Question

The base of S is the triangular region with vertices (0,0),(1,0),(0,1). Cross-sections perpendicular to the y-axis are equilateral triangles.Find the volume V of this solid.

Answer 1

Answer:

$V =\frac{\sqrt{3}}{12}$

Step-by-step explanation:

Diagram relating to the given vertices is attached below

height of equilateral triangle is given as

h = s sin60

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

area of triangle is given as

$A = \frac{1}{2} sh$

$A = \frac{1}{2} s\times \frac{\sqrt{3}}{2} s$

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

equation for line of given figure below that representing  diagonal is given as

y = -x + 1

integrate the from 0 to 1 we will have volume value

$V = \int_{0}^{1} \frac{\sqrt{3}}{4} s^2 dy$

$V = \int_{0}^{1} \frac{\sqrt{3}}{4} (-y+1)^2 dy$

$v = \frac{\sqrt{3}}{4} \int_{0}^{1} (-y+1)^2 dy$

   $=\frac{\sqrt{3}}{4} \int_{0}^{1} (y^2-2y+1)^2 dy$

  $=\frac{3}{4}[\frac{1}{3}y^3 -y^3 + y]_0^1$

$V =\frac{\sqrt{3}}{12}$