the centre). It has:

Curved surface area: $2\pi r^2$ (half of $4\pi r^2$ ).
Total surface area: $3\pi r^2$ (curved $2\pi r^2$ plus circular base $\pi r^2$ ).
Volume: $(2/3) \pi r^3$ (half of $(4/3)\pi r^3$ ).

Worked examples

Example 1. A cone has $r = 7, h = 24$ . Find the slant height, CSA, TSA, and volume.

$\ell = \sqrt{49 + 576} = \sqrt{625} = 25$ .

CSA $= 22/7 \cdot 7 \cdot 25 = 550$ cm $^2$ .

TSA $= 550 + 22/7 \cdot 49 = 550 + 154 = 704$ cm $^2$ .

$V = (1/3) \cdot 22/7 \cdot 49 \cdot 24 = 22 \cdot 7 \cdot 8 = 1232$ cm $^3$ .

Example 2. A cone has slant height $13$ cm and radius $5$ cm. Find the height, CSA, and volume.

$h = \sqrt{169 - 25} = 12$ cm.

CSA $= \pi \cdot 5 \cdot 13 = 65\pi \approx 204.3$ cm $^2$ .

$V = (1/3)\pi(25)(12) = 100\pi \approx 314.3$ cm $^3$ .

Example 3. A sphere has radius $7$ cm. Find $S$ and $V$ .

$S = 4 \cdot 22/7 \cdot 49 = 616$ cm $^2$ .

$V = (4/3)(22/7)(343) = 4 \cdot 22 \cdot 49/3 = 4312/3 \approx 1437.33$ cm $^3$ .

Example 4. A hemisphere has radius $14$ . Find total surface area and volume.

TSA $= 3 \cdot 22/7 \cdot 196 = 1848$ cm $^2$ .

$V = (2/3)(22/7)(2744) = (44 \cdot 2744)/(21) \approx 5749.33$ cm $^3$ .

Example 5. A solid metallic sphere of radius $10$ cm is melted and recast into smaller spheres of radius $2$ cm. How many smaller spheres are formed?

$V_{\text{large}} = (4/3)\pi(1000)$ , $V_{\text{small}} = (4/3)\pi(8)$ . Number $= 1000/8 = 125$ .

Example 6. A cone of radius $6$ and height $8$ is melted and recast into a sphere. Find the sphere's radius.

$V_{\text{cone}} = (1/3)\pi(36)(8) = 96\pi$ .

$(4/3)\pi R^3 = 96\pi \Rightarrow R^3 = 72 \Rightarrow R = \sqrt[3]{72} \approx 4.16$ cm.

Example 7. Calculate the height of a cone whose volume is $924$ cm $^3$ and radius $7$ cm.

$(1/3)(22/7)(49) h = 924 \Rightarrow (1/3)(154) h = 924 \Rightarrow h = 924 \cdot 3/154 = 18$ cm.

Try it yourself

A cone has $r = 6$ , $h = 8$ . Find the slant height, CSA, TSA, and volume.
A sphere has radius $21$ cm. Find $S$ and $V$ .
A hemisphere has radius $10$ cm. Find $S$ (total) and $V$ .
A cone has slant height $25$ cm and radius $7$ cm. Find $h$ and $V$ .
Find the volume of a cone of radius $14$ cm and height $9$ cm.
A sphere has surface area $616$ cm $^2$ . Find its radius and volume.
A solid hemisphere of radius $6$ cm is melted into a cone of height $9$ cm. Find the cone's radius.
A cylindrical container of radius $7$ cm and height $20$ cm is filled with ice cream that is then moulded into cones of radius $3$ cm and height $12$ cm. How many cones can be filled?
A toy is in the shape of a cone of radius $4$ cm and height $6$ cm. Find its slant height and CSA.
A spherical ball is melted to form $64$ equal balls of half the radius. Verify.
A solid cone of radius $3$ and height $4$ . Find its surface area and volume.
Compare the volumes of a sphere of radius $r$ , a cylinder of radius $r$ and height $2r$ , and a cone of radius $r$ and height $2r$ .

Pitfalls / Insight

(1) Slant height vs height. The slant height $\ell$ is not the height. They differ by Pythagoras: $\ell^2 = r^2 + h^2$ . Confusing them is the single most common mistake in cone problems.

(2) CSA vs TSA. "Curved" surface area excludes flat ends; "total" includes them. Read the question carefully.

(3) Hemisphere has two surfaces. The TSA of a hemisphere is $3\pi r^2$ (curved $2\pi r^2$ plus base $\pi r^2$ ). If only the curved is required, use $2\pi r^2$ .

(4) Archimedes ratio. For sphere of radius $r$ , cone of radius $r$ and height $2r$ , and cylinder of radius $r$ and height $2r$ : volumes are in ratio $2 : 1 : 3$ (cone : sphere : cylinder) , actually $1 : 2 : 3$ (cone : sphere : cylinder). The cone is one-third of the cylinder; the sphere is two-thirds.

(5) Recasting problems always equate volumes (the material is conserved). Surface area is not conserved in melting.