Parametric integration negative area? I know there is a question very similar to mine already here Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?
, but I am still a bit unsure after reading the answers as to when the integral gives you a positive area and when it gives you a negative area. My text book says that the area is positive if t traces out the curve clockwise and negative for anticlockwise, however I think there is more going on with regards to whether the curve is above or below the x axis.
Since $A=\int_a^b y\frac{dx}{dt}dt$, my guess is that you get a positive answer if either the graph is above the x axis (y is positive) and the point is moving to the right (dx/dt positive) or the graph is below the x axis and the point is moving to the left. Then you get a negative answer if either the graph is below the x axis but is being traced out moving to the right, or the graph is above the x axis but is being traced out moving to the left.
Was my book wrong? Is my thinking correct?
Thank you in advance :)
 A: Your thinking is correct, but the book is wrong in general.
Assuming that $t\in [a,b]$ with $a<b$, then considering $\displaystyle \int_a^b y \frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y \dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $\dfrac{dx}{dt}$ are both positive, or (b) both are negative. 
(ii) The integrand $y \dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $\dfrac{dx}{dt} < 0$, vice-versa. 
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y \dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point. 
Consider the parametric curve $x=t, y=t^2$ for $t\in [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $\displaystyle \int_a^b y \frac{dx}{dt}dt = \dfrac{1}{3}$, which is positive. 
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=r\cos\theta, y=r\sin\theta$. We find that $r\geq0$ and $0\leq \theta < \dfrac{\pi}{2}$. It's quite easy to show that $\dfrac{d\theta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout. 
Compare this to $x=\cos\theta, y=\sin\theta$ for $\theta\in\left[0,\dfrac{\pi}{2}\right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-\dfrac{\pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y \dfrac{dx}{dt}$, nothing more.
A: Your thinking is correct, and your book is not wrong.
You say that that the integral $\int_a^b y \frac{dx}{dt}\,dt$ is positive when either $y>0$ and $\frac{dx}{dt}>0$ (above the $x$-axis, moving left-to-right), or $y < 0$ and $\frac{dx}{dt} < 0$ (below the $x$-axis, moving right-to-left).
In both of these cases, the particle moves clockwise with respect to the origin.  
A: The integral works because the assumption is that the integral is around an enclosed 2D area where the outside curve does not cross itself.  Thus it really doesn't matter where the area is with respect to the origin of the space, because the sum will always be the same unless we rotate the curve, then things change.  The equation works because as we march to the right (positive X), we are adding the area and as we march to the left (negative X), we are subtracting the area.  Then clockwise curves will yield A greater than zero and counter clockwise will yield A less than zero.
