Question

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

Options

• πa²

• \[\frac{3}{2}\pi a^2\]

• 2πa²

• 3πa²

Answer 1

\[\frac{3}{2}\pi a^2\]

 

\[\begin{array}{l}y^2 \left( 2a -x \right) = x^3 \\ y = \sqrt{\frac{x^3}{2a - x}}\end{array}\]
Let x = 2a sin²θ
    dx = 4a sinθ cosθ dθ
\[\begin{array}{l}\text{ Area }= \hspace{0.167em} \int\limits_0^{2a} \sqrt{\frac{x^3}{2a - x}}dx \\ = \int_0^\frac{\pi}{2} \sqrt{\frac{\left( 8 a^3 \right) \sin^6 \theta}{\left( 2a \right)\hspace{0.167em} \cos^2 \theta}}\hspace{0.167em} \cdot \left( 4a \right)\hspace{0.167em}\sin\theta\hspace{0.167em}\cos\theta\hspace{0.167em}d\theta \\ = 8 a^2 \int_0^\frac{\pi}{2} \sqrt{\sin^6 \theta}\hspace{0.167em}\sin\theta\hspace{0.167em}d\theta \\ = 8 a^2 \left[ \int_0^\frac{\pi}{2} \sin^4 \theta\hspace{0.167em}d\theta \right] \\ = 8 a^2 \left[ \int_0^\frac{\pi}{2} \sin^2 \theta\left( 1- \cos^2 \theta \right)\hspace{0.167em}d\theta \right] \\ = 8 a^2 \left[ \int_0^\frac{\pi}{2} \left( \frac{1 - \cos2\theta}{2} \right) d\theta - \frac{1}{4} \int_0^\frac{\pi}{2} \sin^2  2\theta \text{ d}\theta \right] \\ = 8 a^2 \left[ \frac{1}{2} \left[ \theta \right]_0^\frac{\pi}{2} - \left[ \frac{\sin2\theta}{4} \right]_0^\frac{\pi}{2} \right] - \frac{1}{4}\left[ \int_0^\frac{\pi}{2} \frac{1 - \cos 4\theta}{2}d\theta \right] \\ = 8 a^2 \left[ \left( \frac{\pi}{4} \right) - 0 \right] - \frac{1}{4}\left[ \frac{\pi}{4} - 0 \right] \\ = 8 a^2 \left[ \frac{\pi}{4} - \frac{\pi}{16} \right] = \frac{3}{2}\pi a^2\end{array}\]