3. Calculate the iterated integrals. (a) \( \int_{0}^{\pi / 6} \int_{0}^{\pi / 2}(\sin x+\sin y) d y d x \)

We start by writing the iterated integral: \[ \int_{0}^{\pi/6} \int_{0}^{\pi/2} (\sin x + \sin y) \, dy \, dx. \] **Step 1. Compute the inner integral with respect to \( y \):** For a fixed \( x \), we have \[ \int_{0}^{\pi/2} (\sin x + \sin y) \, dy. \] Since \( \sin x \) is independent of \( y \), it comes out of the inner integral: \[ \int_{0}^{\pi/2} \sin x \, dy + \int_{0}^{\pi/2} \sin y \, dy = \sin x \int_{0}^{\pi/2} dy + \int_{0}^{\pi/2} \sin y \, dy. \] The first part is: \[ \sin x \int_{0}^{\pi/2} dy = \sin x \left(\frac{\pi}{2}\right). \] The second part is: \[ \int_{0}^{\pi/2} \sin y \, dy = \left[-\cos y\right]_{0}^{\pi/2} = \left[-\cos\left(\frac{\pi}{2}\right) + \cos(0)\right] = (0 + 1) = 1. \] Thus, the inner integral becomes: \[ \frac{\pi}{2} \sin x + 1. \] **Step 2. Compute the outer integral with respect to \( x \):** Now, our integral is \[ \int_{0}^{\pi/6} \left(\frac{\pi}{2}\sin x + 1\right) dx. \] This can be split into two simpler integrals: \[ \frac{\pi}{2} \int_{0}^{\pi/6} \sin x \, dx + \int_{0}^{\pi/6} dx. \] 1. Evaluate \( \int_{0}^{\pi/6} \sin x \, dx \): \[ \int \sin x \, dx = -\cos x, \] so \[ \int_{0}^{\pi/6} \sin x \, dx = \left[-\cos x\right]_{0}^{\pi/6} = -\cos\left(\frac{\pi}{6}\right) + \cos(0) = -\frac{\sqrt{3}}{2} + 1 = 1 - \frac{\sqrt{3}}{2}. \] 2. Evaluate \( \int_{0}^{\pi/6} dx \): \[ \int_{0}^{\pi/6} dx = \frac{\pi}{6}. \] Putting these results together: \[ \frac{\pi}{2} \left(1 - \frac{\sqrt{3}}{2}\right) + \frac{\pi}{6}. \] **Step 3. Simplify the expression:** First, distribute \( \frac{\pi}{2} \): \[ \frac{\pi}{2} - \frac{\pi \sqrt{3}}{4} + \frac{\pi}{6}. \] To combine these terms, find a common denominator. The denominators are \(2\), \(4\), and \(6\); a common denominator is \(12\): \[ \frac{\pi}{2} = \frac{6\pi}{12}, \quad \frac{\pi \sqrt{3}}{4} = \frac{3\pi \sqrt{3}}{12}, \quad \frac{\pi}{6} = \frac{2\pi}{12}. \] Now the sum is: \[ \frac{6\pi}{12} - \frac{3\pi \sqrt{3}}{12} + \frac{2\pi}{12} = \frac{(6\pi + 2\pi) - 3\pi\sqrt{3}}{12} = \frac{8\pi - 3\pi\sqrt{3}}{12}. \] Thus, the final answer is: \[ \frac{\pi (8 - 3\sqrt{3})}{12}. \]