Volume and Surface Area of a Capsule—Formula and Examples
A capsule is a solid geometric shape best described as a cylinder with two hemispheres attached at both ends. This is a common shape for tanks and of course medicinal capsules, hence the name. If you know the overall length of a capsule shape from end to end (L) and the diameter (D), you can plug these two measurements into formulas to compute the volume and surface area. Because of the physical restrictions of the shape of a capsule, L cannot be less than D. In fact, when L = D, there is no cylindrical midsection at all, and the shape is actually a sphere. Below are the formulas for capsule volume and surface area, their derivations, and several examples of calculations.
Capsule Volume Formula
For a capsule with a total length of L and a diameter of D, the volume is given by the formula
V = (π/12)(3L - D)D^2
You can see that when L = D you get the familiar formula for the volume of a sphere with diameter D.
How is the capsule volume equation derived? First if we combine the two hemispherical capsule ends into a sphere of radius D/2, then we get a volume of (4π/3)(D/2)^3 = (π/6)D^3. Next, the cylindrical piece in the middle has a radius of D/2 and a height of L - D, therefore its volume is (π/4)(L - D)D^2. If you add the two volumes together, you get
(π/6)D^3 + (π/4)(L - D)D^2 = (π/12)(3L - D)D^2
Capsule Surface Area Formula
For a capsule with an end-to-end length of L and diameter of D, the total surface area is given by the surprisingly simple formula
S = πLD
To derive this formula, we first consider the surface area of the sphere that results from removing the cylinder in the middle. This sphere has a radius of D/2, and so it has a surface area of 4π(D/2)^2 = πD^2. The lateral surface area of the cylinder in the middle is πD(L - D). Adding the two expressions gives
πD^2 + πD(L - D) = πLD
Example 1: Capsule Dimensions Given Volume
A capsule-shaped tank needs to have a volume of 7000 liters (7 cubic meters). If its length must be less than 4.3 meters, what minimum and maximum diameter can the tank must have?
Solution: This needs to be set up as an inequality. If V = 7 and V = (π/12)(3L - D)D^2, then we have
7 = (π/12)(3L - D)D^2
84/(πD^2) = 3L - D
L = D/3 + 28/(πD^2)
And since L < 4.3, we get the following inequality expression for D,
D/3 + 28/(πD^2) < 4.3
We can solve the equation above by looking at the graph of the function L = D/3 + 28/(πD^2) and finding where it dips below the line L = 4.3. These are shown in red and blue in the graph below. Discarding negative values, this tells us the correct range is somewhere between D = 1.5338 meters and D = 12.7351 meters.
But based on the geometry of a capsule, we also have the restriction that D cannot be longer than L. If we graph the line L = D and see where it intersects the curve L = D/3 + 28/(πD^2), we can see that it intersects at D = 2.3734. The line L = D is shown in green below. This means the allowable range of D is narrowed down to (1.5338, 2.3734).
Therefore, the diameter of the capsule taken must be greater than 1.5338 meters but less than 2.3734 meters.
Example 2: Surface Area of a Capsule
The length of a capsule-shaped tank is twice its diameter. If the surface area of the tank is 43.82 square feet, what are the capsule's dimensions?
Solution: Since L = 2D, and S = πLD, we have S = 2πD^2. Plugging in S = 43.82 gives us
43.82 = 2πD^2
21.91/π = D^2
D = sqrt(21.91/π)
D ≈ 2.64
Therefore, the diameter of the capsule is 2.64 feet and the length is 5.28 feet, illustrated below.
Example 3: Percent Volume Increase
A capsule has a length of 6 inches and a diameter of 4 inches. If the length increases by 12% and the diameter increases by 15%, by what percent does the volume increase?
Solution: The starting volume of the capsule is
V = (π/12)(3*6 - 4)4^2
≈ 58.6431 cubic inches
With the new length of 6*1.12 = 6.72 inches and the new diameter of 4*1.15 = 4.6 inches, the new volume is
V = (π/12)(3*6.72 - 4.6)4.6^2
≈ 86.1973 cubic inches
The percent increase is
(100%)(86.1973 - 58.6431)/58.6431 ≈ 46.97%
Example 4: Capsule with Equal Volume and Surface Area
If a capsule has integer length and diameter its volume and surface area are equal, what can be the values of L and D?
Solution: For this problem we first set V = S and find a relation between D and L. This gives us the equation
(π/12)(3L - D)D^2 = πLD
(3L - D)D^2 = 12LD
(3L - D)D = 12L
3LD - 12L = D^2
L = (D^2)/(3D - 12)
We need to find integer values of D that make the expression (2D^2)/(3D - 12) also an integer. This can be done using the guess-and-check method with a table. Not only should D and L be integers, but they should also be positive, and we also have the geometric restriction that L cannot be less than D.
L = (D^2)/(3D - 12)
Is this a solution?
Since (D^2)/(3D - 12) = (D/3)[D/(D-4)], this expression can only take on integer values when D is divisible by 3. But past D = 6, the values of L become less than D, which is physically impossible for a capsule. Therefore, the only solution is L = 6 and D = 6. This shape is actually a sphere of diameter 6, which can be thought of as the degenerate case of a capsule.
Volume and Surface Area of a Capsule Whose Ends Are Not Hemispheres
Many cylindrical tanks have rounded ends that are not hemispherical, but shaped more like a spherical cap, the three dimensional analog of a circular segment. Some may also be ellipsoidal. For such shapes, we can measure the volume and surface area with an alternative pair of formulas. Let M be the length of the cylindrical midsection of the figure, L be the total length of the capsule from end to end, and D be the diameter. (M is not necessarily equal to L - D.) Then the volume and surface area can be approximated with the formulas
V = (π/4)MD^2 + (π/6)(L - M)D^2
= (π/12)(2L + M)D^2
S = πMD + 2π[D^2 + ((L-M)/2)^2]
= (π/2)[2MD + 4D^2 + (L-M)^2]
These are based on the volume of an oblate ellipsoid and surface area of a spherical cap. In general, one does not know the exact mathematical formula that defines the curved surface of the tank ends.