Finding the area by Green's Theorem

Question

If I have

$\ x=\sin^3t$ and $\ y = \cos^3t$

I'm using Green's theorem to calculate the area enclosed by the curve via integration, but at the end I'm getting negative area $-3\pi/8$, which is wrong.

Do I need to calculate by the absolute value? I know that if $\ x=\cos^3t$ and $\ y = \sin^3t$, the answer will be $3\pi/8$, but what to do with this? Why does the sign change value for different paramterization?

This is the formula, which I used; $\frac{1}{2}$ $\int_0^{2π }(x(t).y'(t)-x'(t).y(t))dt $

Answer 1

The reason for obtaining a negative answer is the fact that when you parametrize using the first method, you are moving in the clockwise direction. Thus, the value you will obtain will be negative.

In the second parametrization, the direction you are moving in is the counterclockwise direction, which is the default convention followed while using Green's Theorem. That is why you are obtaining a positive value.

Answer 2

$$P(x,y)=(\sin^3t,\cos^3t)$$ At $t=0$, $P$ is $A(0,1).$
At $t=\frac\pi 2$, $P$ is $B(1,0)$.
At $t=\pi$, $P$ is $C(0,-1)$.
At $t=\frac{3\pi}2$, $P$ is $D(-1,0)$.
At $t=2\pi$, $P$ is $A(1,0)$ again.

We see that orientation is clockwise. In Green's theorem, the contour is oriented counter clockwise. Therefore the integration route will be $BADCB$ not $ABCDA$. So, $$\int_{C}\frac12(xdy-ydx)\\=\int_C\frac{3\cos(4t)-3}{16}dt\\=4 \int_{\overset{\frown}{BA}}\frac{3\cos(4t)-3}{16}dt\\ =\int_{\frac\pi 2}^0\frac{3\cos(4t)-3}{4}dt=\frac{3\pi}{8}. $$