How to Solve "Volume = Surface Area" Geometry Problems
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)
(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.