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 🙂
Best Answer
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.