Question

To calculate the volume of the described solid (triangular region).

Short answer

The volume of the triangular region is \(\frac{{\sqrt 3 }}{{12}}\).

Step-by-step solution

Given data

The triangular region with vertices \((0,0),(1,0)\) and \((0,1)\). Cross-sections perpendicular to the \(y\)-axis are equilateral triangles.

Concept used of Volume

Volumeis a scalar quantity expressed in the amount of three-dimensional space enclosed by a closed surface.

Solve to find the volume

Consider the Equation of line \(x + y = 1\).

Base length of the equilateral triangle \(x = 1 - y\).

The cross-section of the triangular region as shown in Figure 1.

The region lies between \(a = 0\) and \(b = 1\).

The expression to find the volume of the triangular region as shown below.

\(V = \int_a^b A (y)dy\) ….(1)

Find the area of an equilateral triangle by the use of the relation as shown below.

\(\begin{aligned}{}A(y) = \frac{{\sqrt 3 }}{4}{x^2}\\ = \frac{{\sqrt 3 }}{4}{(1 - y)^2}\\ = \frac{{\sqrt 3 }}{4}\left( {1 + {y^2} - 2y} \right)\end{aligned}\)

Substitute 0 for a, 1 for \(b\), and \(\frac{{\sqrt 3 }}{4}\left( {1 + {y^2} - 2y} \right)\) for \(A(y)\) in Equation (1).

\(\begin{aligned}{}V = \int_0^1 {\frac{{\sqrt 3 }}{4}} \left( {1 + {y^2} - 2y} \right)dy\\ = \frac{{\sqrt 3 }}{4}\int_0^1 {\left( {1 + {y^2} - 2y} \right)} dy\\ = \frac{{\sqrt 3 }}{4}\left( {y + \frac{{{y^3}}}{3} - \frac{{2{y^2}}}{2}} \right)_0^1\\ = \frac{{\sqrt 3 }}{4}\left( {1 + \frac{{{1^3}}}{3} - {1^2} - 0} \right)\end{aligned}\)

Which means,

\(\begin{aligned}{}V = \frac{{\sqrt 3 }}{4}\left( {\frac{{3 + 1 - 3}}{3}} \right)\\ = \frac{{\sqrt 3 }}{{12}}\end{aligned}\)

Therefore, the volume of the triangular region is \(\frac{{\sqrt 3 }}{{12}}\).