Area of curve in parametric equation Given the curve defined by the parametric equations:
$$
x=7\cos{3t}\\
y=7\sin{3t}\\
0\le t\le2\pi
$$
What is the area of the region bounded by this curve?
Clearly, 
$$
x^2+y^2=(7\cos{3t})^2+(7\sin{3t})^2=7^2
$$
which is a circle centered at the origin and radius 7. Therefore, the area is $\pi\cdot7^2=49\pi$.
However, it is also true that the area is
$$
A=\int{ydx}=\int_{0}^{2\pi}{y(t)x'(t)dt}=\int_{0}^{2\pi}{7\sin{3t}(-7\sin3t)3dt}=-49\int_{0}^{2\pi}{\sin^2{3t}\;3dt}=-49\cdot(3\pi)=-3(49\pi)
$$
which is the same area as before, but multiplied by $-3$. 
What is answer is correct? The $3t$ in the $\sin$ and $\cos$ terms means that as $t$ goes from $0$ to $2\pi$, the circle is traversed $3$ times, right? Which would explain why the area is multiplied by 3. But what about the minus sign?

In your second way of computing it, you have gone around the circle three times instead of one; thus you have (erroneously) tripled the answer. Also, there is a difference between clockwise and counterclockwise and you clearly got unlucky there, to get a negative area.
