Area enclosed by parametric curve Let $c(t)=(\cos^3t,\sin^3t)$. We want to find the area enclosed by this curve.
$$\begin{align}
\ A & = 4\int_{0} ^{1} {ydx} \\
    & = 4\int_{0} ^{\pi/2} {(\sin^3t)(-3\sin t \cos^2t) dt} \\
    & = ... \\
    & = -\frac{3 \pi}{8} \\
\end{align}$$
So I'm guessing the correct answer should be $\frac{3 \pi}{8}$.
Why is the resulting area negative? Are the limits wrong?
 A: Your first and second integrals aren't exactly equal to each other — they are negatives of each other, and that's why your answer came out negative:
$$\int_{0}^{1}y\,dx\neq\int_{0}^{\pi/2}(\sin^3t)(-3\sin t\cos^2t)\,dt$$
because in fact
$$\int_{0}^{1}y\,dx=-\int_{0}^{\pi/2}(\sin^3t)(-3\sin t\cos^2t)\,dt.$$
The reason is that when $t$ ranges from $0$ to $\pi/2$, the first-quarter portion of the graph is traced from right to left, i.e. in the negative direction, so $dx=(-3\sin t\cos^2t)\,dt$ is negative there. That's why you end up finding the value whose absolute value is equal to the area, but whose sign is negative — which reflects the direction of the curve.
A: Parametric curves are just screaming out to be solved in the complex plane. Consider that
$$
z=\cos^3(t)+i\sin^3(t)\\
A=\frac{1}{2}\int\mathfrak{Im}\{z^* \dot z\}\ dt
$$
$$
z^*=\cos^3(t)-i\sin^3(t)\\
\dot z=3[-\cos^2(t)\sin(t)+i\sin^2(t)\cos(t)]\\
\mathfrak{Im}\{z^* \dot z\}=3\cos^2(t)\sin^2(t)
$$
Therefore (corrected from an earlier version that only considered the $1^{st}$ quadrant)
$$A=\frac{4}{2}\int_0^{\pi/2}3\cos^2(t)\sin^2(t)dt=\frac{12}{2}\frac{1}{32}[4x-\sin(4x)]\big|_0^{\pi/2}=\frac{3\pi}{8}$$
This result has been verified numerically.
