Question

Area enclosed between the curve y² (2a − x) = x³ and the line x = 2a above x-axis is
(a) πa²

(b) $\frac{3}{2}\mathrm{\pi }{a}^2$

(c) 2πa²

(d) 3πa²

Answer 1

$\begin{array}{l}{y}^2\left(2a-x\right)=x^3\\ y=\sqrt{\frac{x^3}{2a-x}}\end{array}$
Let x = 2a sin²θ
dx = 4a sinθ cosθ dθ

$\begin{array}{l}\mathrm{Area}=\hspace{0.17em}\underset{0}{\overset{2a}{\int }}\sqrt{\frac{x^3}{2a-x}}dx\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\frac{\left(8a^3\right){\sin }^6\theta }{\left(2a\right)\hspace{0.17em}{\cos }^2\theta }}\hspace{0.17em}·\left(4a\right)\hspace{0.17em}\sin \theta \hspace{0.17em}\cos \theta \hspace{0.17em}\mathrm{d}\theta \\ =8a^2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{{\sin }^6\theta }\hspace{0.17em}\sin \theta \hspace{0.17em}\mathrm{d}\theta \\ =8a^2\left[{\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^4\theta \hspace{0.17em}\mathrm{d}\theta \right]\\ =8a^2\left[{\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^2\theta \left(1-{\cos }^2\theta \right)\hspace{0.17em}\mathrm{d}\theta \right]\\ =8a^2\left[{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{1-\cos 2\theta }{2}\right)\mathrm{d}\theta -\frac{1}{4}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^{2}2\theta \mathrm{d}\theta \right]\\ =8a^2\left[\frac{1}{2}{\left[\theta \right]}_0^{\frac{\mathrm{\pi }}{2}}-{\left[\frac{\sin 2\theta }{4}\right]}_0^{\frac{\mathrm{\pi }}{2}}\right]-\frac{1}{4}\left[{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1-\cos 4\theta }{2}d\theta \right]\\ =8a^2\left[\left(\frac{\mathrm{\pi }}{4}\right)-0\right]-\frac{1}{4}\left[\frac{\mathrm{\pi }}{4}-0\right]\\ =8a^2\left[\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{16}\right]=\frac{3}{2}\mathrm{\pi }{a}^2\end{array}$