Section 15.2: Integration over general 2d regions

SOLUTION Since \(-1 \le \sin x \le 1\) and \(-1 \le \cos y \le 1\), we have \(-1 \le \sin x \cos y \le 1\) and therefore \[ e^{-1} \le e^{\sin x \cos y} \le e^1 = e \] Thus, using \(m = e^{-1} = 1/e\), \(M = e\), and \(A(D) = \pi(2)^2 = 4\pi\) in Property 11, we obtain \[ \frac{4\pi}{e} \le \iint_D e^{\sin x \cos y} dA \le 4\pi e \]