Find the volume V of the described solid S. The base of S is the triangular region with vertices (0, 0), (3, 0), and (0, 3). Cross-sections perpendicular to the y-axis are equilateral triangles. V

27/2

Step-by-step explanation:

Given

Vertices (0, 0), (3, 0), and (0, 3)

Since the base of the equilateral in the plane perpendicular to the x-axis goes from the x-axis to the line y = 3 - x.

So, the length of each side of the triangle is (3-x)

Calculating the area;

Area = ½bh

Where b = base = 3 - x

height is calculated as;

h² = (3-x) ² + (½ (3-x)) ² - - - from Pythagoras

h² = 9 - 6x + x² + (3/2 - ½x) ²

Let h² = 0

0 = 9 - 6x + x² + (9/4 - 6/4x + ¼x²)

0 = 9 + 9/4 - 6x - 6/4 + x² + ¼x²

0 = 45/4 - 30x/4 + 5x²/4

0. = 5x²/4 - 30x/4 + 45/4

0 = 5x² - 15x/4 - 15x/4 + 45/4

0 = 5x (x/4-¾) - 15 (x/4 - ¾)

0 = (5x - 15) (x/4 - ¾)

5x = 15 or x/4 = 3/4

x = 3 or x = 3

So, h = 3

Area = ½bh

Area = ½ * (3-x) * 3

Area = ½ (9-3x)

Volume = Integral of ½ (9-3x) {3,0}

V = 9/2 - 3x/2 {3,0}

V = 9x/2 - 3x²/4 {3,0}

V = 9 (3) / 2 - 3 (3) ²/4

V = 27/2 - 27/4

V = 27/2