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Express the surface area of a cube as a function of its volume.

Short Answer

Expert verified

The expression for the surface area of a cube as a function of its volume is \({\bf{A = 6}}{{\bf{V}}^{{\bf{2/3}}}}\) and domain of the function is\(\left( {0,\infty } \right)\).

Step by step solution

01

Determine the equation for the volume of a cube

The volume of a cube is given as;

\(\begin{array}{l}V = {L^3}\\L = {\left( V \right)^{1/3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{array}\)

Here, \(L\)is the length of the cube, \(V\) is the volume of a cube.

02

Determine the expression for the surface area of a cube

The surface area of a cube is given as;

\(A = 6{L^2}\)

Substitute the value from equation (1) in the above equation.

\(\begin{array}{l}A = 6{\left( {{{\left( V \right)}^{1/3}}} \right)^2}\\A = 6{V^{2/3}}\end{array}\)

The surface area of the cube is a function of its volume and the volume of the cube cannot be negative. The volume of a cube is always positive, so the domain for the surface area is all positive real numbers.

\(0 < V < \infty \)

Therefore, the expression for the surface area of the cube as a function of its volume is \(A = 6{V^{2/3}}\) and domain of the volume of a cube is\(\left( {0,\infty } \right)\).

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