4.10 - Cones and spheres

4.10 - Cones and spheres

Use the supplied cone and sphere formulae, distinguish curved from total surface area, and find the exposed surfaces of joined solids. Choosing the right height and counting the right surfaces matter as much as substitution.

Volume of a right circular cone (Higher)

A cone of radius 3 cm3\,\mathrm{cm} and height 4 cm4\,\mathrm{cm} is joined to a hemisphere of radius 3 cm3\,\mathrm{cm}. Can we add their volumes and their total surface areas in the same way? We will build the two measures separately.

You already know

A cylinder's volume is πr2h\pi r^2h. A radius is half a diameter. Pythagoras connects the three sides of a right-angled triangle.

A right circular cone has a circular base and a vertex directly above its centre. Its perpendicular height, hh, goes from the vertex to the base centre. Its slant height, ll, runs along the surface to the rim. The central right triangle has sides rr, hh and hypotenuse ll.

[DIAGRAM: asset_slug: m079_cone_heights; caption: Perpendicular height and slant height occupy different sides of a right triangle; description: Two schematic right circular cones show radius/perpendicular height/slant height 3/4/5 cm and 5/12/13 cm. Both central triangles satisfy l squared equals r squared plus h squared. Dashed vertical lines show perpendicular heights.; file: m079_cone_heights.png; recommended_method: matplotlib]
Perpendicular height and slant height occupy different sides of a right triangle

Volume of a right circular cone

V=13πr2hV=\frac13\pi r^2h

Use perpendicular height hh, not slant height. This result is supplied on the Higher sheet; it is absent from the Foundation sheet and is Higher content. In the formulae sheet.

The base area is πr2\pi r^2, but the cone tapers towards its vertex. Its volume is one third of a cylinder with the same base and perpendicular height. It is not half the cylinder: the taper changes the width in two directions.

Worked example 1: a cone's volume

Question. A cone has radius 3 cm3\,\mathrm{cm} and perpendicular height 4 cm4\,\mathrm{cm}. Find its volume exactly and to three significant figures.

Plan. Use the given perpendicular height in the cone-volume formula.

V=13π×32×4substitute radius and height=12π cm3simplify the coefficient≈37.7 cm3three significant figures\begin{aligned} V&=\frac13\pi\times3^2\times4 &&\text{substitute radius and height}\\ &=12\pi\,\mathrm{cm}^3 &&\text{simplify the coefficient}\\ &\approx37.7\,\mathrm{cm}^3 &&\text{three significant figures} \end{aligned}

Why this step? The factor 13\frac13 accounts for the cone's taper.

Answer. 12π cm312\pi\,\mathrm{cm}^3, approximately 37.7 cm337.7\,\mathrm{cm}^3.

Check. The matching cylinder holds π×32×4=36π cm3\pi\times3^2\times4=36\pi\,\mathrm{cm}^3; the cone holds one third of it.

Change only the perpendicular height from 4 cm4\,\mathrm{cm} to 8 cm8\,\mathrm{cm}. The same circular base now supports a taller cone.

Worked example 2: the height changes

Question. A cone has radius 3 cm3\,\mathrm{cm} and perpendicular height 8 cm8\,\mathrm{cm}. Find its volume exactly and to three significant figures.

Plan. Keep the base area fixed and substitute the new height.

V=13π×32×8use perpendicular height=24π cm3simplify≈75.4 cm3round at the end\begin{aligned} V&=\frac13\pi\times3^2\times8 &&\text{use perpendicular height}\\ &=24\pi\,\mathrm{cm}^3 &&\text{simplify}\\ &\approx75.4\,\mathrm{cm}^3 &&\text{round at the end} \end{aligned}

Why this step? With the radius fixed, volume contains only one factor of height.

Answer. 24π cm324\pi\,\mathrm{cm}^3, approximately 75.4 cm375.4\,\mathrm{cm}^3.

Check. Doubling the height has doubled 12π12\pi to 24π24\pi.

A central cross-section helps you identify the height, even when the cone is drawn tilted on the page.

Watch out

For the 33-44-55 cone, V=13π×32×5V=\frac13\pi\times3^2\times5 uses the slant height. Correct it to V=13π×32×4=12π cm3V=\frac13\pi\times3^2\times4=12\pi\,\mathrm{cm}^3.

Change only the perpendicular height once more for this check.

Curved and total cone area (Higher)

Surface area measures material along the cone, so it uses slant height. An open conical shade has a curved surface; a solid cone also has a circular base.

Curved surface area of a cone

Acurved=πrlA_{\text{curved}}=\pi rl

Here ll is slant height. This result is on the Higher sheet; it is absent from the Foundation sheet and is Higher content. In the formulae sheet.

For a cone with a base, add πr2\pi r^2, giving Atotal=πrl+πr2A_{\text{total}}=\pi rl+\pi r^2. This total-area result is not printed on either tier sheet: build it by counting surfaces. There is one base, not two.

If only perpendicular height is given, find the slant height first. Pythagoras gives l=r2+h2l=\sqrt{r^2+h^2}; use the positive root because it is a length. Pythagoras is not printed on either tier sheet: learn it. To find perpendicular height from slant height instead, use h=l2−r2h=\sqrt{l^2-r^2}, with l>rl>r.

Worked example 3: find the slant height

Question. A cone has radius 5 cm5\,\mathrm{cm} and perpendicular height 12 cm12\,\mathrm{cm}. Find its exact total surface area.

Plan. Find the slant height by Pythagoras, then add curved area and one base.

l=52+122=13 cmright triangleAcurved=π×5×13=65π cm2use slant heightAbase=π×52=25π cm2one circleAtotal=65π+25π=90π cm2add the surfaces\begin{aligned} l&=\sqrt{5^2+12^2}=13\,\mathrm{cm} &&\text{right triangle}\\ A_{\text{curved}}&=\pi\times5\times13=65\pi\,\mathrm{cm}^2 &&\text{use slant height}\\ A_{\text{base}}&=\pi\times5^2=25\pi\,\mathrm{cm}^2 &&\text{one circle}\\ A_{\text{total}}&=65\pi+25\pi=90\pi\,\mathrm{cm}^2 &&\text{add the surfaces} \end{aligned}

Why this step? The 12 cm12\,\mathrm{cm} height is inside the cone; 13 cm13\,\mathrm{cm} measures along its surface.

Answer. 90π cm290\pi\,\mathrm{cm}^2.

Check. 13>1213>12, as a hypotenuse should be, and the total exceeds the curved area by 25π cm225\pi\,\mathrm{cm}^2.

Change only the form of the height information: the same cone is described using its slant height. Pythagoras is no longer needed.

Worked example 4: slant height supplied

Question. A cone has radius 5 cm5\,\mathrm{cm} and slant height 13 cm13\,\mathrm{cm}. Find its exact total surface area.

Plan. Use the supplied slant height directly and include the base.

Acurved=π×5×13=65π cm2given slant heightAbase=π×52=25π cm2circular baseAtotal=65π+25π=90π cm2sum\begin{aligned} A_{\text{curved}}&=\pi\times5\times13=65\pi\,\mathrm{cm}^2 &&\text{given slant height}\\ A_{\text{base}}&=\pi\times5^2=25\pi\,\mathrm{cm}^2 &&\text{circular base}\\ A_{\text{total}}&=65\pi+25\pi=90\pi\,\mathrm{cm}^2 &&\text{sum} \end{aligned}

Why this step? The letter ll in πrl\pi rl already matches the given measurement.

Answer. 90π cm290\pi\,\mathrm{cm}^2.

Check. This describes the same cone as Worked example 3, so the total agrees.

Before entering numbers, say which surface each term measures. That prevents a correct formula being used for the wrong task.

Watch out

Atotal=π×5×12+25π=85πA_{\text{total}}=\pi\times5\times12+25\pi=85\pi uses perpendicular height in the curved-area formula. Replace 1212 by 1313 to get 90π cm290\pi\,\mathrm{cm}^2.

Change only the slant height in the next question.

Spheres and hemispheres (Higher)

A sphere is a solid whose surface points are all the same distance from its centre. That distance is its radius. It has no flat base to add.

Sphere measures

A=4πr2,V=43πr3A=4\pi r^2,\qquad V=\frac43\pi r^3

These results are supplied on the Higher sheet; neither is on the Foundation sheet, and both are Higher content. In the formulae sheet.

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