A-LevelMathematicsIntegration9709/12 · Oct/Nov 2015 · Q1011 Marks
The diagram shows part of the curve `y = √(9 – 2x²)`. The point P (2, 1) lies on the curve and the normal to the curve at P intersects the x-axis at A and the y-axis at B. [Figure 4]
- (i)Show that B is the mid-point of AP.[6]
- (ii)Find, showing all necessary working, the exact volume obtained when the shaded region is rotated through 360° about the y-axis.[5]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.
- [B1] dy/dx = 1/(2√(9-2x²)) × (-4x) (Without "-4x" Allow even if B0 above)
- [B1] × (-4x)
- [M1] At P, x = 2, m = -4 Normal grad = 1/4 (For m₁m₂ = -1 calculus needed)
- [M1] Eqn AP y - 1 = 1/4(x – 2) (Equation of normal)
- [A1] → A (-2, 0) or B (0, 2) (Coordinates of A or B)
- [A1] Midpoint AP also (0, 2) (Full justification)
- [M1] ∫x² dy (Attempt to integrate x²)
- [A1] = ∫(9 - y²)/2 dy (Correct integration)
- [B1] Upper limit = 3 (Evaluates upper limit)
- [DM1] Uses limits 1 to 3 (Uses both limits correctly)
- [A1] → volume = 4⅔ π
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