= Solution
The annulus is open and hence <Lebesgue measurable>; multiplying its indicator by the continuous Gaussian gives a measurable function. It is bounded by the indicator of $[-b,b]^2$, whose <Lebesgue integral> is $4b^2$. Thus it is <Lebesgue integrable> by comparison.
Use the <change of variables formula> with <polar coordinates> $x=r\cos\theta$, $y=r\sin\theta$. Removing one radial cut makes this a smooth one-to-one coordinate map; the cut is a <null set> and does not affect the integral. The absolute <Jacobian determinant> is $r$. By <Fubini theorem>,
$$
\int_V e^{-(x^2+y^2)/2}\,d(x,y)=\int_0^{2\pi}\int_a^b e^{-r^2/2}r\,dr\,d\theta
=\boxed{2\pi(e^{-a^2/2}-e^{-b^2/2})}.
$$
The antiderivative is $-e^{-r^2/2}$, which also makes the sign of the answer transparent.
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