SOLUTION If we first integrate with respect to \(x\), we get \[
\iint_R y \sin(xy) dA = \int_0^\pi \int_1^2 y \sin(xy) dx dy =
\int_0^\pi [-\cos(xy)]_{x=1}^{x=2} dy
\] \[
= \int_0^\pi (-\cos 2y + \cos y) dy = \left[ -\frac{1}{2}\sin 2y + \sin
y \right]_0^\pi = 0
\] NOTE If we reverse the order of integration and first
integrate with respect to \(y\) in
Example 6, we get \[
\iint_R y \sin(xy) dA = \int_1^2 \int_0^\pi y \sin(xy) dy dx
\] but this order of integration is much more difficult than the
method given in the example because it involves integration by parts
twice. Therefore, when we evaluate double integrals it is wise to choose
the order of integration that gives simpler integrals.