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.