Calculate: $\int\int_D e^{-(x^2/a^2 + y^2/b^2)}dxdy$ on the exterior of the $x^2/a^2+y^2/b^2=1$, how do I get the exterior of this? I know the parametrization is:
$$x=ar\cos t$$ $$y=br\sin t$$ and the Jacobian $J=abr$
How do I get the exterior of that ellipse?
Is it just that $\frac {x^2}{a^2}+\frac {y^2}{b^2}\geq1?$ and then $r\in[1,\infty], t\in [0,2\pi]?$ In that case the result of the double integral would be: $\boxed{\frac {-ab\pi}{e}}$. Is that correct?
$$\int\int_D e^{-(x^2/a^2 + y^2/b^2)}dxdy=$$
$$=ab\int_0^{2\pi}\int_1^{+\infty} e^{-r^2}r\,\mathbb dr\,\mathbb d\theta=$$
$$=-\dfrac{ab}{2}\int_0^{2\pi}\int_1^{+\infty} e^{-r^2}(-2r)\,\mathbb dr\,\mathbb d\theta=$$
$$=-\dfrac{ab}{2}\int_0^{2\pi}\left[ e^{-r^2}\right]_1^{+\infty}\mathbb d\theta=$$
$$=-\dfrac{ab}{2}\int_0^{2\pi}\left[ e^{-\infty}-e^{-1}\right]\mathbb d\theta=$$ $$=\dfrac{ab}{2e}\int_0^{2\pi}\mathbb d\theta$$