Unit 5, lesson 56

Volume and Surface Area

How to tell whether a question asks for the space inside a solid or the surfaces around it, and how to find volume and surface area for prisms, cylinders, pyramids, cones and spheres.

  • Theory part 111
  • Practice part 112
  • Area, Volume and Scaling

Theory, part 111

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Practice, part 112

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A soup can raises two different questions. How much soup fits inside is a volume, measured in cubic units. How much metal it takes to make the can is a surface area, measured in square units. Many solid geometry questions on the SAT turn on reading which of the two is wanted, and whether the surface includes a top, a bottom or neither.

See also Geometry Formulas and Units in Foundations for Calculus.

Key terms

Volume

Definition The volume of a solid is the number of unit cubes needed to fill it, with fractional cubes allowed. It is measured in cubic units such as cubic centimeters. If a solid is split into pieces that do not overlap, its volume is the sum of their volumes. A rectangular box with length ℓ\ell, width ww and height hh has volume V=ℓwhV = \ell w h.

In plain words Volume is how much a solid holds, or how much space it takes up.

Example A box 55 cm long, 44 cm wide and 33 cm tall holds 5⋅4⋅3=605 \cdot 4 \cdot 3 = 60 cubic centimeters. Each layer on the bottom has 2020 unit cubes, and there are 33 layers.

Surface area

Definition The surface area of a solid is the total area of all the surfaces that bound it, measured in square units. A closed solid includes every face. An open solid, such as a box without a lid or a can without a top, leaves out the missing faces, so a question has to say which surfaces count. The SAT reference sheet lists volume formulas for common solids but no surface area formulas, so surface area is found by adding the areas of the faces and curved surfaces one at a time.

In plain words Surface area is how much wrapping paper covers the outside with no overlap. Ask what is exposed before you add anything.

Example The closed 55 by 44 by 33 box has two faces of each size, so

SA=2(5⋅4+5⋅3+4⋅3)=2(20+15+12)=94 square centimeters.\begin{aligned} SA &= 2(5 \cdot 4 + 5 \cdot 3 + 4 \cdot 3) \\[0.4em] &= 2(20 + 15 + 12) \\[0.4em] &= 94 \text{ square centimeters.} \end{aligned}

Without a lid it loses one 55 by 44 face, leaving 7474 square centimeters.

Net

Definition A net of a solid is a flat figure that can be folded along its edges, without overlapping, to form the surface of the solid. The area of a net equals the surface area of the closed solid it folds into.

In plain words Cut the solid along some edges and lay it flat. Every face shows up once, so adding the areas in the net gives the surface area.

Example A net of the 55 by 44 by 33 box is six rectangles, two of size 5×45 \times 4, two of size 5×35 \times 3 and two of size 4×34 \times 3. A net of a closed cylinder is two circles and one rectangle.

Prism

Definition A prism is a solid with two congruent polygons in parallel planes as bases, joined by faces that are parallelograms. In a right prism those lateral faces are rectangles. The height hh of a prism is the perpendicular distance between the two base planes. With base area BB,

V=Bh.V = B h.

In plain words A prism is the same polygon stacked up to a height. Its volume is the area of one layer times how high the stack is.

Example A right prism has a base that is a right triangle with legs 33 and 44 and hypotenuse 55, and its length is 1010. The base area is B=6B = 6, so V=60V = 60. For the surface area, add the two triangles and the three rectangles.

SA=2⋅6+(3+4+5)⋅10=12+120=132.\begin{aligned} SA &= 2 \cdot 6 + (3 + 4 + 5) \cdot 10 \\[0.4em] &= 12 + 120 = 132. \end{aligned}

Cylinder

Definition A right circular cylinder has two parallel circular bases of radius rr and a curved lateral surface, and its height hh is the perpendicular distance between the bases. Its volume is V=πr2hV = \pi r^2 h. The lateral surface unrolls into a rectangle with width 2πr2\pi r, the circumference, and height hh, so the lateral area is 2πrh2\pi r h and the closed surface area is

SA=2πr2+2πrh.SA = 2\pi r^2 + 2\pi r h.

In plain words A cylinder is a prism with a circle for a base. Peel off the label and you get a rectangle as long as the circumference.

Example A cylinder has radius 33 and height 1010. Its volume is π⋅9⋅10=90π\pi \cdot 9 \cdot 10 = 90\pi. The lateral area is 2π⋅3⋅10=60π2\pi \cdot 3 \cdot 10 = 60\pi, so the closed surface area is 18π+60π=78π18\pi + 60\pi = 78\pi, and with the top open it is 9π+60π=69π9\pi + 60\pi = 69\pi.

Pyramid

Definition A pyramid has a polygon as its base and triangular faces that meet at a single point, the apex. Its height hh is the perpendicular distance from the apex to the plane of the base. With base area BB,

V=13Bh.V = \frac{1}{3} B h.

Its surface area is the base area plus the areas of the triangular faces.

In plain words A pyramid holds one third as much as a prism with the same base and the same height.

Example A pyramid has a square base with side 66 and height 44, with the apex directly over the center of the base. Its volume is 13⋅36⋅4=48\frac{1}{3} \cdot 36 \cdot 4 = 48. For the surface area, each triangular face needs its own height, the slant height from the apex to the midpoint of a base edge. That is the hypotenuse of a right triangle with legs 44 and 33, so it is 55. The surface area is 36+4⋅12⋅6⋅5=9636 + 4 \cdot \frac{1}{2} \cdot 6 \cdot 5 = 96.

Radius, height and slant height

Definition In a cylinder, cone or sphere, the radius is the radius of the circular base or of the sphere. The height of a solid is always a perpendicular distance, between the bases or from the apex to the base plane. In a right circular cone the slant height ℓ\ell is the distance from the apex to a point on the edge of the base. The height, radius and slant height form a right triangle with ℓ\ell as the hypotenuse, so

ℓ2=r2+h2.\ell^2 = r^2 + h^2.

Volume formulas use the height. Lateral area formulas for cones and pyramids use the slant height. In a diagram of a solid, rely on the dimensions that are labeled or stated, not on how long a segment looks.

In plain words Height goes straight up through the middle. Slant height runs down the side. The slant height is always the longer of the two, so do not swap them.

Example A cone has slant height 1010 and radius 66. Its height is h=100−36=8h = \sqrt{100 - 36} = 8, and its volume is

V=13π⋅62⋅8=96π.V = \frac{1}{3} \pi \cdot 6^2 \cdot 8 = 96\pi.

Using the slant height 1010 in place of 88 would give 120π120\pi.

Cone

Definition A right circular cone has a circular base of radius rr and an apex directly above the center of the base at height hh. Its volume is

V=13πr2h.V = \frac{1}{3} \pi r^2 h.

Its lateral surface unrolls into a sector of a circle with radius ℓ\ell, the slant height, and arc length 2πr2\pi r, the circumference of the base. The full circle of radius ℓ\ell has circumference 2πℓ2\pi \ell and area πℓ2\pi \ell^2, so the sector is the fraction 2πr2πℓ=rℓ\frac{2\pi r}{2\pi \ell} = \frac{r}{\ell} of that circle, as in Arc Length and Sector Area. The lateral area and the closed surface area are therefore

rℓ⋅πℓ2=πrℓ,SA=πr2+πrℓ.\begin{gathered} \frac{r}{\ell} \cdot \pi \ell^2 = \pi r \ell, \\[0.4em] SA = \pi r^2 + \pi r \ell. \end{gathered}

The lateral formula uses the slant height ℓ\ell, not the height hh.

In plain words A cone is to a cylinder what a pyramid is to a prism, one third of the volume. Its curved side flattens into a fan shape.

Example A cone has radius 33 and height 44, so its slant height is 55. The volume is 13π⋅9⋅4=12π\frac{1}{3} \pi \cdot 9 \cdot 4 = 12\pi, the lateral area is π⋅3⋅5=15π\pi \cdot 3 \cdot 5 = 15\pi, and the closed surface area is 9π+15π=24π9\pi + 15\pi = 24\pi.

Sphere

Definition A sphere is the set of all points in space at a fixed distance rr, the radius, from a fixed point, the center. Its volume and surface area are

V=43πr3andSA=4πr2.V = \frac{4}{3} \pi r^3 \quad \text{and} \quad SA = 4\pi r^2.

The volume formula is on the reference sheet and the surface area formula is not, so remember it in case a question does not give it. A hemisphere is half a sphere, with half the volume. Its curved surface has area 2πr22\pi r^2, and a closed hemisphere adds the flat disk πr2\pi r^2 for a total of 3πr23\pi r^2.

In plain words Volume grows with the cube of the radius and surface area with the square, so doubling the radius makes a ball hold eight times as much while its skin is only four times as large.

Example A ball of radius 33 inches has volume 43π⋅27=36π\frac{4}{3} \pi \cdot 27 = 36\pi cubic inches and surface area 4π⋅9=36π4\pi \cdot 9 = 36\pi square inches. The numbers match only by coincidence at this radius. The units show that one is a volume and the other an area.

Check yourself

  1. A cylindrical tank with an open top has a radius of 33 meters and a height of 55 meters. How many square meters of sheet metal are needed to make the bottom and the side of the tank? (A) 30π30\pi, (B) 39π39\pi, (C) 45π45\pi, (D) 48π48\pi
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The answer is (B). The bottom has area π⋅32=9π\pi \cdot 3^2 = 9\pi and the side has area 2π⋅3⋅5=30π2\pi \cdot 3 \cdot 5 = 30\pi, which totals 39π39\pi. Choice (A) leaves out the bottom, choice (D) includes a top the tank does not have, and choice (C) is the volume π⋅9⋅5\pi \cdot 9 \cdot 5 in cubic meters.

  1. A right circular cone has a radius of 55 and a slant height of 1313. The volume of the cone is kπk\pi. What is the value of kk?
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The answer is 100100. The height is 132−52=12\sqrt{13^2 - 5^2} = 12, so

V=13π⋅25⋅12=100π.V = \frac{1}{3} \pi \cdot 25 \cdot 12 = 100\pi.

Using the slant height as the height would give 3253\frac{325}{3}, which is not the volume.

  1. A solid is made of a cylinder with radius 22 and height 66 and a hemisphere of radius 22 attached to one base. What is the volume of the solid? (A) 24π24\pi, (B) 88π3\frac{88\pi}{3}, (C) 104π3\frac{104\pi}{3}, (D) 40π40\pi
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The answer is (B). The cylinder has volume π⋅4⋅6=24π\pi \cdot 4 \cdot 6 = 24\pi and the hemisphere has volume

12⋅43π⋅23=16π3,\frac{1}{2} \cdot \frac{4}{3} \pi \cdot 2^3 = \frac{16\pi}{3},

so the total is 72π3+16π3=88π3\frac{72\pi}{3} + \frac{16\pi}{3} = \frac{88\pi}{3}. Choice (A) ignores the hemisphere, choice (C) adds a whole sphere, and choice (D) adds the sphere’s surface area 16π16\pi as if it were a volume.

How this lesson connects