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 by the two parabolas

y^{2} = 4 a x and x^{2} = 4 a y

.

Accepted Answer

Given curves are: $y^{2} = 4 a x$ and $x^{2} = 4 a y$

First we have to find the point of intersection of the two curves.

Point of intersection of the two curves can be calculated as, ${(\frac{x^{2}}{4 a})}^{2} = 4 a x$

$\Rightarrow \frac{x^{4}}{16 a^{2}} = 4 a x$

$\Rightarrow x^{4} = 64 a^{3} x$

$\Rightarrow x^{4} - 64 a^{3} x = 0$

$\Rightarrow x (x^{3} - 64 a^{3}) = 0$

$\Rightarrow x = 0, x = 4 a$

Also, $y = 0, y = 4 a$

The point of intersection of the given two curves are: $(0, 0)$ and $(4 a, 4 a)$

Therefore, the required area $O A B C O$ $= \int_{0}^{4 a} (y_{2} - y_{1}) d x$

$= \int_{0}^{4 a} [\sqrt{4 a x} - \frac{x^{2}}{4 a}] d x$ [Where, $y_{1} = \sqrt{4 a x} and y_{2} = \frac{x^{2}}{4 a}$ ]

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

$= \int_{0}^{4 a} ((2 \sqrt{a}) x^{\frac{1}{2}} - \frac{1}{4 a} (x^{2})) d x$

$= {[\frac{4 \sqrt{a}}{3} x^{\frac{3}{2}} - \frac{x^{3}}{12 a}]}_{0}^{4 a}$

$= [\frac{4 \sqrt{a}}{3} (4 a)^{\frac{3}{2}} - \frac{(4 a)^{3}}{12 a}] - [\frac{4 \sqrt{a}}{3} (0)^{\frac{3}{2}} - \frac{(0)^{3}}{12 a}]$

$= [\frac{4 \sqrt{a}}{3} 8 a^{\frac{3}{2}} - \frac{64 a^{3}}{12 a}] - 0$

$= \frac{32 a^{2}}{3} - \frac{16 a^{2}}{3}$

$= \frac{16 a^{2}}{3}$

Hence, the required area is $\frac{16 a^{2}}{3}$ sq.units.

Embibe Experts Solutions for Chapter: Area Under Plane Curves (Application of Definite Integrals), Exercise 1: Odisha Board-2018

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Area Under Plane Curves (Application of Definite Integrals), Exercise 1: Odisha Board-2018

Questions from Embibe Experts Solutions for Chapter: Area Under Plane Curves (Application of Definite Integrals), Exercise 1: Odisha Board-2018 with Hints & Solutions

Learn and Prepare for any exam you want !