
Area bounded by the curve \[{y^2}\left( {2a - x} \right) = {x^3}\] and the line \[x = 2a\], is
A. \[3\pi {a^2}\]
B. $\dfrac{{3\pi {a^2}}}{2}$
C. $\dfrac{{3\pi {a^2}}}{4}$
D. $\dfrac{{\pi {a^2}}}{4}$
Answer
164.4k+ views
Hint: In this question, we are given the equation of the curve and the line. We have to calculate the area between them (area bounded by the curve and the line). Using the curve equation calculate the value of $y$ and let $x = 2a{\sin ^2}\theta $, differentiate it with respect to $\theta $. Then, integrate $y$ with respect to $dx$ from $x = 0$ to $x = 2a$ because the coordinate of line on $x - axis$ is $2a$ and the curve is starting from $x = 0$. Then, apply the rules of integration and solve it further.
Formula Used: Trigonometric identities –
1. ${\cos ^2}\theta = 1 - {\sin ^2}\theta $
2. ${\sin ^2}\theta = 1 - {\cos ^2}\theta $
3. $\sin 2\theta = 2\sin \theta \cos \theta $
4. $\cos 2\theta = 1 - 2{\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$
Integration formula –
$\int {f\left( y \right)dx = \dfrac{{\int {f\left( y \right)dx} }}{{\left( {\dfrac{{dy}}{{dx}}} \right)}}} $
And the integration of $\sin x$ is $\cos x + c$
Complete step by step Solution:
Given that,
Equation of the curve is \[{y^2}\left( {2a - x} \right) = {x^3}\] and the equation of the line is \[x = 2a\]
So, we have \[{y^2}\left( {2a - x} \right) = {x^3}\]
Also, written as $y = \sqrt {\dfrac{{{x^3}}}{{\left( {2a - x} \right)}}} $
Now, let $x = 2a{\sin ^2}\theta $ -------- (1)
Differentiate $x = 2a{\sin ^2}\theta $ with respect to $\theta $,
We get, $dx = 4a\sin \theta \cos \theta d\theta $ ------------ (2)
Substituting $x = 2a{\sin ^2}\theta $ in $y = \sqrt {\dfrac{{{x^3}}}{{\left( {2a - x} \right)}}} $,
It will be, $y = \sqrt {\dfrac{{{{\left( {2a{{\sin }^2}\theta } \right)}^3}}}{{\left( {2a - 2a{{\sin }^2}\theta } \right)}}} $
By simplifying, we get
$y = \sqrt {\dfrac{{\left( {8{a^3}{{\sin }^6}\theta } \right)}}{{2a\left( {1 - {{\sin }^2}\theta } \right)}}} $
Using, trigonometric identity i.e., ${\cos ^2}\theta = 1 - {\sin ^2}\theta $
It implies that, $y = \sqrt {\dfrac{{\left( {4{a^2}{{\sin }^6}\theta } \right)}}{{{{\cos }^2}\theta }}} $
Here, the square root of the term $\dfrac{{\left( {4{a^2}{{\sin }^6}\theta } \right)}}{{{{\cos }^2}\theta }}$ is $\dfrac{{\left( {2a{{\sin }^3}\theta } \right)}}{{\cos \theta }}$
So, $y = \dfrac{{\left( {2a{{\sin }^3}\theta } \right)}}{{\cos \theta }}$
Now, to calculate the area integrate $y$ with respect to $x$,
Area of the bounded region $ = \int\limits_0^{2a} {\dfrac{{\left( {2a{{\sin }^3}\theta } \right)}}{{\cos \theta }}} dx$
Using equation (2),
Area $ = 2a\int\limits_0^{2a} {\dfrac{{\left( {{{\sin }^3}\theta } \right)}}{{\cos \theta }}} 4a\sin \theta \cos \theta d\theta $
Here, the limits will also change
As we know that, $x = 2a{\sin ^2}\theta $
Therefore, at $x = 0$ we get $\theta = 0$
And at $x = 2a$, $\theta = \dfrac{\pi }{2}$
So, Area $ = 2a\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {{{\sin }^3}\theta } \right)}}{{\cos \theta }}} 4a\sin \theta \cos \theta d\theta $
Now, cancel the like terms
It will be,
Area \[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^4}\theta d\theta } \] ---------- (3)
Using, trigonometric identity i.e., ${\sin ^2}\theta = 1 - {\cos ^2}\theta $
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta \left( {1 - {{\cos }^2}\theta } \right)d\theta } \]
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta - {{\sin }^2}\theta {{\cos }^2}\theta d\theta } \]
Multiply and divide the term by $4$,
It will be, \[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta - \dfrac{{4{{\sin }^2}\theta {{\cos }^2}\theta }}{4}d\theta } \]
Using trigonometric identity i.e., $\sin 2\theta = 2\sin \theta \cos \theta $
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta - \dfrac{1}{4}{{\sin }^2}2\theta d\theta } \]
Using trigonometric identity i.e., $\cos 2\theta = 1 - 2{\sin ^2}\theta $
Also written as, ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{1 - \cos 2\theta }}{2}} \right) - \dfrac{1}{4}\left( {\dfrac{{1 - \cos 4\theta }}{2}} \right)d\theta } \]
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{2} - \dfrac{{\cos 2\theta }}{2} - \dfrac{1}{8} + \dfrac{{\cos 4\theta }}{8}d\theta } \]
\[ = 8{a^2}\left[ {\dfrac{\theta }{2} - \dfrac{{\sin 2\theta }}{4} - \dfrac{\theta }{8} + \dfrac{{\sin 4\theta }}{{32}}} \right]_0^{\dfrac{\pi }{2}}\]
Substituting the limits in the required integration,
\[ = 8{a^2}\left[ {\dfrac{\pi }{4} - \dfrac{{\sin 2\left( {\dfrac{\pi }{2}} \right)}}{4} - \dfrac{\pi }{{16}} + \dfrac{{\sin 4\left( {\dfrac{\pi }{2}} \right)}}{{32}}} \right]\]
As we know that, $\sin \pi = \sin 2\pi = 0$
Therefore,
Area \[ = 8{a^2}\left[ {\dfrac{\pi }{4} - 0 - \dfrac{\pi }{{16}} + 0} \right]\]
L.C.M. of $4$ and $16$ is $16$
\[ = 8{a^2}\left[ {\dfrac{{4\pi - \pi }}{{16}}} \right]\]
\[ = 8{a^2}\left[ {\dfrac{{3\pi }}{{16}}} \right]\]
\[ = \dfrac{{3\pi {a^2}}}{2}\]

Therefore, the correct option is (B).
Note:To solve such a question, one should have a good knowledge of trigonometric and integration formulas. Also, if you are changing the function and then always change the limits too. Put the limits in the same to calculate new limits. Try to open terms using trigonometric identities as much as it is possible. And to solve the limits, firstly solve the integration past then last subtract the terms (put the upper limit values and then lower limits).
Formula Used: Trigonometric identities –
1. ${\cos ^2}\theta = 1 - {\sin ^2}\theta $
2. ${\sin ^2}\theta = 1 - {\cos ^2}\theta $
3. $\sin 2\theta = 2\sin \theta \cos \theta $
4. $\cos 2\theta = 1 - 2{\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$
Integration formula –
$\int {f\left( y \right)dx = \dfrac{{\int {f\left( y \right)dx} }}{{\left( {\dfrac{{dy}}{{dx}}} \right)}}} $
And the integration of $\sin x$ is $\cos x + c$
Complete step by step Solution:
Given that,
Equation of the curve is \[{y^2}\left( {2a - x} \right) = {x^3}\] and the equation of the line is \[x = 2a\]
So, we have \[{y^2}\left( {2a - x} \right) = {x^3}\]
Also, written as $y = \sqrt {\dfrac{{{x^3}}}{{\left( {2a - x} \right)}}} $
Now, let $x = 2a{\sin ^2}\theta $ -------- (1)
Differentiate $x = 2a{\sin ^2}\theta $ with respect to $\theta $,
We get, $dx = 4a\sin \theta \cos \theta d\theta $ ------------ (2)
Substituting $x = 2a{\sin ^2}\theta $ in $y = \sqrt {\dfrac{{{x^3}}}{{\left( {2a - x} \right)}}} $,
It will be, $y = \sqrt {\dfrac{{{{\left( {2a{{\sin }^2}\theta } \right)}^3}}}{{\left( {2a - 2a{{\sin }^2}\theta } \right)}}} $
By simplifying, we get
$y = \sqrt {\dfrac{{\left( {8{a^3}{{\sin }^6}\theta } \right)}}{{2a\left( {1 - {{\sin }^2}\theta } \right)}}} $
Using, trigonometric identity i.e., ${\cos ^2}\theta = 1 - {\sin ^2}\theta $
It implies that, $y = \sqrt {\dfrac{{\left( {4{a^2}{{\sin }^6}\theta } \right)}}{{{{\cos }^2}\theta }}} $
Here, the square root of the term $\dfrac{{\left( {4{a^2}{{\sin }^6}\theta } \right)}}{{{{\cos }^2}\theta }}$ is $\dfrac{{\left( {2a{{\sin }^3}\theta } \right)}}{{\cos \theta }}$
So, $y = \dfrac{{\left( {2a{{\sin }^3}\theta } \right)}}{{\cos \theta }}$
Now, to calculate the area integrate $y$ with respect to $x$,
Area of the bounded region $ = \int\limits_0^{2a} {\dfrac{{\left( {2a{{\sin }^3}\theta } \right)}}{{\cos \theta }}} dx$
Using equation (2),
Area $ = 2a\int\limits_0^{2a} {\dfrac{{\left( {{{\sin }^3}\theta } \right)}}{{\cos \theta }}} 4a\sin \theta \cos \theta d\theta $
Here, the limits will also change
As we know that, $x = 2a{\sin ^2}\theta $
Therefore, at $x = 0$ we get $\theta = 0$
And at $x = 2a$, $\theta = \dfrac{\pi }{2}$
So, Area $ = 2a\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {{{\sin }^3}\theta } \right)}}{{\cos \theta }}} 4a\sin \theta \cos \theta d\theta $
Now, cancel the like terms
It will be,
Area \[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^4}\theta d\theta } \] ---------- (3)
Using, trigonometric identity i.e., ${\sin ^2}\theta = 1 - {\cos ^2}\theta $
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta \left( {1 - {{\cos }^2}\theta } \right)d\theta } \]
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta - {{\sin }^2}\theta {{\cos }^2}\theta d\theta } \]
Multiply and divide the term by $4$,
It will be, \[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta - \dfrac{{4{{\sin }^2}\theta {{\cos }^2}\theta }}{4}d\theta } \]
Using trigonometric identity i.e., $\sin 2\theta = 2\sin \theta \cos \theta $
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}\theta - \dfrac{1}{4}{{\sin }^2}2\theta d\theta } \]
Using trigonometric identity i.e., $\cos 2\theta = 1 - 2{\sin ^2}\theta $
Also written as, ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{1 - \cos 2\theta }}{2}} \right) - \dfrac{1}{4}\left( {\dfrac{{1 - \cos 4\theta }}{2}} \right)d\theta } \]
\[ = 8{a^2}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{2} - \dfrac{{\cos 2\theta }}{2} - \dfrac{1}{8} + \dfrac{{\cos 4\theta }}{8}d\theta } \]
\[ = 8{a^2}\left[ {\dfrac{\theta }{2} - \dfrac{{\sin 2\theta }}{4} - \dfrac{\theta }{8} + \dfrac{{\sin 4\theta }}{{32}}} \right]_0^{\dfrac{\pi }{2}}\]
Substituting the limits in the required integration,
\[ = 8{a^2}\left[ {\dfrac{\pi }{4} - \dfrac{{\sin 2\left( {\dfrac{\pi }{2}} \right)}}{4} - \dfrac{\pi }{{16}} + \dfrac{{\sin 4\left( {\dfrac{\pi }{2}} \right)}}{{32}}} \right]\]
As we know that, $\sin \pi = \sin 2\pi = 0$
Therefore,
Area \[ = 8{a^2}\left[ {\dfrac{\pi }{4} - 0 - \dfrac{\pi }{{16}} + 0} \right]\]
L.C.M. of $4$ and $16$ is $16$
\[ = 8{a^2}\left[ {\dfrac{{4\pi - \pi }}{{16}}} \right]\]
\[ = 8{a^2}\left[ {\dfrac{{3\pi }}{{16}}} \right]\]
\[ = \dfrac{{3\pi {a^2}}}{2}\]

Therefore, the correct option is (B).
Note:To solve such a question, one should have a good knowledge of trigonometric and integration formulas. Also, if you are changing the function and then always change the limits too. Put the limits in the same to calculate new limits. Try to open terms using trigonometric identities as much as it is possible. And to solve the limits, firstly solve the integration past then last subtract the terms (put the upper limit values and then lower limits).
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Difference Between Natural and Whole Numbers: JEE Main 2024

Hess Law of Constant Heat Summation: Definition, Formula & Applications

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Advanced 2025 Notes
