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 bounded by the curve

y^{2} = 2 x

and

x^{2} + y^{2} = 4 x

.

Accepted Answer

Given $y^{2} = 2 x$ .......(i) and $x^{2} + y^{2} = 4 x$ ........(ii)

For finding the values of $x$ , substitute (i) in (ii)

$\Rightarrow x^{2} + 2 x = 4 x$

$\Rightarrow x^{2} - 2 x = 0$

$\Rightarrow x (x - 2) = 0$

$\Rightarrow x = 0, 2$

Also, $x^{2} + y^{2} = 4 x$

$\Rightarrow x^{2} - 4 x = - y^{2}$

$\Rightarrow x^{2} - 4 x + 4 = - y^{2} + 4$

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

Area enclosed between curves $y_{1} = f (x)$ and $y_{2} = g (x)$ and lines $x = a & x = b$ , $b > a$ is given by $|\int_{a}^{b} [f (x) - g (x)] d x|$ .

$∴$ Area of the shaded region $= 2 \cdot \int_{0}^{2} [\sqrt{2^{2} - (x - 2)^{2}} - \sqrt{2 x}] d x$

$= 2 [[\frac{x - 2}{2} \cdot \sqrt{2^{2} - (x - 2)^{2}} {+ \frac{2^{2}}{2} \sin^{- 1} (\frac{x - 2}{2})]}_{0}^{2} - {[\sqrt{2} \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{2}]$ $[∵ \int \frac{1}{\sqrt{a^{2} - {(b x + c)}^{2}}} d x = \frac{1}{b} \sin^{- 1} (\frac{b x + c}{a}) + C]$

$= 2 [(0 + 0 - 1.0 + 2 . \frac{π}{2}) - \frac{2 \sqrt{2}}{3} (2^{\frac{3}{2}} - 0)]$

$= \frac{4 π}{2} - \frac{8}{3}$

$= 2 π - \frac{8}{3}$

$= 2 (π - \frac{4}{3})$ sq.units

Therefore, required area is $2 (π - \frac{4}{3})$ sq.units.

NCERT Solutions for Chapter: Application of Integrals, Exercise 1: EXERCISE

NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Application of Integrals, Exercise 1: EXERCISE

Questions from NCERT Solutions for Chapter: Application of Integrals, Exercise 1: EXERCISE with Hints & Solutions

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