The area (in sq units) enclosed by the loop o | vedclass

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(D) The given curve is $ay^2 = x^2(a - x)$,where $a > 0$.Since the curve is symmetric about the $x$-axis,the total area is twice the area above the $x

Vedclass · Accepted Answer

$\frac{8}{15} a^2$

Vedclass · Answer

(D) The given curve is $ay^2 = x^2(a - x)$,where $a > 0$.Since the curve is symmetric about the $x$-axis,the total area is twice the area above the $x$-axis.From the equation,$y^2 = \frac{x^2(a - x)}{a}$,so $y = \pm x \sqrt{\frac{a - x}{a}}$.The loop exists for $x \in [0, a]$.The required area $A$ is given by:$A = 2 \int_0^a y \, dx = 2 \int_0^a x \sqrt{\frac{a - x}{a}} \, dx$$A = \frac{2}{\sqrt{a}} \int_0^a x \sqrt{a - x} \, dx$Let $a - x = t^2$,then $dx = -2t \, dt$. When $x = 0, t = \sqrt{a}$ and when $x = a, t = 0$.$A = \frac{2}{\sqrt{a}} \int_{\sqrt{a}}^0 (a - t^2) t (-2t \, dt) = \frac{4}{\sqrt{a}} \int_0^{\sqrt{a}} (at^2 - t^4) \, dt$$A = \frac{4}{\sqrt{a}} \left[ \frac{at^3}{3} - \frac{t^5}{5} \right]_0^{\sqrt{a}}$$A = \frac{4}{\sqrt{a}} \left( \frac{a(\sqrt{a})^3}{3} - \frac{(\sqrt{a})^5}{5} \right) = \frac{4}{\sqrt{a}} \left( \frac{a^2 \sqrt{a}}{3} - \frac{a^2 \sqrt{a}}{5} \right)$$A = 4 \left( \frac{a^2}{3} - \frac{a^2}{5} \right) = 4 \left( \frac{5a^2 - 3a^2}{15} \right) = 4 \left( \frac{2a^2}{15} \right) = \frac{8}{15} a^2$.Thus,the correct option is $(d)$.

The area (in sq units) enclosed by the loop of the curve $ay^2 = x^2(a - x), (a > 0)$ is

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