Q.3 (B) (ii) Find the area of the region bounded by the curve x²=16y and the line y=4 .

Answer: 128/3 sq. units The parabola x²=16y (vertex at the origin, opening upwards, symmetric about the Y-axis) meets y=4 where x²=64 , i.e. x=± 8 . By symmetry, required area = 2 × area in the first quadrant between the Y-axis, the parabola and y=4 . Taking horizontal strips, x=4√(y) : A=2∫₀⁴ x dy=2∫₀⁴ 4√(y) dy=8[2/3y^(3/2)]₀⁴=8·2/3· 8=128/3 Required area =128/3 sq. units.

Find the area of the region bounded by the curve \(x^2=16y\) and the line \(y=4\).

Q.3 (B) (ii) Find the area of the region bounded by the curve \(x^2=16y\) and the line \(y=4\).

Doubtly Asked 28 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: \(\frac{128}{3}\) sq. units

The parabola \(x^2=16y\) (vertex at the origin, opening upwards, symmetric about the Y-axis) meets \(y=4\) where \(x^2=64\), i.e. \(x=\pm 8\).

By symmetry, required area = 2 \(\times\) area in the first quadrant between the Y-axis, the parabola and \(y=4\). Taking horizontal strips, \(x=4\sqrt{y}\):

\(A=2\int_0^4 x\,dy=2\int_0^4 4\sqrt{y}\,dy=8\left[\frac{2}{3}y^{3/2}\right]_0^4=8\cdot\frac{2}{3}\cdot 8=\frac{128}{3}\)

Required area \(=\frac{128}{3}\) sq. units.

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Find the area of the region bounded by the curve \(x^2=16y\) and the line \(y=4\).

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