Question

The area enclosed by the curves $x=a\sin 2t$, $y=a\sin t$ is given by

• A. $\frac{2a^3}{3}$
• B. $\frac{8a^2}{3}$
• C. $\frac{5a^2}{3}$
• D. $\frac{4a^2}{3}$

Answer 1

The correct option is B $\frac{8a^2}{3}$

Given $x=a\sin 2t$, $y=a\sin t$
$\therefore$ $x=2a\sin t\cos t$
$=2y\sqrt{(1-\frac{y^2}{a^2})}\Rightarrow a^2{x}^2=4a^2(a^2-y^2)$(Note:curve is symmetrical about both the axes)
$\therefore$ Curve intersect y-axis at $(0,\pm a)$, passes through origin (0, 0) and tangents origin (0, 0) are $x=\pm 2y$. Curve has two loops, the upper loops is traced as t varies 0 to $\pi$ ($\because$ y=x=0 when t=0 or $t=\pi$)
Now $x\frac{dy}{dt}-y\frac{dx}{dt}=\left(a\sin 2t\right)\left(a\cos t\right)-\left(a\sin t\right)\left(2a\cos 2t\right)=2a^2\sin t{\cos }^{2}t-2a^2\sin t\cos 2t$
$=2a^2\sin t[{\cos }^{2}t-\cos 2t]=2a^2\sin t[{\cos }^{2}t-2{\cos }^{2}t+1]=2a^2{\sin }^{3}t$
Now required area of one loop is
$=\frac{1}{2}{\int }_0^{\pi }(x\frac{dy}{dt}-y\frac{dx}{dt})dt=\frac{1}{2}{\int }_0^{\pi }2a^2{\sin }^{3}tdt$
$=a^2{\int }_0^{\pi }{\sin }^{3}tdt=a^2\left(2\right){\int }_0^{\pi /2}{\sin }^{3}tdt$
$=2a^2\times \frac{2}{3}=\frac{4a^2}{3}$ $(\because {\int }_0^{\pi /2}{\sin }^{3}tdt=\frac{2}{3})$
$\therefore$ Total area of both loops $=2\times \frac{4a^2}{3}=\frac{8a^2}{3}$ ($\because$ curve is symmetrical about both axes)