The area enclosed between the curve {y^2}(2a | vedclass

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(B) The given curve is ${y^2}(2a - x) = {x^3}$.Since the power of $y$ is even,the curve is symmetric about the $x$-axis.For $y$ to be real,we must hav

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$\frac{3\pi {a^2}}{2}$

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(B) The given curve is ${y^2}(2a - x) = {x^3}$.Since the power of $y$ is even,the curve is symmetric about the $x$-axis.For $y$ to be real,we must have $\frac{{x^3}}{{2a - x}} \ge 0$. This implies $0 \le x The line $x = 2a$ is an asymptote to the curve.The area above the $x$-axis is given by $A = \int_0^{2a} y \, dx = \int_0^{2a} \sqrt{\frac{x^3}{2a - x}} \, dx$.Let $x = 2a \sin^2 	heta$,then $dx = 4a \sin 	heta \cos 	heta \, d	heta$.When $x = 0, 	heta = 0$ and when $x = 2a, 	heta = \frac{\pi}{2}$.$A = \int_0^{\pi/2} \sqrt{\frac{8a^3 \sin^6 	heta}{2a \cos^2 	heta}} \cdot 4a \sin 	heta \cos 	heta \, d	heta = \int_0^{\pi/2} \frac{2a \sqrt{2a} \sin^3 	heta}{\sqrt{2a} \cos 	heta} \cdot 4a \sin 	heta \cos 	heta \, d	heta = 8a^2 \int_0^{\pi/2} \sin^4 	heta \, d	heta$.Using Wallis' formula,$\int_0^{\pi/2} \sin^4 	heta \, d	heta = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$.Thus,$A = 8a^2 \cdot \frac{3\pi}{16} = \frac{3\pi {a^2}}{2}$.Therefore,the correct option is $B$.

The area enclosed between the curve ${y^2}(2a - x) = {x^3}$ and the line $x = 2a$ above the $x$-axis is

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