Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 846: 15

$$0$$

Given $$\iint_{\mathcal{R}} x^{3} d A, \quad \mathcal{R}=[-4,4] \times[0,5]$$ Then \begin{align*} \iint_{\mathcal{R}} x^{3} d A&=\int_{-4}^{4}\int_{0}^{5}x^3dydx\\ &= \int_{-4}^{4} x^3y\bigg|_{0}^{5}dx\\ &=5\int_{-4}^{4} x^3dx\\ &= 0 \end{align*}