Question

set up integrals for both orders of integration. use the more convenient order to evaluate the integral over the plane region r.

\\( \int \int _ { r } \sin x \sin y d a \\)

r: rectangle with vertices \\( ( - \pi, 0 ), ( \pi, 0 ), ( \pi, \pi / 6 ), ( - \pi, \pi / 6 ) \\)

note that the given region r is bounded by the lines \\( x = \\)

, \\( x = \\)

, \\( y = 0 \\), and \\( y = \\)

.

Explanation:

Step1: Identify x - bounds

The vertices are \((-\pi, 0)\), \((\pi, 0)\), \((\pi, \pi/6)\), \((-\pi, \pi/6)\). For the \(x\) - coordinates, the minimum \(x\) - value is \(-\pi\) and the maximum \(x\) - value is \(\pi\). So \(x=-\pi\) and \(x = \pi\).

Step2: Identify y - bounds

For the \(y\) - coordinates, the minimum \(y\) - value is \(0\) and the maximum \(y\) - value is \(\pi/6\). So \(y = 0\) and \(y=\pi/6\).

Answer:

The given region \(R\) is bounded by the lines \(x = -\pi\), \(x=\pi\), \(y = 0\), and \(y=\frac{\pi}{6}\). So the blanks should be filled with \(-\pi\), \(\pi\), and \(\frac{\pi}{6}\) respectively.