Question

first set up the dx dy order of integration. for this, consider a horizontal rectangle in the given region. in this case, the outer bounds are given by y = 0 and y = \frac{\pi}{6}, and the inner bounds are x = -\pi and x = \pi. therefore, the iterated integral is given by \int_{0}^{\frac{\pi}{6}} \int_{-\pi}^{\pi} \sin x \sin y dx dy.

Explanation:

Step1: Identify the integral bounds for \( y \)

The outer integral (with respect to \( y \)) has bounds from \( y = 0 \) to \( y=\frac{\pi}{6} \). So the limits for \( y \) are \( 0 \) and \( \frac{\pi}{6} \).

Step2: Identify the integral bounds for \( x \)

The inner integral (with respect to \( x \)) has bounds from \( x = -\pi \) to \( x=\pi \). So the limits for \( x \) are \( -\pi \) and \( \pi \).

Step3: Write the iterated integral

The integrand is \( \sin x \sin y \), and we integrate with respect to \( x \) first (inner integral) then with respect to \( y \) (outer integral). So the iterated integral is \( \int_{y = 0}^{\frac{\pi}{6}} \int_{x=-\pi}^{\pi} \sin x \sin y \, dx \, dy \).

Answer:

The iterated integral in \( dx \, dy \) order is \( \boldsymbol{\int_{0}^{\frac{\pi}{6}} \int_{-\pi}^{\pi} \sin x \sin y \, dx \, dy} \)