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

Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line

y = x

, and the circle

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

Accepted Answer

Given the boundaries of the area to be found are,

the circle, $x^{2} + y^{2} = 32$

the line, $y = x . . . . . . . . . (2)$

the $x$ -axis

And, area should be in first quadrant.

From the equation, of the first circle, $x^{2} + y^{2} = 32$

the vertex at $(0, 0)$ i.e. the origin

the radius is $4 \sqrt{2} unit$

Now to find the point of intersection of $(1)$ and $(2)$ substitute $y = x$ in $(1)$

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

$2 x^{2} = 32$

$x^{2} = 16$

$x = \pm 4$

Substituting $x$ in $(2),$ we get $y = \pm 4$

So the two points, $A$ and $B$ where $(1)$ and $(2)$ meet are $A = (4, 4)$ and $B = (- 4, - 4)$

As $x$ and $y$ have even powers for both the circles, they will be symmetrical about the $x -$ axis and $y -$ axis.

Consider the circle, $x^{2} + y^{2} = 32$ can be re-written as

$y^{2} = 32 - x^{2}$

$y = \sqrt{(32 - x^{2})}$

Let us drop a perpendicular from $A$ on to $x -$ axis. The base of the perpendicular is $C = (4, 0)$

Area of the required region $=$ Area of $O A D O .$

Area of $O A D O =$ Area of $C A D C +$ Area of $O A C$

Area of $OADO = \int_{4}^{4 \sqrt{2}} \sqrt{32 - x^{2}} dx + \int_{0}^{4} xdx$

$= {[\frac{x \sqrt{32 - x^{2}}}{2} + \frac{32}{2} \sin^{- 1} (\frac{x}{4 \sqrt{2}})]}_{4}^{4 \sqrt{2}} + {[\frac{x^{2}}{2}]}_{0}^{4}$

Using the formula, $\int \sqrt{a^{2} - x^{2}} dx = \frac{x \sqrt{a^{2} - x^{2}}}{2} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a}) + c$ and $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$

$=$ $[\frac{4 \sqrt{2} \sqrt{32 - 32}}{2} + 16 \sin^{- 1} (\frac{4 \sqrt{2}}{4 \sqrt{2}}) - (\frac{4 \sqrt{32 - 16}}{2} + 16 \sin^{- 1} (\frac{1}{\sqrt{2}}))] + [\frac{16}{2} - 0]$

$= [0 + 16 \times \frac{π}{2} - (8 + 16 \times \frac{π}{4})] + 8$

$= 4 π$

Hence, area of the required region is $4 π$ square units.

Embibe Experts Solutions for Chapter: Application of Integrals, Exercise 1: CBSE-2018

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Application of Integrals, Exercise 1: CBSE-2018

Questions from Embibe Experts Solutions for Chapter: Application of Integrals, Exercise 1: CBSE-2018 with Hints & Solutions

Learn and Prepare for any exam you want !