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 bounded by curves

x^{2} + y^{2} = 1

and

y = |x|

.

Accepted Answer

Given curves $x^{2} + y^{2} = 1 . . . . . (i)$ and $y = |x| = x . . . . . . (i i)$

Substitute $(i i)$ in $(i)$ , we get

$x^{2} + x^{2} = 1 \Rightarrow 2 x^{2} = 1$

$\Rightarrow x^{2} = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}$

The coordinates of M are $(\frac{1}{\sqrt{2}}, 0)$

$∴$ Required Area = 2[Area under circle $y = \sqrt{1 - x^{2}}$ in first quadrant - area under the line $y = |x|$ ]

$\Rightarrow$ Required Area = $2 \int_{0}^{\frac{1}{\sqrt{2}}} (\sqrt{1 - x^{2}} - x) d x$

$= 2 {[\frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} \frac{x}{1} - \frac{x^{2}}{2}]}_{0}^{\frac{1}{\sqrt{2}}}$ [Using $\sqrt{a^{2} - x^{2}} = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C$ ]

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

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

$= 2 [\frac{1}{4} + \frac{π}{8} - \frac{1}{4}] = 2 \times \frac{π}{8}$

$= \frac{π}{4}$ sq.units

Therefore, the area bounded by curves $x^{2} + y^{2} = 1$ and $y = |x|$ is $\frac{π}{4}$ sq.units

Find the area bounded by curves $x^{2} + y^{2} = 1$ and $y = |x|$ .

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area bounded by curves x2+y2=1 and y=x.

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area bounded by curves $x^{2} + y^{2} = 1$ and $y = |x|$ .