Question

What is the area bounded by the curve $y^2\left(2a-x\right)=x^3$ and the line $x=2a$

• A. $3{\mathrm{\pi a}}^2$ sq units
• B. $\frac{3{\mathrm{\pi a}}^2}{2}$ sq units
• C. $\frac{3{\mathrm{\pi a}}^2}{4}$ sq units
• D. $\frac{6{\mathrm{\pi a}}^2}{5}$ sq units

Answer 1

The correct option is B

$\frac{3{\mathrm{\pi a}}^2}{2}$ sq units

Explanation of correct answer :

Finding area within the curves:

The graph of the curves is as shown:

Given, $y^2\left(2a-x\right)=x^3$

$\Rightarrow y=\pm \sqrt{\frac{x^3}{2a-x}}$

Let, $x=2a{\sin }^2\varnothing$

so, $dx=4a\sin \varnothing \cos \varnothing d\varnothing$

Required area $={\int }_0^{2a}\sqrt{\frac{x^3}{2a-x}}dx$

$={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\frac{{\left(2a{\sin }^2\varnothing \right)}^3}{2a-2a{\sin }^2\varnothing }}4a\sin \varnothing \cos \varnothing d\varnothing$

$={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\frac{{\left(2a{\sin }^2\varnothing \right)}^3}{2a(1-{\sin }^2\varnothing )}}4a\sin \varnothing \cos \varnothing d\varnothing$

$$\begin{gathered} ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\frac{8a^3{\sin }^6\varnothing }{2a{\cos }^2\varnothing }}4a\sin \varnothing \cos \varnothing d\varnothing \mathbf{}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{\varnothing }\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\varnothing }\mathbf{)} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{4a^2{\sin }^6\varnothing }}{{\cos }^2\varnothing }4a\sin \varnothing \cos \varnothing d\varnothing \\ =8a^2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{{\sin }^6\varnothing }\sin \varnothing d\varnothing \\ =8a^2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{{\sin }^6\varnothing }\sqrt{{\sin }^2\varnothing d\varnothing } \\ =8a^2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{{\sin }^8\varnothing }\mathbf{}\mathbf{[}\mathbf{\because }{\mathit{a}}^{\mathbf{n}}\mathbf{\times }{\mathit{a}}^{\mathbf{m}}\mathbf{=}{\mathit{a}}^{\mathbf{n}\mathbf{+}\mathbf{m}}\mathbf{]} \\ =8a^2{\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^4\varnothing d\varnothing \\ =\frac{3{\mathrm{\pi a}}^2}{2}\mathbf{}\mathbf{(}{\mathbf{\int }}_{\mathbf{0}}^{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{sin}}^{\mathbf{n}}\mathbf{x}\mathbf{d}\mathbf{x}\mathbf{=}\{\begin{array}{l}\frac{\left(n-1\right)\left(n-3\right)\left(n-5\right)...1}{n\left(n-2\right)\left(n-4\right)...2}\left(\frac{\pi }{2}\right)n=\mathrm{even}\\ \frac{\left(n-1\right)\left(n-3\right)\left(n-5\right)...2}{n\left(n-2\right)\left(n-4\right)...3}n=\mathrm{odd}\end{array}\mathbf{)} \end{gathered}$$

Thus, the area within the curves is $\frac{3{\mathrm{\pi a}}^2}{2}$sq units.

Hence, option (B) is the correct answer.