Question

Find the volume v of the described solid s. The base of a solid s is the triangular region with vertices (0, 0), (4, 0), and (0, 4). Cross-sections perpendicular to the y-axis are equilateral triangles.

Answer 1

The volume of the solid S in the given question is 5.48unit³.

What is volume?

A three-dimensional space's occupied volume is measured.

It is frequently expressed numerically in a variety of imperial or US-standard units as well as SI-derived units.

The definition of length and volume are connected.

So, the volume of the solid S:
An equilateral triangle's sides are shown as a cross-section.

An equilateral triangle's height is determined by:

$h = sSin60 = \frac{\sqrt{3} }{2} s$

Consequently, one triangle's area is:

$A=\frac{1}{2} s h=\frac{1}{2} s \cdot \frac{\sqrt{3}}{2} s=\frac{\sqrt{3}}{4} s^2$

The line equation that depicts the diagonal is:

$\begin{aligned}& x+y=1 \\& y=-x+1 \\& x=-y+1\end{aligned}$

This will indicate the s value integrate from 0 to 2 if we integrate along the y-axis.

$\begin{aligned}& V=\int_0^2 \frac{\sqrt{3}}{4} s^2 d x \\& =\frac{\sqrt{3}}{4} \int_0^2(-y+1)^2 d x \\& =\frac{\sqrt{3}}{4} \int_0^2\left(y^2-2 y+1\right) d x \\& =\frac{\sqrt{3}}{4}\left[\frac{1}{3} y^3-y^2+y\right] \\& \left.=\frac{\sqrt{3}}{4}\left[\frac{1}{3}(2)^3-(2)^2+2\right)\right] \\& =5.48\end{aligned}$

Therefore, the volume of the solid S in the given question is 5.48unit³.

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Correct question:
Find the volume V of the described solid S. The base of S is the triangular region with vertices (0, 0), (2, 0), and (0, 2). Cross-sections perpendicular to the y-axis are equilateral triangles.