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Composite and Hollow Solids: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The stepped casting

A right stepped prism has constant depth 4 cm. Its lower rectangular block is 9 cm wide and 2 cm high. An upper block covers the leftmost 4 cm of the width and adds 3 cm of height. The interiors of the two blocks do not overlap. Find the total volume.

The dashed line separates two nonoverlapping blocks; the section is shown directly.

Text description of this figure

Schematic side section of a right stepped prism. Full width 9 cm, lower height 2 cm, upper width 4 cm and added height 3 cm. Constant perpendicular depth 4 cm. All corners in the section are right angles.
1. Hint 1
  Separate the two blocks without counting any region twice.
2. Hint 2
  The upper block uses the added height, not the full height from the bottom.
Answer

$120\text{ cm}^3$ .

Full solution

The lower block uses all 9 cm of width and has volume 72 cubic centimeters.

The upper block has width 4 cm, depth 4 cm, and added height 3 cm.
Its volume is 48 cubic centimeters.

Add the two nonoverlapping volumes.
$V=72+48$
The total volume is 120 cubic centimeters.

Answer

$120\text{ cm}^3$ .

Key idea

A nonoverlapping split makes volume addition count every part once.
Problem 2 One rounded end

A solid consists of a right circular cylinder of radius 3 cm and length 7 cm, with one solid hemisphere of the same radius attached flush to one end. The other end is a flat exposed disk. The solid has no other contacts.

Find its total volume.

Find its exterior area, including the flat disk.

A semicircle in this axial section represents a hemisphere in space; each dashed diameter marks an internal join.

Text description of this figure

True axial section of a cylinder with one hemispherical end and one flat end. The common radius is 3 cm. The cylinder length is 7 cm. Dashed diameters mark internal joins, not exposed disks.
1. Hint 1
  List the two volume pieces and the three exposed surface types.
2. Hint 2
  The flat disks at the join are internal. Only the opposite cylinder base remains exposed.
Answer

Volume $81\pi\text{ cm}^3$ ; exterior area $69\pi\text{ cm}^2$ .

Full solution

The cylinder volume is $63\pi$ cubic centimeters.
The hemisphere volume is $18\pi$ cubic centimeters.
Their interiors do not overlap.
$V=81\pi\text{ cm}^3$

The cylinder curved side contributes $42\pi$ square centimeters.
The hemisphere curved side contributes $18\pi$ square centimeters.
The one exposed flat base contributes $9\pi$ square centimeters.

Add just these exposed surfaces.
$S=42\pi+18\pi+9\pi$
Thus the exterior area is $69\pi$ square centimeters.

Answer

Volume $81\pi\text{ cm}^3$ ; exterior area $69\pi\text{ cm}^2$ .

Key idea

Joining disks disappear from the exterior even though the component volumes add.
Problem 3 The open sleeve

A right circular pipe has outer radius 6 cm, inner radius 3 cm, and length 4 cm. Both ends are open. Find its material volume and the total area of every exposed material surface, including the inner wall and both annular ends. Give exact answers.

The ring is material; the middle of the axial section is empty. The dashed inner circle distinguishes the bore.

Text description of this figure

A right circular pipe open at both ends. The true end section has outer radius 6 cm and inner radius 3 cm; the axial section shows an empty bore through the full length 4 cm. Radii are measured from the common center. No computed measurements are shown.
1. Hint 1
  The void is a coaxial cylinder of the full pipe length.
2. Hint 2
  For area, count both curved walls and both end rings.
Answer

Material volume $108\pi\text{ cm}^3$ ; exposed area $126\pi\text{ cm}^2$ .

Full solution

Subtract the bore volume from the outer cylinder volume.
$V=\pi(6^2-3^2)(4)$
The material volume is $108\pi$ cubic centimeters.

The outer curved area is $48\pi$ square centimeters.
The inner curved area is $24\pi$ square centimeters.
Each annulus has area $27\pi$ square centimeters.

Count each surface once.
$S=48\pi+24\pi+2(27\pi)$
The exposed area is $126\pi$ square centimeters.

Answer

Material volume $108\pi\text{ cm}^3$ ; exposed area $126\pi\text{ cm}^2$ .

Key idea

An open bore leaves no end disk, but it does expose a wall and annular rims.
Problem 4 A drilled panel

A rectangular block is 14 cm by 10 cm by 4 cm. A cylindrical hole of radius 3 cm passes straight through the 4 cm thickness, perpendicular to and strictly inside the 14 cm by 10 cm faces.

Find the retained material volume.

Find all exposed surface area, including the inside of the hole and the underside.

The dashed circle bounds the removed region. Depth is perpendicular to the page, not inferred from the drawing.

Text description of this figure

True section perpendicular to a cylindrical through hole in a rectangular block. The outer rectangular face is 14 cm by 10 cm. The centered circular opening has radius 3 cm. The block thickness and bore length is 4 cm, perpendicular to this face. The hole fits strictly inside the face.
1. Hint 1
  Track removed volume separately from changes to the boundary.
2. Hint 2
  For area, remove two circular patches and add the cylindrical bore wall.
Answer

Volume $(560-36\pi)\text{ cm}^3$ ; area $(472+6\pi)\text{ cm}^2$ .

Full solution

The whole block volume is 560 cubic centimeters.
The through cylinder removes $36\pi$ cubic centimeters.
Thus
$V=(560-36\pi)\text{ cm}^3$

The original box area includes three opposite face pairs.
$S_0=2(140+56+40)$
Its total is 472 square centimeters.

The two openings remove $18\pi$ square centimeters.
The inner wall adds $24\pi$ square centimeters.
$S=472-18\pi+24\pi$
The total exposed area is $(472+6\pi)$ square centimeters.

Answer

Volume $(560-36\pi)\text{ cm}^3$ ; area $(472+6\pi)\text{ cm}^2$ .

Key idea

A drilled hole subtracts material while creating a new inside boundary.
Problem 5 The full length

A solid capsule consists of a right cylinder with a hemisphere attached at each end. All three parts have radius 4 cm. The full tip-to-tip length is 22 cm.

Find the cylinder length.

Find the whole volume and exterior area exactly. Neither joining disk is exposed.

A semicircle in this axial section represents a hemisphere in space; each dashed diameter marks an internal join. The dimension line measures the full tip-to-tip length.

Text description of this figure

True axial section of a cylinder with two hemispherical ends. The common radius is 4 cm. The full tip-to-tip length is 22 cm. Dashed diameters mark internal joins, not exposed disks.
1. Hint 1
  Each end extends one radius beyond the cylinder.
2. Hint 2
  Combine the two hemisphere volumes as one ball, and their curved areas as one sphere.
Answer

Cylinder length 14 cm; volume $\frac{928\pi}{3}\text{ cm}^3$ ; exterior area $176\pi\text{ cm}^2$ .

Full solution

Subtract the two end extensions.
$h=22-2(4)$
The cylinder length is 14 cm.

The cylinder volume is $224\pi$ cubic centimeters, and the two hemispheres together contribute $\tfrac{256\pi}{3}$ cubic centimeters.
$V=224\pi+\frac{256\pi}{3}$
Thus the whole volume is $\tfrac{928\pi}{3}$ cubic centimeters.

The cylinder side and the two curved ends are the entire exterior.
$S=2\pi(4)(14)+4\pi(4)^2$
The exterior area is $176\pi$ square centimeters.

Answer

Cylinder length 14 cm; volume $\frac{928\pi}{3}\text{ cm}^3$ ; exterior area $176\pi\text{ cm}^2$ .

Key idea

Full capsule length includes the two radius-length end extensions.
Problem 6 A straight corner brace

A rectangular box has perpendicular edge lengths 8 cm, 9 cm, and 12 cm. A straight brace joins opposite vertices through its interior.

Find the brace length exactly. Explain the two right triangles that justify your calculation.

The dashed segments locate a base diagonal and a space diagonal. No diagonal length is evaluated.

Text description of this figure

A rectangular box with perpendicular edge lengths 8, 9, and 12 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. No diagonal length is evaluated.
1. Hint 1
  A base diagonal connects the first pair of opposite base vertices.
2. Hint 2
  That base diagonal is perpendicular to a vertical edge through its endpoint.
Answer

17 cm. The two right triangles are a base-face triangle and the triangle containing its diagonal and a vertical edge.

Full solution

Name successive base vertices A, B, C, D and let G be directly above C.
The first right triangle is ABC in the base.
$AC^2=8^2+9^2$
Thus the base diagonal squared is 145 square centimeters.

Edge CG is perpendicular to the base plane and therefore to AC.
The second right triangle is ACG.
$AG^2=145+12^2$
Its squared hypotenuse is 289 square centimeters, so AG is 17 cm.

The brace is the space diagonal AG.
Both right angles come from the rectangular box, rather than the appearance of its perspective sketch.

Each equation comes from one of these true right triangles.

Answer

17 cm. The two right triangles are a base-face triangle and the triangle containing its diagonal and a vertical edge.

Key idea

A space diagonal is the hypotenuse of a second right triangle.
Problem 7 The hemispherical shell

A hollow hemispherical bowl is formed by removing a concentric radius-3 cm hemisphere from a solid radius-5 cm hemisphere. The two flat circular mouths lie in the same plane. The inner region is empty and open at the mouth.

Find the material volume.

Find the total exposed area, including the outer curved surface, inner curved surface, and flat annular rim.

The true section shows the two radii and wall profile. The open mouth is not a material disk.

Text description of this figure

True axial section through a hollow hemispherical bowl. Outer radius 5 cm and inner radius 3 cm share center O on the mouth plane. The semicircular boundaries represent curved walls; the two straight segments in the mouth plane represent the annular rim. The central mouth is open.
1. Hint 1
  Subtract half-ball volumes for the material.
2. Hint 2
  The inner curved wall is exposed too; area is not found by subtracting the two curved areas.
Answer

Material volume $\frac{196\pi}{3}\text{ cm}^3$ ; exposed area $84\pi\text{ cm}^2$ .

Full solution

Subtract the concentric empty hemisphere.
$V=\frac23\pi(5^3-3^3)$
The material volume is $\tfrac{196\pi}{3}$ cubic centimeters.

The outer curved surface has area $50\pi$ square centimeters.
The inner curved surface has area $18\pi$ square centimeters.
The flat rim annulus has area $16\pi$ square centimeters.

All three are exposed.
$S=50\pi+18\pi+16\pi$
The total exposed area is $84\pi$ square centimeters.

Answer

Material volume $\frac{196\pi}{3}\text{ cm}^3$ ; exposed area $84\pi\text{ cm}^2$ .

Key idea

Removed volume is subtracted, but an exposed inner surface is included positively.
Problem 8 A well that stops inside

A solid rectangular block is 9 cm by 7 cm by 6 cm. A rectangular well with a 3 cm by 2 cm horizontal opening extends 4 cm down from its top. The opening is strictly inside the top face and the well does not reach the bottom.

Find the remaining material volume.

Find the total exposed area, including the well walls, well floor, and the block underside. Explain how the opening and well floor affect the area count.

The blind well has a floor. Its walls are new boundary surfaces.

Text description of this figure

Top and vertical section of a rectangular block 9 cm by 7 cm by 6 cm, with an interior rectangular blind well 3 cm by 2 cm and depth 4 cm. The well does not reach the bottom and does not touch any side. Both views are schematic.
1. Hint 1
  A blind well has an exposed floor inside it.
2. Hint 2
  The removed top patch and new floor have equal areas. The well walls still add area.
Answer

Volume $354\text{ cm}^3$ ; exposed area $358\text{ cm}^2$ . The opening and well floor areas cancel in the boundary count.

Full solution

The full block occupies 378 cubic centimeters.
The well removes 24 cubic centimeters.
$V=378-24$
Thus 354 cubic centimeters remain.

The original block area is 318 square centimeters.
The opening removes 6 square centimeters, while the well floor adds the same 6 square centimeters.

The four well walls total 40 square centimeters.
$S=318-6+6+2(3+2)(4)$
The exposed area is 358 square centimeters.

The cavity depth is 4 cm, shorter than the 6 cm block height, so the retained bottom is 2 cm thick and the stated blind well is possible.

Answer

Volume $354\text{ cm}^3$ ; exposed area $358\text{ cm}^2$ . The opening and well floor areas cancel in the boundary count.

Key idea

A blind cavity exposes a floor as well as side walls.
Problem 9 The floor-mounted display

An ideal solid display consists of a rectangular plinth 6 cm wide, 5 cm deep, and 3 cm high, with a solid hemisphere of radius 2 cm centered on top. Its entire flat disk is attached to the plinth, with no gap or overlapping interiors. The plinth bottom is flush against a floor.

Find the material volume.

A coating covers the exterior above the floor, including the uncovered part of the plinth top and the curved hemisphere. It cannot reach the bottom or the internal join. Find the exact area coated. Ignore seams and coating thickness.

The dashed diameter is an internal join. The lower dashed line locates the floor that hides the bottom.

Text description of this figure

True central vertical section of a solid rectangular plinth with base width 6 cm and height 3 cm. A hemisphere of radius 2 cm is attached by its entire disk at the center of the top. The plinth depth is 5 cm perpendicular to the section. The bottom rectangle is flush against the floor. A dashed join marks the internal hemisphere base.
1. Hint 1
  Decide which top patch becomes internal and which bottom patch the floor hides.
2. Hint 2
  Start with the box total, remove its bottom and the covered circular top patch, then add the hemisphere curve.
Answer

Volume $(90+\frac{16\pi}{3})\text{ cm}^3$ ; coating area $(96+4\pi)\text{ cm}^2$ .

Full solution

The box volume is 90 cubic centimeters.
The hemisphere adds $\tfrac{16\pi}{3}$ cubic centimeters.
Thus
$V=\left(90+\frac{16\pi}{3}\right)\text{ cm}^3$

The box total area is 126 square centimeters.
Its 30-square-centimeter bottom is hidden by the floor.

The hemisphere hides a top disk of area $4\pi$ square centimeters, then contributes curved area $8\pi$ square centimeters.
$S=126-30-4\pi+8\pi$
The coating area is $(96+4\pi)$ square centimeters.

The hemisphere diameter is 4 cm, below both the 6 cm width and 5 cm depth, so its whole joining disk fits on the top as stated.

Answer

Volume $(90+\frac{16\pi}{3})\text{ cm}^3$ ; coating area $(96+4\pi)\text{ cm}^2$ .

Key idea

An area specification determines which otherwise exterior surfaces are counted.
Problem 10 Recover the pipe dimensions

An ideal right circular pipe is 7 cm long. Its radial wall thickness is 2 cm, constant around the bore. Its material volume is $224\pi$ cubic centimeters. Both ends are open, the bore is coaxial with the outside, and all dimensions are positive.

Find the outer radius and inner radius.

Find the empty bore volume.

Find the area of every exposed material surface, including both curved walls and both end annuli. Verify the recovered radii against the thickness and material volume.

The ring is material; the middle of the axial section is empty. The dashed inner circle distinguishes the bore.

Text description of this figure

A right circular pipe open at both ends. The true end section has outer radius R and inner radius r; the axial section shows an empty bore through the full length 7 cm. Radii are measured from the common center. No computed measurements are shown.
1. Hint 1
  The thickness gives the difference of the radii. The material volume gives the difference of their squares.
2. Hint 2
  Factor the squared difference as a product of radius difference and radius sum.
3. Hint 3
  Once both radii are known, list two walls and two annuli before finding area.
Answer

Outer radius 9 cm; inner radius 7 cm; bore volume $343\pi\text{ cm}^3$ ; exposed area $288\pi\text{ cm}^2$ .

Full solution

Let R and r be the outer and inner radii in centimeters.
Divide the material volume by the 7 cm length and by pi.
$R^2-r^2=32$

The radial thickness supplies the difference.
$R-r=2$
Factor the squared difference.
$2(R+r)=32$
Thus the radius sum is 16 cm.

Combining the sum 16 cm and difference 2 cm gives outer radius 9 cm and inner radius 7 cm.
Both are positive, with the inner smaller than the outer.

The empty bore uses the inner radius.
$V_{\text{bore}}=\pi(7)^2(7)$
Its volume is $343\pi$ cubic centimeters.

The outside wall contributes $126\pi$ square centimeters; the inside wall contributes $98\pi$ square centimeters.
Each annulus has area $32\pi$ square centimeters.
$S=126\pi+98\pi+2(32\pi)$
The exposed area is $288\pi$ square centimeters.

The radii differ by 2 cm as required.
Their material-volume check is
$\pi(9^2-7^2)(7)=224\pi\text{ cm}^3$
Both supplied records agree.

Answer

Outer radius 9 cm; inner radius 7 cm; bore volume $343\pi\text{ cm}^3$ ; exposed area $288\pi\text{ cm}^2$ .

Key idea

Thickness and material volume together determine the two radii before boundary area is counted.

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