How does this double integral make sense? Am I right in thinking that $dt^2$ essentially just makes the integral $0$ no matter what the limits are?

Question

I came across this interesting question that I'm having some difficulty in conceptually understanding:

Say we are integrating a region to find the area enclosed by the curve $C$ given by $x = \sin(2t)$ and $y = \sin(t)$, where $t$ is a parameter in the range $0 \le t \le \pi$. Trying to do this in Cartesian coordinates by using the double integral $\iint_C \ dx dy$ isn't the right way, so I tried to switch coordinates by differentiating $x$ and $y$ with respect to $t$. But then I inevitably end up with the integral $\iint_C 2\cos(2t)cos(t) \ dt^2$. How exactly does this make sense? Am I right in thinking that $dt^2$ essentially just makes the integral $0$ no matter what the limits are (since $dt^2$ is even smaller than $dt$)?

Any comments would be much appreciated. Thanks.

EDIT : to make the substitutions I used the fact that $dx/dt = 2cos(2t)$ and $dy/dt = cos(t)$ and then just subbed in dx and dy from there.

Answer 1

You make a confusion between your double integral and the fact $\int_a^b f(x)dx$ gives you the area between the graph of $f$ and the $x$ axis. You cannot plug the expression for the boundary of your domain into the integral for the interior of the domain. There is a device for that which is Green-Riemann formula. Here it is much simpler to make the following observations.

1. plot the boundary of $C$ it is not difficult here and it is interesting.
2. remark that $y\geq 0$ and that $x= 2 \sin(t) \cos(t)=\pm2 y \sqrt{1-y^2}$. Therefore, for $x\geq 0$, $C$ is the graph of the function $x=f(y)=2 y \sqrt{1-y^2}$.

Now you should be able to see that the area that you are looking for is given by $4 \int_0^1 y \sqrt{1-y^2}dy=4/3$.