Alternative formula for parametric integration

Question

Given a circle $x^2+y^2=R^2$ for which we would like to calculate its area

One way is using the function: $y=\sqrt{R^2-x^2}$ and integrating as follows:

$2\int_{-R}^R\sqrt{R^2-x^2}\,dx=\pi R^2$

The second way is to use parametric equation $x=R\cos t$ and $y=R\sin t$ with $0\le t \le 2\pi$ and integrate as follows:

$2\int_{\pi}^0y\frac{dx}{dt}\,dt=-2R^2\int_{\pi}^0\sin^2 t\,dt=\pi R^2$

Note that you need to integrate from $t=\pi$ to $t=0$ which corresponds to the integration from $x=-R$ to $x=R$.

However, I found another formula for parametric integration in one of my notebooks, and I don't know where does it come from, but it works like magic:

$\frac{1}{2}\int_{t_{1}}^{t_{2}} (x\frac{dy}{dt}-y\frac{dx}{dt})\,dt$

Not only that it works, with this formula you need to integrate from $t=0$ to $t=2\pi$ to get the right answer:

$\frac{1}{2}\int_{0}^{2\pi} R^2(\cos^2t+\sin^2t)\,dt=\pi R^2$.

Can someone prove this formula? And why do the integration can be done over the whole circle and not half of it?

Thank you

Answer 1

Note that when $x=R\cos t$ and $y=R\sin t$, we can see that $x=\frac{dy}{dt}$ and $y=-\frac{dx}{dt}$. Substituting these results in the third integral form gives us: $$\frac{1}{2}\int_{0}^{2\pi} (x^2+y^2) dt =\frac{1}{2}\int_{0}^{2\pi} R^2 dt = \int_{0}^{\pi} R^2 dt$$ since, $\int_{0}^{a} f(t) dt = \frac{1}{2}\int_{0}^{2a} f(t) dt$ when $f(2a-t)=f(t)$.

Answer 2

The following formula:

$$\frac{1}{2}\int_{0}^{2\pi} R^2(\cos^2t+\sin^2t)\,dt=\pi R^2$$

is simply obtained by integration from $0 \to 2\pi$ of the infinitesimal slice:

$$\frac12 \rho \cdot \rho \cdot d\theta=\frac12 (R^2 \cos^2 \theta+R^2 \sin^2 \theta)\cdot d\theta=\frac12 R^2 (\cos^2 \theta+\sin^2 \theta) \cdot d\theta=\frac12 R^2 \cdot d\theta$$