The area of the curve xy^2 = a^2(a - x) bound | vedclass

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(A) The equation of the curve is $xy^2 = a^2(a - x)$,which implies $y^2 = \frac{a^2(a - x)}{x}$.Since the curve is symmetrical about the $x$-axis,the

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$\pi a^2$

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(A) The equation of the curve is $xy^2 = a^2(a - x)$,which implies $y^2 = \frac{a^2(a - x)}{x}$.Since the curve is symmetrical about the $x$-axis,the total area $A$ is twice the area above the $x$-axis.$A = 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{a} a \sqrt{\frac{a - x}{x}} \, dx$.Let $x = a \sin^2 	heta$,then $dx = 2a \sin 	heta \cos 	heta \, d	heta$.When $x = 0, 	heta = 0$ and when $x = a, 	heta = \frac{\pi}{2}$.$A = 2 \int_{0}^{\pi/2} a \sqrt{\frac{a - a \sin^2 	heta}{a \sin^2 	heta}} (2a \sin 	heta \cos 	heta) \, d	heta$$A = 4a^2 \int_{0}^{\pi/2} \frac{\cos 	heta}{\sin 	heta} \sin 	heta \cos 	heta \, d	heta$$A = 4a^2 \int_{0}^{\pi/2} \cos^2 	heta \, d	heta$Using the property $\int_{0}^{\pi/2} \cos^2 	heta \, d	heta = \frac{\pi}{4}$,we get:$A = 4a^2 	imes \frac{\pi}{4} = \pi a^2$.

The area of the curve $xy^2 = a^2(a - x)$ bounded by the $y$-axis is

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