{
  "id": "surface-area-to-volume-ratio",
  "version": "feda43913118",
  "status": "published",
  "name": "Surface Area to Volume Ratio Calculator",
  "question": "Find the surface area to volume ratio",
  "summary": "Finds the surface area, the volume and the surface area to volume ratio (SA:V) of a cube, a sphere, a cylinder or a rectangular box, with the steps.",
  "category": "math",
  "subcategory": "geometry",
  "url": "https://www.acalculator.org/math/surface-area-to-volume-ratio-calculator",
  "markdown": "https://www.acalculator.org/math/surface-area-to-volume-ratio-calculator.md",
  "kind": "function",
  "method": "Cube: SA = 6a², V = a³, SA:V = 6 ÷ a. Sphere: SA = 4πr², V = 4/3 πr³, SA:V = 3 ÷ r. Cylinder: SA = 2πr² + 2πrh, V = πr²h, SA:V = 2(r + h) ÷ (rh). Box: SA = 2(lw + lh + wh), V = lwh.",
  "assumptions": [
    "All lengths are in the same unit; the ratio is per that unit of length, so it changes when the unit changes.",
    "Typed lengths are read as the exact decimals you typed."
  ],
  "inputs": {
    "$schema": "https://json-schema.org/draft/2020-12/schema",
    "type": "object",
    "properties": {
      "shape": {
        "title": "Shape",
        "description": "The shape: a cube, a sphere, a cylinder or a rectangular box.",
        "type": "string",
        "enum": [
          "cube",
          "sphere",
          "cylinder",
          "box"
        ]
      },
      "a": {
        "title": "Side (a)",
        "description": "The length of one edge of the cube.",
        "type": "number",
        "minimum": 1e-9,
        "maximum": 1000000000
      },
      "r": {
        "title": "Radius (r)",
        "description": "The radius: half the diameter.",
        "type": "number",
        "minimum": 1e-9,
        "maximum": 1000000000
      },
      "l": {
        "title": "Length (l)",
        "description": "The length of the box.",
        "type": "number",
        "minimum": 1e-9,
        "maximum": 1000000000
      },
      "w": {
        "title": "Width (w)",
        "description": "The width of the box.",
        "type": "number",
        "minimum": 1e-9,
        "maximum": 1000000000
      },
      "h": {
        "title": "Height (h)",
        "description": "The height of the cylinder or the box.",
        "type": "number",
        "minimum": 1e-9,
        "maximum": 1000000000
      },
      "u": {
        "title": "Length unit",
        "description": "The unit of the lengths you typed; the area is in its square and the volume in its cube.",
        "type": "string",
        "enum": [
          "um",
          "mm",
          "cm",
          "m",
          "in",
          "ft"
        ]
      }
    }
  },
  "outputs": {
    "result": {
      "label": "Surface area to volume ratio",
      "description": "SA:V as a ratio to 1, and per unit of length.",
      "format": "text"
    },
    "ratio": {
      "label": "Surface area to volume ratio (per unit of length)",
      "description": "The surface area divided by the volume.",
      "format": "number"
    },
    "area": {
      "label": "Surface area (unit²)",
      "description": "The total area of the outside, in the typed unit squared.",
      "format": "number"
    },
    "volume": {
      "label": "Volume (unit³)",
      "description": "The space inside, in the typed unit cubed.",
      "format": "number"
    },
    "steps": {
      "label": "Steps",
      "description": "The surface area, the volume and the ratio worked out.",
      "format": "text"
    }
  },
  "defaultAnswer": {
    "inputs": {
      "shape": "cube",
      "a": 2,
      "r": 3,
      "l": 3,
      "w": 2,
      "h": 1,
      "u": "cm"
    },
    "outputs": {
      "result": "3 : 1 (3 per cm)",
      "ratio": 3,
      "area": 24,
      "volume": 8,
      "steps": "Surface area = 6a² = 6 × 2² = 24 cm²; Volume = a³ = 2³ = 8 cm³; Ratio = 6 ÷ a = 6 ÷ 2 = 3 per cm"
    },
    "text": "The surface area to volume ratio is 3 : 1 (3 per cm)."
  },
  "examples": [
    {
      "given": {
        "shape": "cube",
        "a": 1,
        "u": "mm"
      },
      "expect": {
        "area": 6,
        "volume": 1,
        "ratio": 6,
        "result": "6 : 1 (6 per mm)"
      },
      "source": "OpenStax, Biology 2e, §4.2 Prokaryotic Cells (Cell Size: a sphere’s surface area is 4πr² and its volume 4πr³/3; Figure 4.7: cells of 1 mm³ and 6 mm² and of 8 mm³ and 24 mm² have surface area to volume ratios of 6 to 1 and 3 to 1), https://openstax.org/books/biology-2e/pages/4-2-prokaryotic-cells (retrieved 2026-10-19); OpenStax, Prealgebra 2e, §9.6 Solve Geometry Applications: Volume and Surface Area (rectangular solid V = LWH, S = 2LH + 2LW + 2WH; cube V = s³, S = 6s²; sphere V = 4/3 πr³, S = 4πr²; cylinder V = πr²h, S = 2πr² + 2πrh), https://openstax.org/books/prealgebra-2e/pages/9-6-solve-geometry-applications-volume-and-surface-area (retrieved 2026-10-19)"
    },
    {
      "given": {
        "shape": "cube",
        "a": 2,
        "u": "mm"
      },
      "expect": {
        "area": 24,
        "volume": 8,
        "ratio": 3,
        "steps": "Surface area = 6a² = 6 × 2² = 24 mm²; Volume = a³ = 2³ = 8 mm³; Ratio = 6 ÷ a = 6 ÷ 2 = 3 per mm"
      },
      "source": "OpenStax, Biology 2e, §4.2 Prokaryotic Cells (Cell Size: a sphere’s surface area is 4πr² and its volume 4πr³/3; Figure 4.7: cells of 1 mm³ and 6 mm² and of 8 mm³ and 24 mm² have surface area to volume ratios of 6 to 1 and 3 to 1), https://openstax.org/books/biology-2e/pages/4-2-prokaryotic-cells (retrieved 2026-10-19); OpenStax, Prealgebra 2e, §9.6 Solve Geometry Applications: Volume and Surface Area (rectangular solid V = LWH, S = 2LH + 2LW + 2WH; cube V = s³, S = 6s²; sphere V = 4/3 πr³, S = 4πr²; cylinder V = πr²h, S = 2πr² + 2πrh), https://openstax.org/books/prealgebra-2e/pages/9-6-solve-geometry-applications-volume-and-surface-area (retrieved 2026-10-19)"
    },
    {
      "given": {
        "shape": "cube",
        "a": 2.5,
        "u": "in"
      },
      "expect": {
        "area": 37.5,
        "volume": 15.625,
        "ratio": 2.4
      },
      "source": "OpenStax, Prealgebra 2e, §9.6 Solve Geometry Applications: Volume and Surface Area (rectangular solid V = LWH, S = 2LH + 2LW + 2WH; cube V = s³, S = 6s²; sphere V = 4/3 πr³, S = 4πr²; cylinder V = πr²h, S = 2πr² + 2πrh), https://openstax.org/books/prealgebra-2e/pages/9-6-solve-geometry-applications-volume-and-surface-area (retrieved 2026-10-19); hand calculation in content.mdx: 37.5 ÷ 15.625 = 2.4"
    },
    {
      "given": {
        "shape": "sphere",
        "r": 3,
        "u": "um"
      },
      "expect": {
        "area": 113.09733552923255,
        "volume": 113.09733552923255,
        "ratio": 1,
        "result": "1 : 1 (1 per µm)"
      },
      "source": "OpenStax, Biology 2e, §4.2 Prokaryotic Cells (Cell Size: a sphere’s surface area is 4πr² and its volume 4πr³/3; Figure 4.7: cells of 1 mm³ and 6 mm² and of 8 mm³ and 24 mm² have surface area to volume ratios of 6 to 1 and 3 to 1), https://openstax.org/books/biology-2e/pages/4-2-prokaryotic-cells (retrieved 2026-10-19); hand calculation in content.mdx: 4π × 9 = 36π and 4/3 π × 27 = 36π, so SA:V = 3 ÷ 3 = 1",
      "tolerance": 1e-15
    },
    {
      "given": {
        "shape": "cylinder",
        "r": 1,
        "h": 2,
        "u": "cm"
      },
      "expect": {
        "area": 18.84955592153876,
        "volume": 6.283185307179586,
        "ratio": 3
      },
      "source": "OpenStax, Prealgebra 2e, §9.6 Solve Geometry Applications: Volume and Surface Area (rectangular solid V = LWH, S = 2LH + 2LW + 2WH; cube V = s³, S = 6s²; sphere V = 4/3 πr³, S = 4πr²; cylinder V = πr²h, S = 2πr² + 2πrh), https://openstax.org/books/prealgebra-2e/pages/9-6-solve-geometry-applications-volume-and-surface-area (retrieved 2026-10-19); hand calculation in content.mdx: SA = 2π × 1 × 3 = 6π, V = π × 1 × 2 = 2π, SA:V = 3",
      "tolerance": 1e-15
    },
    {
      "given": {
        "shape": "box",
        "l": 3,
        "w": 2,
        "h": 1,
        "u": "in"
      },
      "expect": {
        "area": 22,
        "volume": 6,
        "ratio": 3.6666666666666665,
        "result": "3.66667 : 1 (3.66667 per in)"
      },
      "source": "OpenStax, Prealgebra 2e, §9.6 Solve Geometry Applications: Volume and Surface Area (rectangular solid V = LWH, S = 2LH + 2LW + 2WH; cube V = s³, S = 6s²; sphere V = 4/3 πr³, S = 4πr²; cylinder V = πr²h, S = 2πr² + 2πrh), https://openstax.org/books/prealgebra-2e/pages/9-6-solve-geometry-applications-volume-and-surface-area (retrieved 2026-10-19); hand calculation in content.mdx: SA = 2 × (6 + 3 + 2) = 22, V = 6, SA:V = 22 ÷ 6"
    }
  ],
  "sources": [
    "OpenStax, Biology 2e, §4.2 Prokaryotic Cells, Cell Size: a sphere’s surface area is 4πr² and its volume 4πr³/3; Figure 4.7: cells of 1 mm³ with 6 mm² and of 8 mm³ with 24 mm² have surface area to volume ratios of 6 to 1 and 3 to 1, retrieved 2026-10-19. https://openstax.org/books/biology-2e/pages/4-2-prokaryotic-cells",
    "OpenStax, Prealgebra 2e, §9.6 Solve Geometry Applications: Volume and Surface Area: rectangular solid V = LWH, S = 2LH + 2LW + 2WH; cube V = s³, S = 6s²; sphere V = 4/3 πr³, S = 4πr²; cylinder V = πr²h, S = 2πr² + 2πrh; Example 9.49 (a cube of 2.5 in: V = 15.625 in³, S = 37.5 in²), retrieved 2026-10-19. https://openstax.org/books/prealgebra-2e/pages/9-6-solve-geometry-applications-volume-and-surface-area"
  ],
  "related": [
    "surface-area",
    "volume",
    "sphere",
    "cylinder-volume"
  ],
  "changelog": []
}