JEE Advanced 2022 Paper 1 · Mathematics · Area Under Curves · hard

Let R be the region common to the interior of the parabola y^2=4x and the disc x^2+y^2≤12. Find the area of R.

Numerical answer

Let

R

be the region common to the interior of the parabola

y^2=4x

and the disc

x^2+y^2\le12

. Find the area of

R

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 13.35

The curves meet where

x^2+4x=12

, i.e.

x=2

y=\pm2\sqrt2

. By symmetry about the

x

-axis, area

=2\left[\int_0^2 2\sqrt x\,dx+\int_2^{2\sqrt3}\sqrt{12-x^2}\,dx\right]

. The first integral is

\frac43\cdot2^{3/2}=\frac{8\sqrt2}{3}

. The second is

\left[\frac x2\sqrt{12-x^2}+6\sin^{-1}\frac{x}{2\sqrt3}\right]_2^{2\sqrt3}=3\pi-\sqrt2-6\sin^{-1}\frac1{\sqrt3}

. Therefore area

=2\left[\frac{5\sqrt2}{3}+3\pi-6\sin^{-1}\frac1{\sqrt3}\right]=6\pi+\frac{10\sqrt2}{3}-12\sin^{-1}\frac{1}{\sqrt3}\approx13.35

. (Numerically checked.)

Original question written in the JEE Advanced 2022 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake