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 :)
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.