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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?

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  • $\begingroup$ This is the negative of the area of the ellipse $x^2 + (y/2)^2 = 1$. $\endgroup$ – Tob Ernack Jul 12 '17 at 11:03
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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}

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