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The curve y = k x² is a parabola. (i) Find the area enclosed between this parabola, the x-axis and the horizontal line y = y0. (ii) The parabola is then revolved about the y-axis to generate a cup-shaped solid (a paraboloid). Find the volume of this paraboloid-cup.

1. Area = (4/3)*y0*sqrt(y0/k); Volume = pi*y0²/(2k)
2. Area = (2/3)*y0*sqrt(y0/k); Volume = pi*y0²/k
3. Area = (4/3)*y0*sqrt(y0/k); Volume = pi*y0²/k
4. Area = (2/3)*y0*sqrt(y0/k); Volume = pi*y0²/(2k)

Correct answer: Area = (4/3)*y0*sqrt(y0/k); Volume = pi*y0²/(2k)

Solution

The area between the line and the parabola is (4/3)*y0*sqrt(y0/k); revolving about the y-axis with disks of radius x = sqrt(y/k) gives V = pi*y0²/(2k).

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