Chapter 15 - Multiple Integrals - 15.2 Exercises - Page 1012: 37

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$\iint_R\frac{xy}{1+x^4}dA=\int_{-1}^1\int_0^1\frac{xy}{1+x^4}dydx$ $=\int_{-1}^1[\frac{xy^2}{2(1+x^4)}]_0^1dx$ $=\int_{-1}^1\frac{x}{2(1+x^4)}dx$ Note that $g(x)=\frac{x}{2(1+x^4)}$ is an odd function since $g(-x)=\frac{-x}{2(1+(-x)^4)}=\frac{-x}{2(1+x^4)}=-g(x)$ So, $\iint_R \frac{xy}{1+x^4}dA=\int_{-1}^1\frac{x}{2(1+x^4)}dx=0$