Name: Area Between Curve and X-axis
Uploaded: 2019-07-19
Duration: 1 min 50 s
Description: As you might have seen, definite integration is used to find the area bounded by a curve. In this video, we will learn and understand the area bounded by a c...

Question

Find the area of the region lying between the parabolas.

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

Accepted Answer

Given,

For finding the points of intersection of the two parabolas, we equate the values of $y^{2}$ from their equations.

From the equation $x^{2} = 4 y \Rightarrow y = \frac{x^{2}}{4}$

$∴ y^{2} = {(\frac{x^{2}}{4})}^{2} = \frac{x^{4}}{16}$

$\frac{x^{4}}{16} = 4 x$

$x^{4} = 64 x$

$x^{4} - 64 x = 0$

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

$x = 0, x^{3} = 64$

i.e. $x = 0, x = 4$

When $x = 0, y = 0$

When $x = 4, y = \frac{4^{2}}{4} = \frac{16}{4} = 4$

$∴$ Point of intersection are $(0, 0)$ and $(4, 4)$ .

Required area $=$ Area of the region $O B A C O =$ $[$ Area of the region $O D A C O$ $]$ $-$ $[$ Area of the region $O D A B O$ $]$

Now, area of the region $O D A C O$ $=$ Area under the parabola $y^{2} = 4 x$

i.e. $y = 2 \sqrt{x}$ between $x = 0$ and $x = 4$

Then, area of the region $O D A B O =$ Area under the parabola $x^{2} = 4 y$

i.e. $y = \frac{x^{2}}{4}$ between $x = 0$ and $x = 4$

Required area $= \int_{0}^{4} (y_{1} - y_{2}) . d x$ , where $y_{1} = 2 \sqrt{x}$ and $y_{2} = \frac{x^{2}}{4}$

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

$= 2 \int_{0}^{4} x^{\frac{1}{2}} . d x - \frac{1}{4} \int_{0}^{4} x^{2} . d x$

$= 2 {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{4} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{0}^{4}$

$= \frac{4}{3} {[x^{\frac{3}{2}}]}_{0}^{4} - \frac{1}{12} {[x^{3}]}_{0}^{4}$

$= \frac{4}{3} [4^{\frac{3}{2}} - 0] - \frac{1}{12} [4^{3} - 0]$

$= \frac{4}{3} [{(2^{2})}^{\frac{3}{2}}] - \frac{64}{12}$

$= (\frac{32}{3} - \frac{64}{12}) = (\frac{32}{3} - \frac{16}{3}) = \frac{16}{3}$

Hence, the area of the required region is $\frac{16}{3}$ sq.units.

Maharashtra Board Solutions for Chapter: Application of Definite Integration, Exercise 2: MISCELLANEOUS EXERCISE 5

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Application of Definite Integration, Exercise 2: MISCELLANEOUS EXERCISE 5

Questions from Maharashtra Board Solutions for Chapter: Application of Definite Integration, Exercise 2: MISCELLANEOUS EXERCISE 5 with Hints & Solutions

Learn and Prepare for any exam you want !