Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?

Question

I have the following parametric equation:

$$x=t^2-2t$$ $$y=\sqrt{t}$$

I'm interested finding the area of the region bounded by this curve and the y-axis (i.e. $0 \leq t \leq 2$).

We have:

$$\frac{\mathrm{d}x}{\mathrm{d}t}=2t-2$$

$$\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{1}{2\sqrt{t}}$$

Originally, I solved it like this:

$$A = -\int _a^bx\mathrm{d}y\Rightarrow -\int _0^2\left(t^2-2t\right)\left(\frac{1}{2\sqrt{t}}\right)\mathrm{d}t=\frac{8\sqrt{2}}{15}\approx 0.7542$$

I then happened to try it the other way, and was surprised that I got the same result:

$$A = \int _a^by\mathrm{d}x\Rightarrow \int _0^2\sqrt{t}\left(2t\:-2\right)\mathrm{d}t=\frac{8\sqrt{2}}{15}\approx 0.7542$$

I tried to figure out why this is giving me the same result and realized what's going on:

Let A = the region enclosed between my curve and the y-axis.
Let B = the region enclosed between my curve on $1 \le t \le 2$ and the x-axis.
Let C = the region enclosed between my curve on $0 \le t \le 1$ and the x-axis.

Therefore, we have $A = B - C$.

If we calculate B and C:

$$B=\int _1^2\sqrt{t}\left(2t-2\right)\mathrm{d}t=\frac{8\left(\sqrt{2}+1\right)}{15}\approx 1.2879$$

$$C=\int _0^1\sqrt{t}\left(2t-2\right)\mathrm{d}t=\frac{-8}{15}\approx -0.5333$$

My question is: why is C returning negative area, while B is returning positive area? They are both in the same quadrant, and both are calculating y-values from the curve to the x-axis.

Answer 1

This may or may not help:

Think of your curve as a path traced out by a moving point. The point starts at the origin at time $t=0$. As $t$ goes from $0$ to $1$, the point moves in the second quadrant, moving up and to the left until it reaches the point $(-1,1)$ at time $t=1$. Then from $t=1$ to $t=2$ it moves up and to the right until it reaches the point $(0,\sqrt2)$ at time $t=2$. The path resembles a semicircle in the second quadrant that touches the origin and the point $(2,\sqrt2)$.

Let's denote the portion of the curve from $t=0$ to $t=1$ by $C'$ and the portion from $t=1$ to $t=2$ by $B'$.

Your integral $C$ is negative because the point is traveling to the left as $t$ ranges from $t=0$ to $t=1$ ($dx/dt$ is negative). $C$ gives the negative of the area under $C'$.

Your integral $B$ is positive because the point is traveling to the right as $t$ ranges from $t=1$ to $t=2$ ($dx/dt$ is positive). $B$ gives the area under $B'$. But (draw the picture), the area under $B'$ includes the area under $C'$. Things cancel when you add the integrals $B$ and $C$ to give the correct area bounded by the "semicircle" and the $y$-axis.

Answer 2

You can see that:

$2t-2 < 0$ for $t \in (0,1)$, and $\sqrt t>0$ for $t \in (0,1)$. This is why the second integral is negative - because the curve is always below the $t$ axis.

Similarily, for the first integral, the curve is always above the $t$ axis, so the area is algebraically positive.

EDIT: David gives a physical explanation which is much more enlightning, I think. Plotting your function, which is the inverse of $y=x^4-2x^2$ will also prove useful. It will help follow his explanation.