Math by First Principles

Composite and Hollow Solids

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Learning goals

  • Calculate composite volumes by separating added pieces from removed regions
  • Identify exposed surfaces while excluding internal joins and including cavity walls
  • Measure hollow cylinders and solids with hemispherical ends using specified boundaries
  • Recover missing dimensions from linked volume, surface, and thickness information
  • Find rectangular box diagonals by applying the Pythagorean theorem twice

Add pieces, subtract empty regions

A composite solid is built from simpler solids. We can use their known volume formulas once we identify pieces with nonoverlapping interiors.

For example, a cylinder of radius 2 cm and height 5 cm has volume 20π20\pi cubic centimeters. Add a solid hemisphere of the same radius, touching only at their circular bases. The hemisphere contributes 16π3\tfrac{16\pi}{3} cubic centimeters. Their shared disk has no thickness and contributes no volume twice.

Exploded cylinder and hemisphere of common radius 2 cm, followed below by the assembled solid and its true axial section. Cylinder height 5 cm. Separated circular joining disks are dashed; they become internal after assembly. The lower flat cylinder base remains exposed.Exploded cylinder and hemisphere of common radius 2 cm, followed below by the assembled solid and its true axial section. Cylinder height 5 cm. Separated circular joining disks are dashed; they become internal after assembly. The lower flat cylinder base remains exposed.Exploded parts5 cmr 2 cmAssembledSchematic axial sectionInternal join
Exploded, assembled, and section views show the same cylinder and hemispherical end. Dashed joining disks become internal.
V=20π+16π3V=76π3 cm3.\begin{aligned} V&=20\pi+\frac{16\pi}{3}\\ V&=\frac{76\pi}{3}\text{ cm}^3. \end{aligned}

In the true axial section, a plane through the cylinder axis cuts a rectangle and a semicircular region. That flat section helps locate the parts; its area is not the solid’s volume.

When pieces meet only at their boundaries, add their volumes. When material is removed, subtract the removed volume from the original volume. If two chosen pieces overlap in a region with positive volume, adding both counts that region twice; choose a nonoverlapping split or subtract the overlap once.

Worked example 1 A stepped block

Problem. A right stepped prism has constant depth 5 cm. Its lower rectangular block is 8 cm wide and 3 cm high. An upper block covers the leftmost 3 cm of that width and adds 4 cm of height. The blocks have no gap or overlap. Find the total volume.

Schematic side section of a right stepped prism. Full width 8 cm, lower height 3 cm, upper width 3 cm and added height 4 cm. Constant perpendicular depth 5 cm. All corners in the section are right angles.Schematic side section of a right stepped prism. Full width 8 cm, lower height 3 cm, upper width 3 cm and added height 4 cm. Constant perpendicular depth 5 cm. All corners in the section are right angles.8 cm3 cm34Depth 5 cm
The dashed line separates two nonoverlapping blocks; the section is shown directly.

Solution. The lower block uses the whole width. The upper block uses only the 3 cm width and the added height.

Vlower=8(5)(3)Vlower=120 cm3Vupper=3(5)(4)Vupper=60 cm3V=180 cm3.\begin{aligned} V_{\text{lower}}&=8(5)(3)\\ V_{\text{lower}}&=120\text{ cm}^3\\ V_{\text{upper}}&=3(5)(4)\\ V_{\text{upper}}&=60\text{ cm}^3\\ V&=180\text{ cm}^3. \end{aligned}

The upper block’s height is 4 cm. Using the full 7 cm height for that block would include part of the lower block a second time.

Check your understanding

A composite solid has two pieces with nonoverlapping interiors. Their volumes are 60 and 12 cubic centimeters. What is the whole volume?

Follow the exposed boundary

An exposed surface is part of the boundary included in the stated measurement. A surface can be hidden behind the object in a drawing and still be exposed. Conversely, a clearly drawn joining face becomes internal when the pieces are attached.

Suppose separate pieces have total surface areas 50 and 38 square centimeters and meet over one matching patch of area 9 square centimeters. Both copies of the patch are inside the assembly.

Sexposed=50+38−2(9)Sexposed=70 cm2.\begin{aligned} S_{\text{exposed}}&=50+38-2(9)\\ S_{\text{exposed}}&=70\text{ cm}^2. \end{aligned}

Thus, when two solids have disjoint interiors and exactly one shared patch, subtract twice the patch area from their separate total surface areas. Each original total counted a copy. If a support hides another surface, exclude that surface separately when the task asks for exposed area.

Removing material can also create a new boundary. A drilled hole removes disks from the original outside but exposes a curved wall inside the hole.

Worked example 2 A hole through a block

Problem. A rectangular block is 10 cm by 8 cm by 6 cm. A circular cylindrical hole of radius 2 cm is drilled straight through the 6 cm thickness, perpendicular to the 10 cm by 8 cm faces. The hole lies strictly inside those faces. Find the remaining material volume and the total exposed area, including the hole wall and the underside.

Schematic section perpendicular to a cylindrical through hole in a rectangular block. The outer rectangular face is 10 cm by 8 cm. The centered circular opening has radius 2 cm. The block thickness and bore length is 6 cm, perpendicular to this face. The hole fits strictly inside the face.Schematic section perpendicular to a cylindrical through hole in a rectangular block. The outer rectangular face is 10 cm by 8 cm. The centered circular opening has radius 2 cm. The block thickness and bore length is 6 cm, perpendicular to this face. The hole fits strictly inside the face.2 cm10 cm8Thickness 6 cm
The dashed circle bounds the removed region. Depth is perpendicular to the page, not inferred from the drawing.

Solution. The cylinder removed has the full 6 cm length.

V=10(8)(6)−π(2)2(6)V=(480−24π) cm3.\begin{aligned} V&=10(8)(6)-\pi(2)^2(6)\\ V&=(480-24\pi)\text{ cm}^3. \end{aligned}

The original box has six faces. Compute their area before changing its boundary.

Sbox=2(80+60+48).S_{\text{box}}=2(80+60+48).

The two circular openings remove two disks. The inner cylindrical wall adds a new exposed area.

S=376−2π(2)2+2π(2)(6)S=(376+16π) cm2.\begin{aligned} S&=376-2\pi(2)^2+2\pi(2)(6)\\ S&=(376+16\pi)\text{ cm}^2. \end{aligned}

The remaining volume decreased, while the total exposed area increased. Area follows the actual boundary, rather than the amount of material alone.

Check your understanding

Two solids with disjoint interiors have separate total surface areas 82 and 58 square centimeters. They are glued over one matching patch of area 6 square centimeters, with no other contact. What is the exterior area?

Thick pipes have inner surfaces and end rings

An open hollow cylinder has a bore passing all the way through it. Take an outer cylinder of radius 5 cm and length 8 cm, then remove a coaxial cylinder of radius 3 cm and the same length. Coaxial means that the cylinders share one axis.

A right circular pipe open at both ends. The true end section has outer radius 5 cm and inner radius 3 cm; the axial section shows an empty bore through the full length 8 cm. Radii are measured from the common center. No computed measurements are shown.A right circular pipe open at both ends. The true end section has outer radius 5 cm and inner radius 3 cm; the axial section shows an empty bore through the full length 8 cm. Radii are measured from the common center. No computed measurements are shown.R 5 cmr 3 cmAxial sectionLength 8 cm
The ring is material; the middle of the axial section is empty. The dashed inner circle distinguishes the bore.

Worked example 3 Measure the pipe material

Problem. A right circular pipe has outer radius 5 cm, inner radius 3 cm, and length 8 cm. Both ends are open. Find its material volume and the area of all its exposed surfaces, including the bore wall and both end rings.

Solution. The end section is an annulus, the ring between two concentric circles. Its area is the outer disk area minus the inner disk area.

V=π(52−32)(8)V=128π cm3.\begin{aligned} V&=\pi(5^2-3^2)(8)\\ V&=128\pi\text{ cm}^3. \end{aligned}

The outside curved wall has area 80π80\pi square centimeters. The inside curved wall has area 48π48\pi square centimeters. Each annular end has area 16π16\pi square centimeters.

S=80π+48π+2(16π)S=160π cm2.\begin{aligned} S&=80\pi+48\pi+2(16\pi)\\ S&=160\pi\text{ cm}^2. \end{aligned}

An open end has no disk across the bore. Its material still has a flat annular end surface.

For a right circular pipe of positive length hh, outer radius RR, and inner radius rr, with the inner radius positive and smaller than the outer radius, the same subtraction gives

Vmaterial=π(R2−r2)h.V_{\text{material}}=\pi(R^2-r^2)h.

When both ends are open and all four surface types are included,

S=2πRh+2πrh+2π(R2−r2).\begin{aligned} S&=2\pi Rh+2\pi rh\\ &\quad+2\pi(R^2-r^2). \end{aligned}

The wall thickness is R−rR-r, not the difference of the diameters. The empty bore volume uses the inner radius. A sealed end, a covering, or an attached part changes the boundary list, so this surface formula must then be reconsidered.

Check your understanding

A right circular pipe has outer radius 4 cm, inner radius 1 cm, and length 6 cm. Both ends are open. What is its material volume?

Hemispherical ends and missing lengths

A hemispherical end contributes one radius to the length along the axis. For example, a cylinder 7 cm long with two radius-2 cm hemispherical ends has total length 11 cm. The two hemispheres together have the volume and curved area of a whole ball and sphere. Their flat disks are internal joins.

Worked example 4 A solid capsule

Problem. A solid consists of a right circular cylinder with a solid hemisphere attached at each end. All three parts have radius 3 cm, and the full tip-to-tip length is 16 cm. Find the cylinder length, total volume, and exterior surface area. There are no exposed flat disks.

True axial section of a cylinder with two hemispherical ends. The common radius is 3 cm. The full tip-to-tip length is 16 cm. Dashed diameters mark internal joins, not exposed disks.True axial section of a cylinder with two hemispherical ends. The common radius is 3 cm. The full tip-to-tip length is 16 cm. Dashed diameters mark internal joins, not exposed disks.r 3 cm16 cmTip-to-tip length (schematic)
A semicircle in this axial section represents a hemisphere in space; each dashed diameter marks an internal join.

Solution. Each hemisphere extends 3 cm beyond a cylinder end. Remove both extensions from the full length.

h=16−2(3).h=16-2(3).

The cylinder length is 10 cm. The two hemispheres have the volume of one radius-3 cm ball.

V=π(3)2(10)+43π(3)3V=126π cm3.\begin{aligned} V&=\pi(3)^2(10)+\frac43\pi(3)^3\\ V&=126\pi\text{ cm}^3. \end{aligned}

The exterior consists of the cylinder’s curved side and two curved hemispheres.

S=2π(3)(10)+4π(3)2S=96π cm2.\begin{aligned} S&=2\pi(3)(10)+4\pi(3)^2\\ S&=96\pi\text{ cm}^2. \end{aligned}

Adding the separate closed total areas would include all four copies of the joining disks. None belongs to this exterior.

You can also work backward from volume. For a cylinder with one solid hemispherical end, subtract the hemisphere volume before dividing the remainder by the cylinder base area. Check that the recovered cylinder length is positive and returns the given whole volume. Throughout this lesson, an end called a cap is a hemisphere; no formula for a general spherical cap is needed.

Check your understanding

A cylinder has a radius-2 cm hemisphere attached at each end. Its full tip-to-tip length is 14 cm. How long is the cylinder alone?

A diagonal through a rectangular box

The straight segment between opposite box vertices is a space diagonal. Its projection onto a rectangular base is that base’s diagonal. These two diagonals belong to different right triangles.

Worked example 5 Two right triangles sharing a diagonal

Problem. A rectangular box has perpendicular edge lengths 6 cm, 8 cm, and 24 cm. Find the exact length of a space diagonal. Show the two right triangles used.

A rectangular box with perpendicular edge lengths 6, 8, and 24 cm. Bottom A B C D and upper vertices E F G H are in matching order. AC is a base diagonal; AG is the space diagonal. Hidden back edges are dashed. The right triangles ABC and ACG are highlighted; CG is perpendicular to the base plane.A rectangular box with perpendicular edge lengths 6, 8, and 24 cm. Bottom A B C D and upper vertices E F G H are in matching order. AC is a base diagonal; AG is the space diagonal. Hidden back edges are dashed. The right triangles ABC and ACG are highlighted; CG is perpendicular to the base plane.ABCG6 cm8 cm24
AC lies in the base plane. CG is perpendicular to that plane, so triangle ACG is right at C.

Solution. First find the base diagonal ACAC in the 6 cm by 8 cm rectangle.

AC2=62+82AC=10 cm.\begin{aligned} AC^2&=6^2+8^2\\ AC&=10\text{ cm}. \end{aligned}

The vertical edge CGCG is perpendicular to the entire base plane, so it is perpendicular to ACAC. In right triangle ACGACG, the space diagonal AGAG is the hypotenuse.

AG2=102+242AG=26 cm.\begin{aligned} AG^2&=10^2+24^2\\ AG&=26\text{ cm}. \end{aligned}

For any rectangular box with positive perpendicular edge lengths aa, bb, and cc, the first right triangle gives the base diagonal squared as a2+b2a^2+b^2. The second adds c2c^2. Nothing depended on the particular lengths, so

d=a2+b2+c2.d=\sqrt{a^2+b^2+c^2}.

This shortcut depends on the rectangular box’s perpendicular edges. A perspective sketch alone does not establish those right angles or supply a missing depth.

Check your understanding

A rectangular box has perpendicular edge lengths 2 cm, 3 cm, and 6 cm. What is its space diagonal length?

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More resources (optional)

Other explanations of this lesson, if you want a second take.

A bit of history (optional)

When Filippo Brunelleschi’s dome rose over Florence’s cathedral between 1420 and 1436, its outside and inside were different structures. The cathedral’s official account describes two masonry shells, joined by ribs, with a space between them.

The dome was built without the great wooden supporting framework normally used while constructing a vault. Its octagonal structure was made in stone and brick; the bricks followed a herringbone pattern. The builders also needed ways to carry material to the growing worksite above the city.

That space between the shells did not simply vanish from the finished building. It contains the stairway toward the lantern at the top. A visitor climbing there passes through a part of the building that someone looking only at the outside would never see.

Opera di Santa Maria del Fiore describes the dome and its construction.

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