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The base of s is the region enclosed by the parabola y = 3 − 2x2 and the x−axis. Cross-sections perpendicular to the y−axis are squares.

Sagot :

The volume of the base of the region enclosed by the parabola is; V = 9

Integral Volume of a Solid

The length of each cross-section is determined by the horizontal distance (parallel to the x-axis) from one end of the parabola to the other.

Thus;

y = 3 - 2x²

Making x the subject gives us;

x = ±√[(3 - y)/2]

Thus;

The horizontal distance is;

√[(3 - y)/2] - (-√[(3 - y)/2])

⇒ 2√[(3 - y)/2]

The area of each cross-section is simply the square of the section's side length, so the area would be;

A = (2√[(3 - y)/2])²

A = 6 - 2y

Thus, volume is;

V = [tex]\int\limits^3_0 {6 - 2y} \, dy[/tex]

Solving this gives V = 9

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