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 lying above

X -

axis and included between the circle

x^{2} + y^{2} = 8 x

and inside of parabola

y^{2} = 4 x

Accepted Answer

We have, given equations

$x^{2} + y^{2} = 8 x$ ......(i)

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

Equations (i) can be written as

$(x - 4)^{2} + y^{2} = (4)^{2}$

So equations (i) represents a circle with centre $(4, 0)$ and radius $4$ .

From (i) and (ii), we get: $x^{2} + 4 x = 8 x or x^{2} - 4 x = 0 \Rightarrow x (x - 4) = 0$

$∴ x = 0, 4$

Therefore, from (ii), $y = 0, 4$

Therefore, points of intersection of circle (i) and parabola (ii) are $O (0, 0) and P (4, 4)$

Required area $=$ area of region $O P Q C O$

$= (area of region O C P O) + (area of region P C Q P)$

$= \int_{0}^{4} y d x + \int_{4}^{8} y d x$

$= 2 \int_{0}^{4} \sqrt{x} d x + \int_{4}^{8} \sqrt{(4)^{2} - (x - 4)^{2}} d x$

$= 2 {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{4} + {[\frac{(x - 4)}{2} \sqrt{(4)^{2} - (x - 4)^{2}} + \frac{(4)^{2}}{2} \cdot \sin^{- 1} (\frac{x - 4}{4})]}_{4}^{8}$

$= \frac{4}{3} [(4)^{\frac{3}{2}} - 0] + [(\frac{8 - 4}{2}) \sqrt{16 - 16} + 8 \sin^{- 1} (1)] - [0 + 8 \sin^{- 1} (0)]$

$= \frac{4}{3} [(4)^{\frac{3}{2}}] + (0 + 8 \times \frac{π}{2}) - (0 + 0)$

$= \frac{4}{3} \times 8 + 4 π$

$= \frac{32}{3} + 4 π$

$= \frac{4}{3} (8 + 3 π)$ sq.units

Hence, the required area is $\frac{4}{3} (8 + 3 π)$ sq.units.

Find the area of the region lying above $X -$ axis and included between the circle $x^{2} + y^{2} = 8 x$ and inside of parabola $y^{2} = 4 x$

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area of the region lying above X- axis and included between the circle x2+y2=8x and inside of parabola y2=4x

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area of the region lying above $X -$ axis and included between the circle $x^{2} + y^{2} = 8 x$ and inside of parabola $y^{2} = 4 x$