# Chapter 16 - Multiple Integration - 16.2 Double Integrals over More General Regions - Exercises - Page 858: 10

$\dfrac{1}{96}$

#### Work Step by Step

The domain $D$ for given region can be expressed as: $x^2 \leq y \leq x-x^2$ and $0 \leq x \leq \dfrac{1}{2}$ The iterated integral can be calculated as: $\iint_{D} 2y d A=\int_{0}^{1/2} \int_{x^2}^{x-x^2} 2y dy dx\\=\int_{0}^{1/2} [y^2]_{x^2}^{x-x^2} \ dx\\=\int_{0}^{1/2} (x^2+x^4-2x^3-x^4) \ dx \\=[\dfrac{x^3}{3} - \dfrac{x^4}{4}|_{0}^{1/2} \\=\dfrac{1}{96}$

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