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

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KollegeApply · Accepted Answer

$\frac{3}{2}$$\pi$ a^2 \begin{array}{l}y^2 \left( 2a -x \right) = x^3 \ y = $\sqrt$(x^$\frac{3}{2}$a - x)\end{array} Let x = 2a sin2$\theta$ dx = 4a sin$\theta$ cos$\theta$ d$\theta$ \begin{array}{l}\text{ Area }= \hspace{0.167em} \int\limits_0^{2a} $\sqrt$(x^$\frac{3}{2}$a - x)dx \ = \int_0^$\pi$/2 \sqrt{\frac{\left( 8 a^3 \right) \sin^6 $\theta$}{\left( 2a \right)\hspace{0.167em} \cos^2 $\theta$}}\hspace{0.167em} · \left( 4a \right)\hspace{0.167em}\sin$\theta$\hspace{0.167em}\cos$\theta$\hspace{0.167em}d$\theta$ \ = 8 a^2 \int_0^$\pi$/2 $\sqrt$(\sin^6 $\theta$)\hspace{0.167em}\sin$\theta$\hspace{0.167em}d$\theta$ \ = 8 a^2 \left[ \int_0^$\pi$/2 \sin^4 $\theta$\hspace{0.167em}d$\theta$ \right] \ = 8 a^2 \left[ \int_0^$\pi$/2 \sin^2 $\theta$\left( 1- \cos^2 $\theta$ \right)\hspace{0.167em}d$\theta$ \right] \ = 8 a^2 \left[ \int_0^$\pi$/2 \left( \frac{1 - \cos2$\theta$}{2} \right) d$\theta$ - $\frac{1}{4}$ \int_0^$\pi$/2 \sin^2 2$\theta$ \text{ d}$\theta$ \right] \ = 8 a^2 \left[ $\frac{1}{2}$ \left[ $\theta$ \right]_0^$\pi$/2 - \left[ \sin2$\theta$/4 \right]_0^$\pi$/2 \right] - $\frac{1}{4}$\left[ \int_0^$\pi$/2 \frac{1 - \cos 4$\theta$}{2}d$\theta$ \right] \ = 8 a^2 \left[ \left( $\pi$/4 \right) - 0 \right] - $\frac{1}{4}$\left[ $\pi$/4 - 0 \right] \ = 8 a^2 \left[ $\pi$/4 - $\pi$/16 \right] = $\frac{3}{2}$$\pi$ a^2\end{array} [Solution diagram: 44112_solution_1.png]

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