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MathChapter 20: Surface Areas and Volumes
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Cylinders extend a circular base through a perpendicular height, while spheres measure every point from one center. Radius must be identified before any area or volume formula is used, especially when the problem supplies diameter or circumference instead.

Right-cylinder anatomy

Right-cylinder anatomyA right cylinder has congruent parallel circular bases of radius 5 units, perpendicular height 12 units between their centers, and a curved lateral surface.
Right-cylinder anatomyA right cylinder has congruent parallel circular bases of radius 5 units, perpendicular height 12 units between their centers, and a curved lateral surface.r = 5h = 12circular baselateral surfaceFigure not drawn to scale.
Right-cylinder anatomyA right cylinder has congruent parallel circular bases of radius 5 units, perpendicular height 12 units between their centers, and a curved lateral surface.

Figure not drawn to scale.

Right cylinder
A solid with two congruent parallel circular bases whose centers are joined by a perpendicular altitude h. Each base has radius r.
Cylinder quantities
QuantityFormulaWhat is includedUnits
Lateral area\(2\pi rh\)Curved surface onlysquare units
Total area\(2\pi rh+2\pi r^2\)Curved surface and two basessquare units
Volume\(\pi r^2h\)Circular base area times heightcubic units
Cylinder lateral area
\[2\pi rh\]

The curved surface unwraps to a rectangle of width 2πr and height h.

Cylinder total surface area
\[2\pi rh+2\pi r^2\]

Add two exposed circular bases to lateral area.

Cylinder volume
\[V=\pi r^2h\]

This is B times h with circular base area B=πr².

Why lateral area is 2πrh

Cylinder and its netA cylinder of radius r and height h unwraps into a rectangle of height h and width equal to circumference 2 pi r, plus two circular bases.
Cylinder and its netA cylinder of radius r and height h unwraps into a rectangle of height h and width equal to circumference 2 pi r, plus two circular bases.height hcircumference = 2πrr
Cylinder and its netA cylinder of radius r and height h unwraps into a rectangle of height h and width equal to circumference 2 pi r, plus two circular bases.
Worked example

Cylinder area audit

A closed cylinder has radius 5 cm and height 12 cm. Find its lateral and total surface areas.

  1. Lateral surface

    LA=2π(5)(12)=120π cm².

  2. Two bases

    Two circular bases contribute 2π(5²)=50π cm².

Lateral area is 120π cm²; total surface area is 170π cm².

From circumference to volume

Circumference-first workflow

  1. Recover radius

    Use r=C/(2π).

  2. Square radius

    Calculate circular base area B=πr².

  3. Multiply by height

    Use V=πr²h and attach cubic units.

Spheres and great circles

Planes intersecting a sphereThree panels show a tangent plane meeting a sphere at one point, an offset plane producing a smaller circular section, and a plane through the center producing a great circle.
Planes intersecting a sphereThree panels show a tangent plane meeting a sphere at one point, an offset plane producing a smaller circular section, and a plane through the center producing a great circle.tangent: one pointsmaller circlegreat circle through center
Planes intersecting a sphereThree panels show a tangent plane meeting a sphere at one point, an offset plane producing a smaller circular section, and a plane through the center producing a great circle.
Sphere
The set of all points in space at one fixed distance r from a center.
Great circle
A circular cross-section whose plane passes through the sphere's center. Its radius equals the sphere's radius.
Cylinder and sphere formula comparison
SolidSurface areaVolumeCritical input
Cylinder\(2\pi rh+2\pi r^2\)\(\pi r^2h\)r and perpendicular h
Sphere\(4\pi r^2\)\(\tfrac43\pi r^3\)radius r
Sphere surface area
\[SA=4\pi r^2\]
Sphere volume
\[V=\frac{4}{3}\pi r^3\]
Worked example

Sphere from diameter

A sphere has diameter 12 m. Find its exact volume.

  1. Convert diameter

    Radius is 6 m, not 12 m.

  2. Use the volume formula

    V=(4/3)π(6³)=288π m³.

The sphere volume is 288π m³.

Hemispheres and joined solids

Cylinder with a hemisphere capA right cylinder of radius 4 units and height 10 units is capped by a hemisphere with the same radius; the circular contact surface is internal.
Cylinder with a hemisphere capA right cylinder of radius 4 units and height 10 units is capped by a hemisphere with the same radius; the circular contact surface is internal.r = 4h = 10hemisphere capcylinderFigure not drawn to scale.
Cylinder with a hemisphere capA right cylinder of radius 4 units and height 10 units is capped by a hemisphere with the same radius; the circular contact surface is internal.

Figure not drawn to scale.

Hemisphere volume
\[V_{\text{hemisphere}}=\frac12\left(\frac43\pi r^3\right)=\frac23\pi r^3\]

A hemisphere is half a sphere by volume.

Worked example

Cylinder with hemisphere cap

A cylinder of radius 4 ft and height 10 ft is capped by a hemisphere of radius 4 ft. Find total volume.

  1. Cylinder

    π(4²)(10)=160π ft³.

  2. Hemisphere

    (2/3)π(4³)=128π/3 ft³.

Total volume is 608π/3 ft³.

Sphere inscribed in a cylinder

Sphere inscribed in a cylinderA sphere of radius r touches both circular bases and the lateral surface of its cylinder, so cylinder radius is r and cylinder height is the sphere diameter 2r.
Sphere inscribed in a cylinderA sphere of radius r touches both circular bases and the lateral surface of its cylinder, so cylinder radius is r and cylinder height is the sphere diameter 2r.r = rheight = 2rsphere touches both bases and the lateral surface
Sphere inscribed in a cylinderA sphere of radius r touches both circular bases and the lateral surface of its cylinder, so cylinder radius is r and cylinder height is the sphere diameter 2r.

Exact inscribed relationship

Sphere

Radius r and volume (4/3)πr³

Cylinder

Radius r, height 2r, and volume 2πr³

Ratio

Sphere volume divided by cylinder volume is 2/3.

Hollow and drilled cylinders

Coaxial cylindrical holeA cylinder of outer radius 6 units and height 9 units has a coaxial cylindrical hole of radius 2 units drilled through its entire height.
Coaxial cylindrical holeA cylinder of outer radius 6 units and height 9 units has a coaxial cylindrical hole of radius 2 units drilled through its entire height.R = 6r = 2h = 9V = πh(R² − r²)Figure not drawn to scale.
Coaxial cylindrical holeA cylinder of outer radius 6 units and height 9 units has a coaxial cylindrical hole of radius 2 units drilled through its entire height.

Figure not drawn to scale.

Coaxial cylindrical hole
\[V_{\text{remaining}}=\pi R^2h-\pi r^2h=\pi h(R^2-r^2)\]

R is outer radius; r is the smaller hole radius; both share height h.

Cubic units and scaling

Scaling a cylinder
ChangeVolume factorReason
Radius ×a\(a^2\)Radius is squared in πr²h.
Height ×b\(b\)Height is first power.
Both changes\(a^2b\)Multiply the independent factors.

Net diagonals and water transfer

Diagonal of an unrolled tube
\[d^2=h^2+C^2\]

The cut cylinder becomes a rectangle with side lengths height h and circumference C.

Mini check

Cylinder and sphere check

A cylinder has circumference 12π cm and height 5 cm. What is its exact volume?

  1. 60π cm³
  2. 120π cm³
  3. 180π cm³
  4. 720π cm³
Show answer and explanation

Answer: 180π cm³.

r=12π/(2π)=6, then V=π(6²)(5)=180π cm³.

Key takeaways

Cylinder and sphere essentials

  • Cylinder lateral area is 2πrh; total area adds two πr² bases; volume is πr²h.
  • Sphere surface area is 4πr² and volume is (4/3)πr³.
  • A plane through a sphere's center makes a great circle.
  • Composite volume adds components; removed cylindrical holes subtract.
  • Radius changes affect cylinder volume quadratically, and cubic conversions cube the linear factor.
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Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.