Question

The area enclosed by the curve $x = a\, cos^3\, t,\,, y = b\,sin^3\, t$ and the positive directions of x-axis and y-axis is

• A. $\frac{\pi ab}{4}$
• B. $\frac{\pi ab}{32}$
• C. $\frac{3\pi ab}{32}$
• D. $\frac{5\pi ab}{32}$

Answer 1

Answer: (C)

$y = 0$, when $t = 0$ and then $x = a$
So desire area
$A = \int\limits^{a}_{0} ydx = \int\limits^{0}_{\pi/2}b\,sin^{3}\,t\left(-3a\,cos^{2}\,t\,sin\,tdt\right)$
$A = 3ab \int\limits^{\pi/2}_{0}sin^{4} \,t \,cos^{2}\, t \,dt = 3ab \int\limits^{\pi /2}_{0}cos^{4} \,t \,sin^{4}\, tdt $
$\therefore 2A = 3ab \int\limits^{\frac{\pi}{2}}_{0}cos^{2}\, t \,sin^{2}\, tdt = 3ab\cdot\frac{\pi}{16} \Rightarrow A = \frac{3\pi ab}{32}$