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10 June, 14:15

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

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  1. 10 June, 14:23
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    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
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