JEE Advanced 2019 Paper 1 · Mathematics · Area Under Curves · easy

Find the area of the region \(x,y): x^2+y^2≤ 8, y≥ (x^2/(2)\.

Numerical answer

Find the area of the region

\{(x,y):\ x^2+y^2\le 8,\ y\ge \tfrac{x^2}{2}\}

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 7.62

The curves meet where

x^2+\frac{x^4}{4}=8

, i.e.

x^4+4x^2-32=0

, so

x^2=4

and

x=\pm2

with

y=2

. Between

x=-2

and

x=2

the upper boundary is the circle

y=\sqrt{8-x^2}

and the lower boundary is the parabola

y=\frac{x^2}2

. The area is

2\int_0^2\left(\sqrt{8-x^2}-\frac{x^2}{2}\right)dx

. Using

\int\sqrt{8-x^2}\,dx=\frac x2\sqrt{8-x^2}+4\arcsin\frac{x}{2\sqrt2}

, we get

\int_0^2\sqrt{8-x^2}\,dx=2+\pi

, while

\int_0^2\frac{x^2}2dx=\frac43

. So the area is

2\left(2+\pi-\frac43\right)=\frac43+2\pi\approx1.3333+6.2832=7.62

Original question written in the JEE Advanced 2019 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