How can the following integral be evaluated without using Green's Theorem?

Question

How can the following integral be solved without using Green's Theorem and without converting it into a line integral?
$\iint_{R}(-1)dxdy$
where R is the region enclosed by $x=\cos(t)$, $y=2\sin(t)$, and $t$ varies from
$t=0$ to $t=2\pi$

How can the Jacobian be evaluated in this case?

Answer 1

No Jacobian, it is just \begin{align} \iint_R (-1)\,dx\,dy=\int_{-1}^1\left(\int_{-2\sqrt{1-x^2}}^{2\sqrt{1-x^2}}(-1)\,dy\right)\,dx=-4\int_{-1}^1\sqrt{1-x^2}\, dx=-2\pi \end{align}