The area between the curve y^2 (a + x) = (a - | vedclass

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(C) The given curve is $y^2 = \frac{(a-x)^3}{a+x}$.As $x 	o -a$,$y 	o \infty$,so $x = -a$ is the vertical asymptote.The curve intersects the $x$-axi

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$3\pi a^2$

Vedclass · Answer

(C) The given curve is $y^2 = \frac{(a-x)^3}{a+x}$.As $x 	o -a$,$y 	o \infty$,so $x = -a$ is the vertical asymptote.The curve intersects the $x$-axis at $x = a$ (where $y=0$).The area $A$ is given by $2 \int_{-a}^{a} y \, dx = 2 \int_{-a}^{a} \sqrt{\frac{(a-x)^3}{a+x}} \, dx$.Let $x = a \cos 	heta$,then $dx = -a \sin 	heta \, d	heta$.When $x = -a, 	heta = \pi$. When $x = a, 	heta = 0$.$A = 2 \int_{\pi}^{0} \sqrt{\frac{(a - a \cos 	heta)^3}{a + a \cos 	heta}} (-a \sin 	heta) \, d	heta = 2a^2 \int_{0}^{\pi} \sqrt{\frac{a^3(1-\cos 	heta)^3}{a(1+\cos 	heta)}} \sin 	heta \, d	heta$$A = 2a^2 \int_{0}^{\pi} \frac{(1-\cos 	heta) \sqrt{1-\cos 	heta}}{\sqrt{1+\cos 	heta}} \sin 	heta \, d	heta = 2a^2 \int_{0}^{\pi} (1-\cos 	heta) \frac{\sin(	heta/2)}{\cos(	heta/2)} (2 \sin(	heta/2) \cos(	heta/2)) \, d	heta$$A = 4a^2 \int_{0}^{\pi} (1-\cos 	heta) \sin^2(	heta/2) \, d	heta = 4a^2 \int_{0}^{\pi} (2 \sin^2(	heta/2)) \sin^2(	heta/2) \, d	heta = 8a^2 \int_{0}^{\pi} \sin^4(	heta/2) \, d	heta$.Using $\int_{0}^{\pi} \sin^4(	heta/2) \, d	heta = 2 \int_{0}^{\pi/2} \sin^4 \phi \, d\phi = 2 	imes \frac{3 	imes 1}{4 	imes 2} 	imes \frac{\pi}{2} = \frac{3\pi}{8}$.Thus,$A = 8a^2 	imes \frac{3\pi}{8} = 3\pi a^2$.

The area between the curve $y^2 (a + x) = (a - x)^3$ and its vertical asymptote is

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