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 :
$$\frac{{3\pi ab}}{{32}}$$
Solution :
$$y = 0,$$ when $$t = 0$$ and then $$x = a$$
So desire area
$$\eqalign{
& A = \int\limits_0^a {y\,dx} \cr
& \Rightarrow A = \int\limits_{\frac{\pi }{2}}^0 {b\,{{\sin }^3}t\left( { - 3a\,{{\cos }^2}t\,\sin \,t\,dt} \right)} \cr
& \Rightarrow A = 3ab\int\limits_0^{\frac{\pi }{2}} {{{\sin }^4}t\,{{\cos }^2}t\,dt} \cr
& \Rightarrow A = 3ab\int\limits_0^{\frac{\pi }{2}} {{{\cos }^4}t\,{{\sin }^2}t\,dt} \cr
& \therefore \,2A = 3ab\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}t\,{{\sin }^2}t\,dt} \cr
& \Rightarrow 2A = 3ab.\frac{\pi }{{16}} \cr
& \Rightarrow A = \frac{{3\pi ab}}{{32}} \cr} $$