# Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 846: 23

$$\frac{4}{3}$$

#### Work Step by Step

\begin{aligned} \int_{-1}^{1} \int_{0}^{\pi} x^{2} \sin y d y d x &=\left.\int_{-1}^{1} x^{2}(-\cos y)\right|_{y=0} ^{\pi} d x\\ &=\int_{-1}^{1} x^{2}(-\cos \pi+\cos 0) d x \\ &=\int_{-1}^{1} x^{2}(1+1) d x=\int_{-1}^{1} 2 x^{2} d x\\ &=\left.\frac{2}{3} x^{3}\right|_{-1} ^{1}=\frac{2}{3}\left(1^{3}-(-1)^{3}\right)\\ &=\frac{4}{3} \end{aligned}

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