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

To tackle the iterated integral \( \int_{0}^{\pi / 6} \int_{0}^{\pi / 2}(\sin x+\sin y) d y d x \), we first evaluate the inner integral: \[ \int_{0}^{\pi / 2} (\sin x + \sin y) dy = \int_{0}^{\pi / 2} \sin x \, dy + \int_{0}^{\pi / 2} \sin y \, dy. \] The first part, \( \int_{0}^{\pi / 2} \sin x \, dy \), simplifies to \( \sin x \cdot \left( \frac{\pi}{2} \right) \) since \(\sin x\) is constant with respect to \(y\). For the second part, we have: \[ \int_{0}^{\pi / 2} \sin y \, dy = -\cos y \bigg|_{0}^{\pi / 2} = -\cos(\pi/2) + \cos(0) = 0 + 1 = 1. \] So, the inner integral becomes: \[ \int_{0}^{\pi / 2} (\sin x + \sin y) dy = \frac{\pi}{2} \sin x + 1. \] Now, we evaluate the outer integral: \[ \int_{0}^{\pi / 6} \left( \frac{\pi}{2} \sin x + 1 \right) dx = \frac{\pi}{2} \int_{0}^{\pi / 6} \sin x \, dx + \int_{0}^{\pi / 6} 1 \, dx. \] Calculating the first part: \[ \int_{0}^{\pi / 6} \sin x \, dx = -\cos x \bigg|_{0}^{\pi / 6} = -\cos(\pi/6) + \cos(0) = -\frac{\sqrt{3}}{2} + 1 = 1 - \frac{\sqrt{3}}{2}. \] Then: \[ \frac{\pi}{2} \left( 1 - \frac{\sqrt{3}}{2} \right). \] Now we compute the second integral: \[ \int_{0}^{\pi / 6} 1 \, dx = \frac{\pi}{6}. \] Putting it all together: \[ \int_{0}^{\pi / 6} \left( \frac{\pi}{2} \sin x + 1 \right) dx = \frac{\pi}{2} \left( 1 - \frac{\sqrt{3}}{2} \right) + \frac{\pi}{6}. \] This simplifies to: \[ \frac{\pi}{2} - \frac{\pi \sqrt{3}}{4} + \frac{\pi}{6}. \] Finding a common denominator (12): \[ \frac{6\pi}{12} - \frac{3\pi \sqrt{3}}{12} + \frac{2\pi}{12} = \frac{8\pi}{12} - \frac{3\pi \sqrt{3}}{12} = \frac{8\pi - 3\pi \sqrt{3}}{12}. \] Thus, the evaluated iterated integral is: \[ \frac{8\pi - 3\pi \sqrt{3}}{12}. \]