# Cell Size Limitations Cell size is limited by the cell's surface-area-to-volume ratio, such that cells cannot become so large that the surface of the cell is insufficient to take in nutrients and remove wastes.
Most cells are too small to be seen with the naked eye (the few exceptions include some egg cells, some of which are up to 13 centimeters in diameter, like the ostrich egg). Cells are so small because of the constraints of surface-area-to-volume ratio. Although cells come in many shapes, the easiest to consider is a sphere. The measurement of the outside of the sphere is its surface area, and the space contained within it is its volume. For a cell, the surface area is made up by the cell membrane, and the space within it is the cytoplasm (and the nucleus, in the case of eukaryotic cells). The cell membrane is selectively permeable, allowing some substances to pass through it while blocking others. This ensures nutrients, water, and other vital materials can enter the cell and signals and waste materials can exit it.
Step-By-Step Example
Calculating Surface Area-to-Volume of a Sphere
To determine the surface-area-to-volume ratio, calculate the surface area of a sphere then calculate the volume of the sphere and divide.
Step 1
To calculate the surface area of a sphere (A), multiply 4 by pi ($\pi$) by the square of the radius (r)
$A=4\pi r^2$
Step 2
The volume (V) is calculated by multiplying $\frac43$ by $\pi$ by the cube of the radius (r).
$V=\frac43\pi r^3$
Solution

Solution for a cell with a radius of 1 millimeter (mm):

A and V are calculated as follows:
\begin{aligned}A&=4\pi1^2=4\pi\\V&=\frac{4}{3}\pi1^3=\frac{4}{3}\pi\end{aligned}
The surface-area-to-volume ratio is given as $\frac AV$. For the sphere with radius 1 mm, this ratio is $\frac{\displaystyle4\pi}{\displaystyle\frac43\pi}$, or 3.0.

Solution for a cell with a radius of 2 mm:

A and V are calculated as follows:
\begin{aligned}A&=4\pi2^2=16\pi\\V&=\frac{4}{3}\pi2^3=\frac{32}{3}\pi\end{aligned}
The surface-area-to-volume ratio of this cell is $\frac{\displaystyle16\pi}{\displaystyle\frac{32}3\pi}$, or 1.5.

Solution for a cell with a radius of 3 mm:

A and V are calculated as follows:
\begin{aligned}A&=4\pi3^2=36\pi\\V&=\frac{4}{3}\pi3^3=\frac{108}{3}\pi\end{aligned}
The surface-area-to-volume ratio of this cell is $\frac{\displaystyle36\pi}{\displaystyle\frac{108}3\pi}$, or 1.0.
As the volume of the cell grows, the surface-area-to-volume ratio shrinks. The surface of the cell is made up of the cell membrane, which is tasked with regulating transport of materials into and out of the cell. Cells rely on this transport to bring in nutrients and remove wastes. Eventually, there is not enough cell membrane available to meet the demands of the cell. Cells cannot be maintained larger than this maximum size.

#### Surface-to-Volume Ratio As cell size increases, the surface-to-volume ratio of the cell decreases. Eventually, the cell membrane is not sufficient to keep up with the needs of the cell, and the cell can no longer grow.