Question

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 1

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).