Name: Area Enclosed by a Curve and a Line: Illustration
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: How would you find the area enclosed between a parabola and a straight line? Watch the video to see the method as an application of definite integrals.

Question

Find the area of the region enclosed in first quadrant by the circle

x^{2} + y^{2} = 32

, line

y = x

and

x

-axis.

Accepted Answer

Given circle $x^{2} + y^{2} = 32$ , line $y = x$ .

Finding intersection point of circle and line,

$\Rightarrow x^{2} + y^{2} = 32$

$\Rightarrow x^{2} + x^{2} = 32$

$\Rightarrow 2 x^{2} = 32$

$\Rightarrow x^{2} = 16$

$\Rightarrow x = \pm 4$

$\Rightarrow x = y = \pm 4$

So, intersection points are $(4, 4) and (- 4, - 4)$ .

Circle $x^{2} + y^{2} = 32 \Rightarrow x^{2} + y^{2} = {(4 \sqrt{2})}^{2}$

Circle intersects the positive $x$ -axis at $(4 \sqrt{2}, 0)$ .

Required area $=$ area of the region $B C O B$ $+$ area of the region $B A C B$

Area of the region $B C O B$ $= \int_{0}^{4} x d x$

$= {[\frac{x^{2}}{2}]}_{0}^{4}$

$= [\frac{4^{2}}{2} - 0] = 8$ square units

Area of the region $B A C B$ $= \int_{4}^{4 \sqrt{2}} (\sqrt{{(4 \sqrt{2})}^{2} - x^{2}}) d x$

$= {[\frac{x}{2} \sqrt{32 - x^{2}} + \frac{32}{2} \sin^{- 1} \frac{x}{4 \sqrt{2}}]}_{4}^{4 \sqrt{2}}$

$= [\frac{4 \sqrt{2}}{2} \sqrt{32 - {(4 \sqrt{2})}^{2}} + \frac{32}{2} \sin^{- 1} \frac{4 \sqrt{2}}{4 \sqrt{2}} - \frac{4}{2} \sqrt{32 - {(4)}^{2}} - \frac{32}{2} \sin^{- 1} \frac{4}{4 \sqrt{2}}]$

$= [0 + 16 \sin^{- 1} (1) - 2 \times 4 - 16 \sin^{- 1} \frac{1}{\sqrt{2}}]$

$= [16 \times \frac{π}{2} - 8 - 16 (\frac{π}{4})]$

$= [8 π - 8 - 4 π]$

$= 4 π - 8$ square units

Hence, required area $= 8 + 4 π - 8 = 4 π$ square units

Embibe Experts Solutions for Chapter: Application of Integrals, Exercise 1: Uttar Pradesh Board-2020

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Application of Integrals, Exercise 1: Uttar Pradesh Board-2020

Questions from Embibe Experts Solutions for Chapter: Application of Integrals, Exercise 1: Uttar Pradesh Board-2020 with Hints & Solutions

Learn and Prepare for any exam you want !