Chapter 14 - Multiple Integrals - 14.2 Double Integrals Over Nonrectangular Regions - Exercises Set 14.2 - Page 1016: 20

$$\pi$$

We find: $\int_{0}^{\pi} \int_{0}^{x} x \cos y d y d x=\iint_{R} x \cos y d A$ $=\int_{0}^{\pi}[x \sin y]_{0}^{x} d x$ $=\int_{0}^{\pi} x \sin x d x$ $=\left[x \int \sin x d x-\int\left[\frac{d x}{d x} \int \sin x d x\right] d x\right]_{0}^{\pi}$ $=\left[-x \cos x+\int \cos x d x\right]_{0}^{\pi}$ $=[\sin x-x \cos x]_{0}^{\pi}=\pi$