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How to Solve "Volume = Surface Area" Geometry Problems

Updated on March 19, 2017
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TR Smith is a product designer and former teacher who uses math in her work every day.

Some 3-dimensional geometric shapes have the curious property that their surface areas and volumes are equal. For example, the diagram above shows a cylinder of radius 3 and height 3 capped by a hemisphere of radius 3. This solid has a surface area of 45π square units and a volume of 45π cubic units. As it turns out, if you choose the scale of measurement correctly, any solid shape's volume and surface area can be made equal.

A classic problem in algebra and geometry is to find the dimensions of a particular shape so that the volume and surface area are equal. These problems may have infinitely many solutions, a finite number of valid solutions, or only one unique solution depending on how many free variables there are and whether the solutions are restricted to the integers. Here are several worked out examples to help you solve these kinds of problems.

(1) Find Dimensions of Integer Box with Equal Volume and Surface Area

A classic problem is to find the side lengths of a rectangular box so that the surface area and volume are equal, with the restriction that the side lengths be integers. To solve this problem, let's call the dimensions of the box x, y, and z. If the box's surface area equals its volume, then

2xy + 2xz + 2yz = xyz

To find integer values of x, y, and z that solve this equation, we can first simplify the expression to

xyz - 2xz - 2yz = 2xy
z = 2xy / (xy - 2x - 2y)
z = 2xy / (xy - 2x - 2y + 4 - 4)
z = 2xy / [(x - 2)(y - 2) - 4]

If we find integers x and y such that (x - 2)(y - 2) - 4 divides evenly into 2xy we will get our integer solutions. Using guess-and-check or trial-and-error we get exactly ten solutions

(x, y, z) =

(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)
(4, 5, 20), (4, 6, 12), (4, 8, 8), (5, 5, 10), (6, 6, 6)

For example, you can check that a box with a width of 3 inches, length of 10 inches, and height of 15 inches has a volume of 3*10*15 = 450 cubic inches, and a surface area of 2(3*10 + 3*15 + 10*15) = 450 square inches.

(2) Sphere Whose Surface Area Equals Its Volume

What is the radius of a sphere whose volume equals its surface area? Using the volume formula V = (4π/3)R^3 and the surface area formula S = 4πR^2, we get

(4π/3)R^3 = 4πR^2
(1/3)R = 1
R = 3

A sphere with a radius of 3 units has a volume of 36π cubic units and a surface area of 36π square units.

(3) Integer Cylinder With Equal Volume and Surface Area

Suppose a cylinder's height and diameter are integers. If the cylinder's volume and surface area are equal, what can the height and diameter be?

The volume of a cylinder with diameter D and height H is (π/4)HD^2. The surface area of the cylinder is (π/2)D^2 + πHD. Setting the two expressions equal to each other gives us

(π/4)HD^2 = (π/2)D^2 + πHD
HD/4 = D/2 + H
HD = 2D + 4H
HD - 4H = 2D
H = 2D/(D - 4)

The only integer values of D that yield integer values of H are D = 5, 6, 8, and 12. This gives the four solutions

(H, D) = (10, 5), (6, 6), (4, 8), (3, 12)

Four cylinders whose surface areas equal their volumes.
Four cylinders whose surface areas equal their volumes.

(4) Integer Cone with Equal Volume and Surface Area

Suppose a cone has integer diameter and height, and suppose its surface area and volume are equal. What can its diameter and height be?

The total surface area of a cone with a height of H and a diameter of D is (π/4)D^2 + (π/4)D*sqrt(4H^2 + D^2). The volume of the cone is (π/12)HD^2. Setting these two expressions equal givesus

(π/4)D^2 + (π/4)D*sqrt(4H^2 + D^2) = (π/12)HD^2
(1/4)D + (1/4)sqrt(4H^2 + D^2) = (1/12)HD
3D + 3*sqrt(4H^2 + D^2) = HD
3*sqrt(4H^2 + D^2)= HD - 3D
36H^2 + 9D^2 = (HD)^2 - 6HD^2 + 9D^2
36H^2 = (HD)^2 - 6HD^2
36H = HD^2 - 6D^2
D = sqrt(36H/(H-6))
H = (6D^2)/(D^2 - 36)

The only integer solution to this equation is H = 8 and D = 12. You can check that a cone with a height of 8 inches and a diameter of 12 inches has a volume of 96π cubic inches and a total surface area (base plus lateral area) of 96π square inches.

(5) Square Pyramid with Equal Volume and Surface Area

A square pyramid with a height of H and a base length of B has equal volume and surface area. What is the relationship between H and B, and are there any integer solutions?

The volume of a square pyramid whose base length is B and height H is given by the formula (1/3)HB^2

The surface area of a square pyramid is the sum of the areas of its square base and four triangular sides. The area of a triangular face is (B/4)sqrt(4H^2 + B^2). Therefore, the total surface area of the pyramid is B^2 + B*sqrt(4H^2 + B^2).

Setting the two expressions equal and simplifying gives us

(1/3)HB^2 = B^2 + B*sqrt(4H^2 + B^2)
HB = 3B + 3*sqrt(4H^2 + B^2)

Notice that this is the same as the relation between height and diameter for the previous cone problem. Using the same algebraic steps, we can see that the necessary relation between B and H is

B = sqrt(36H/(H - 6))

As before, the only integer solution to this equation is H = 8 and B = 12.


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    • profile image

      homeschooling parent 6 days ago

      my self study material has this algebra challenge problem about finding the dimensions of a triangular prism that satisfies the conditions

      *all side lengths are integers

      *surface area (sum of the areas of the five faces) equals the volume

      so i call the sides triangle x, y and z the length of the prism n. i got as far as the surface area is

      n(x+y+z) + 2*area of triangle

      and the area is

      n*area of triangle

      i don't know where to go from here and i don't know if this is too advanced for eighth grade. thanks for any help.

    • calculus-geometry profile image

      TR Smith 6 days ago from Eastern Europe

      Thanks for the challenging problem, HomeP. To help answer part of your question, the formula for the area of a triangle with side lengths {x, y, z} is

      area = (1/4)sqrt[(x^2 + y^2 + z^2)^2 - 2(x^4 + y^4 + z^4)]

      which gives you

      prism surface area =

      n(x+y+z) + (1/2)sqrt[(x^2 + y^2 + z^2)^2 - 2(x^4 + y^4 + z^4)]

      prism volume =

      (n/4)sqrt[(x^2 + y^2 + z^2)^2 - 2(x^4 + y^4 + z^4)]

      Setting these two expressions equal and doing some algebraic manipulations to find integer solutions is probably too advanced for even high school, these sorts of problems are usually solved with mathematical software.

      But, there is a simpler approach that is more suitable for middle school and high school. Start with a triangle that is know to have integer side lengths and integer area (that is, start with some promising values of x, y, and z) and then use trial and error to find a value of n that produces a triangular prism with equal volume and surface area.

      Let's try the most basic Pythagorean triangle with sides 3, 4, and 5. If these are the {x, y, z} values and n is still unknown, then we get

      surface area = 12n + 12

      volume = 6n

      Setting these equal and solving for n gives a negative value, so let's try something else. We could scale the triangle's sides by a factor of 3, that is, use x=9, y=12, and z=15. This gives us

      surface area = 36n + 108

      volume = 54n

      Now setting these equal gives us an integer solution n=6, so we have a winner. There are many more solutions to be found just by starting with any triangle that has integer side lengths, area and perimeter, and seeing if you get an integer value of n.

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      homeschooling parent 3 days ago

      thank you for the thorough explanation. i see now that i may have to adapt this to an easier problem.

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