Math by First Principles

Spheres and Hemispheres

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Learning goals

  • Distinguish a sphere, its enclosed ball, and a great circle through its center
  • Explain the equal section areas in the hemisphere and cylinder comparison
  • Calculate sphere and hemisphere volumes using radius or diameter measurements
  • Describe the approximations in the informal radial pyramid argument for sphere area
  • Recover missing dimensions and choose curved or total hemisphere area for a stated boundary

The surface and the space inside

A sphere is the surface consisting of all points a fixed positive distance from a center in three-dimensional space. That distance is its radius. The ball is the sphere together with the region it encloses. For a sphere of radius 3 cm, every surface point is 3 cm from the center; points inside the ball are closer.

We conventionally say “volume of a sphere” to mean the volume of its enclosed ball. Surface area measures the sphere itself. Area uses square units; volume uses cubic units.

A plane through the center cuts the sphere in a great circle, whose radius equals the sphere’s radius. It cuts the ball in the filled disk bounded by that circle. A plane farther from the center makes a smaller circle, provided it still passes through the ball’s interior.

Central and off-center circlesThe sphere outline is circular. Its central circle and an upper smaller circle are drawn as ellipses only because of perspective. A radius R runs from O to the right end of the central circle.ORgreat circle at center
The central circle is a great circle. The upper circle is smaller. Dashed arcs are the hidden halves of circles drawn in perspective.

Worked example 1 A circle away from the center

Problem. A sphere has radius 10 cm. A plane is 6 cm from its center, measured perpendicularly. Find the section circle’s radius and the area of the filled section of the ball.

Solution. Let O be the sphere center, M the perpendicular foot in the plane, and P a point on the circle. Then MP is the section radius. The right triangle OMP has hypotenuse OP of 10 cm and leg OM of 6 cm.

The radius of a plane sectionO is below M. P is to the right of M. OM is perpendicular to the cutting plane, whose trace is the line MP. OP joins the sphere center to a surface point.OMP6 cm10 cmsection radius
This is a true vertical section through O, M, and P. The right angle is at M; the unknown section radius is MP.

Apply Pythagoras, with lengths in centimeters.

MP2=102−62MP2=64MP=8.\begin{aligned} MP^2&=10^2-6^2\\ MP^2&=64\\ MP&=8. \end{aligned}

The circle has radius 8 cm. Its filled disk has area

A=64π cm2.A=64\pi\text{ cm}^2.

Check your understanding

A sphere has radius 9 cm. Which condition makes a section circle a great circle?

Match slices to find volume

A hemisphere is half a sphere, cut by a plane through the center. For volume, we mean the half-ball it bounds, including its equator-level base disk.

The hemisphere has the same volume as a cylinder of the same radius and height with a particular cone removed. The cone’s apex is at the center of the cylinder’s bottom base. Its circular base fills the cylinder’s top. The height of both the cylinder and the cone is the hemisphere radius.

Before reading the calculation, predict the sections at the bottom and the top. At the bottom the hemisphere has its full equatorial disk, while the cone removes just its apex. At the top the hemisphere ends at a point, while the cone removes the cylinder’s whole top disk.

The cone points down to the base levelThe upper panel shows an upper semicircle with a horizontal base and an interior horizontal cut. The lower panel shows a rectangle for the cylinder and a dashed triangle for the removed cone. The triangle apex is at the bottom center and its wide base is across the top. The kept material lies outside this triangle.hemispherezRbase z = 0Rcylinder with cone removedRzRapex z = 0removedtop z = R
Central vertical views. Height z is measured upward from the bottom in both panels. The removed cone widens upward from its apex at z = 0; its top radius is R.

Hemisphere volume by equal sections#

First take radius 4 cm. At height 3 cm above the equator-level base, the hemisphere’s section radius has square 7, by Pythagoras. Its disk area is therefore 7 pi square centimeters. In the comparison cylinder, the full section area is 16 pi square centimeters. Similar triangles give the removed cone’s section radius as 3 cm, so its removed disk area is 9 pi square centimeters. The remaining annulus has area

16π−9π=7π cm2.16\pi-9\pi=7\pi\text{ cm}^2.

At any height z from 0 to 4, both section areas are the same difference of squares. Thus equality holds throughout the 4 cm height, not just at our sample level. Cavalieri gives the hemisphere the cylinder’s volume minus the cone’s volume:

VH=π(4)2(4)−13π(4)2(4)VH=128π3 cm3.\begin{aligned} V_H&=\pi(4)^2(4)-\frac13\pi(4)^2(4)\\ V_H&=\frac{128\pi}{3}\text{ cm}^3. \end{aligned}

Nothing depended on choosing 4. Now let the common radius and height be any positive R. Measure z upward from the equator-level base, with z between 0 and R. If the hemisphere section radius is ρ\rho, Pythagoras gives

ρ2+z2=R2AH(z)=π(R2−z2).\begin{aligned} \rho^2+z^2&=R^2\\ A_H(z)&=\pi(R^2-z^2). \end{aligned}

At positive heights in the cone, section radius divided by height above its apex equals R divided by R. Hence its section radius is z. At height zero the section is the apex, also of radius zero. The cylinder-minus-cone section area is

AC(z)=πR2−πz2AC(z)=π(R2−z2).\begin{aligned} A_C(z)&=\pi R^2-\pi z^2\\ A_C(z)&=\pi(R^2-z^2). \end{aligned}

The hemisphere and the remaining cylinder region lie between parallel planes R units apart. Every corresponding parallel section has equal area, including the limiting endpoint cases. Cavalieri therefore gives

VH=πR3−13πR3VH=23πR3.\begin{aligned} V_H&=\pi R^3-\frac13\pi R^3\\ V_H&=\frac23\pi R^3. \end{aligned}

Two congruent half-balls fill the ball without overlapping interiors, so

Vball=43πR3.V_{\text{ball}}=\frac43\pi R^3.

For the numerical comparison, the section areas below are exact. A few matching rows illustrate the equality; the formula for every height is what justifies Cavalieri.

Height above baseHemisphere disk areaCylinder minus cone area
0 cm16π cm216\pi\text{ cm}^2(16−0)π cm2(16-0)\pi\text{ cm}^2
2 cm12π cm212\pi\text{ cm}^2(16−4)π cm2(16-4)\pi\text{ cm}^2
3 cm7π cm27\pi\text{ cm}^2(16−9)π cm2(16-9)\pi\text{ cm}^2
4 cm0 cm20\text{ cm}^2(16−16)π cm2(16-16)\pi\text{ cm}^2
Equal areas need not have equal shapesThe two sections use the same length scale. The hemisphere section is a filled disk. The comparison section is a larger disk with a concentric smaller disk removed. Its outer and inner radii are labeled 4 cm and 3 cm. Both areas are 7 pi square centimeters.hemisphere diskarea 7π cm²remaining annulus4 cm3 cmarea 7π cm²
True horizontal sections when R is 4 cm and z is 3 cm. The hemisphere disk and the cylinder-minus-cone annulus have equal areas, despite their different shapes.

Check your understanding

In this comparison, the common radius and height are 5 cm. At height 2 cm above the base, what is the area left in the cylinder after the cone section is removed?

Use the volume formulas

Use the radius, even when a problem gives the diameter. The two volume formulas, for positive radius r, are

Vball=43πr3VH=23πr3.\begin{aligned} V_{\text{ball}}&=\frac43\pi r^3\\ V_H&=\frac23\pi r^3. \end{aligned}

Worked example 2 A diameter and two volumes

Problem. A ball has diameter 9 cm. Find its exact volume and the volume of either half when it is cut through the center.

Halve the whole widthA sphere is drawn with its central circle in perspective. The full horizontal diameter through O is marked 9 cm. No computed radius is shown.Odiameter 9 cm
The marked segment crosses the center and joins opposite surface points, so 9 cm is the diameter.

Solution. Halve the diameter first. The radius is 4.5 cm.

Vball=43π(4.5)3Vball=243π2 cm3.\begin{aligned} V_{\text{ball}}&=\frac43\pi(4.5)^3\\ V_{\text{ball}}&=\frac{243\pi}{2}\text{ cm}^3. \end{aligned}

The central cut gives equal half-balls, so each has volume

VH=243π4 cm3.V_H=\frac{243\pi}{4}\text{ cm}^3.

Check your understanding

A hemisphere encloses 18 pi cubic centimeters. What is its radius?

Surface area from small radial pieces

Many narrow pieces pointing toward the center suggest a connection between a ball’s volume and its surface area. This is an informal limiting argument, rather than an exact dissection into a finite number of pyramids.

For a radius of 3 cm, the ball volume is 36 pi cubic centimeters. Imagine dividing its surface into very small curved patches and joining each patch to the center. Replace each patch by a small flat polygon to approximate an ordinary pyramid. Its height is near 3 cm, and its flat base area is near the curved patch area. Adding the pyramid approximations gives

36π≈13(3)S,36\pi\approx\frac13(3)S,

where S is the whole sphere area in square centimeters. This suggests an area of 36 pi square centimeters.

A curved patch and its flat approximationTwo rays from O reach the ends of the curved patch. A solid arc joins their endpoints; a dashed straight segment joins the same endpoints below the arc. A dotted perpendicular runs from O to the straight segment and is labeled height near r. This sketch is schematic, not an exact finite pyramid decomposition.curved patchflat base approximates patchheight ≈ rO
One radial piece is enlarged schematically. The curved patch is replaced by a dashed flat base. The flat pyramid's perpendicular height is near the radius, rather than exactly equal to it.

For any positive radius r, finer patches make the flat bases better approximations to the curved patches and make their perpendicular heights approach r. The sums of the approximating volumes approach the ball volume. This motivates the limiting relation

V=13rS.V=\frac13rS.

It does not claim that a finite curved patch is a flat pyramid base, or that a finite flat pyramid has height exactly r. Making all approximation errors precise would require a fuller proof. At this informal level, use the established volume to obtain the exact sphere-area formula:

43πr3=13rSS=4πr2.\begin{aligned} \frac43\pi r^3&=\frac13rS\\ S&=4\pi r^2. \end{aligned}

Division by r is valid because the radius is positive. A sphere has four times the area of its great-circle disk; it does not consist of four flat disks.

Worked example 3 Recover a radius from surface area

Problem. A sphere has surface area 196 pi square centimeters. Find its radius and the circumference of a great circle.

Solution. Surface area determines the square of the radius.

4πr2=196πr2=49r=7 cm.\begin{aligned} 4\pi r^2&=196\pi\\ r^2&=49\\ r&=7\text{ cm}. \end{aligned}

A great circle has the same radius, so its circumference is

C=14π cm.C=14\pi\text{ cm}.

Check your understanding

Which statement describes the small radial pyramid argument accurately?

A hemisphere has two kinds of boundary

Cut a sphere of radius 2 cm into equal halves. Each curved half has area 8 pi square centimeters, half the sphere’s area. The new flat disk has area 4 pi square centimeters. Including it gives 12 pi square centimeters for the whole boundary of one solid hemisphere.

For any positive radius r, the curved area excludes the flat disk; the total area includes it:

Scurved=2πr2Stotal=3πr2.\begin{aligned} S_{\text{curved}}&=2\pi r^2\\ S_{\text{total}}&=3\pi r^2. \end{aligned}
Separate the curved cap from the base diskA semicircular outer contour rises above an elliptical base. The hidden half of the rim is dashed. The flat disk is tinted lightly and the radius lies in it. Total area includes both the cap and the disk.curved area 2πr²rdisk area πr²
The curved cap contributes two disk areas; the flat base contributes one more. The base is drawn as an ellipse because of perspective.

Worked example 4 Include the specified surfaces

Problem. A solid hemisphere has radius 5 cm. Find each measurement exactly.

Case A: Its curved boundary area.

Case B: Its whole boundary area, including the flat base.

Case C: Its volume.

Solution. Case A uses only half the sphere surface.

Scurved=50π cm2.S_{\text{curved}}=50\pi\text{ cm}^2.

Case B adds one base disk of area 25 pi square centimeters.

Stotal=75π cm2.S_{\text{total}}=75\pi\text{ cm}^2.

Case C is a space measurement, so cube the radius.

VH=250π3 cm3.V_H=\frac{250\pi}{3}\text{ cm}^3.

Worked example 5 Recover dimensions and an exposed area

Problem. A solid hemisphere has total boundary area 108 pi square centimeters, including its flat base. Find its radius and volume. Then find the area exposed when the flat base rests flush against an opaque table, with the whole curved surface uncovered.

Count the exposed capThe hemisphere’s semicircular profile rises above a horizontal table line. A short label beneath the line identifies the hidden flat base. There are no computed dimensions or areas in the drawing.uncovered curved capflat base hidden
The table covers the flat base. The entire curved cap remains exposed.

Solution. The given total includes three great-circle disk areas.

3πr2=108πr2=36r=6 cm.\begin{aligned} 3\pi r^2&=108\pi\\ r^2&=36\\ r&=6\text{ cm}. \end{aligned}

Use this radius in the half-ball volume formula.

VH=23π(6)3VH=144π cm3.\begin{aligned} V_H&=\frac23\pi(6)^3\\ V_H&=144\pi\text{ cm}^3. \end{aligned}

The table hides the flat disk. Only the curved cap is exposed, so

Sexposed=72π cm2.S_{\text{exposed}}=72\pi\text{ cm}^2.

The hidden base has area 36 pi square centimeters; adding it to the exposed area recovers the given total.

Check your understanding

A solid hemisphere has curved area 98 pi square centimeters. What extra area must be included to count its entire boundary?

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A bit of history (optional)

Archimedes, a Greek mathematician from Syracuse in Sicily, studied the sphere and its enclosing cylinder in a work called On the Sphere and Cylinder. He established that the sphere occupies two-thirds of the cylinder’s volume when it fits exactly inside.

The picture also became part of a story about his grave. The later writer Plutarch reported that Archimedes wanted a sphere and cylinder represented there. The Roman writer Cicero described searching for the tomb while in Sicily in 75 BCE. He said he recognized a small column bearing those shapes above the bushes, and had the overgrowth cleared so he could approach it. In his account, part of the inscription had already worn away.

MacTutor’s biography of Archimedes records the mathematical result and Cicero’s account. The tomb story comes to us through these later writings, rather than an inscription we can inspect here.

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