1. **Problem statement:** Evaluate the double integral $$\iint_R \frac{xy}{x^2 + 1} \, dA$$ where $$R = \{(x,y) \mid -1 \leq x \leq 1, 0 \leq y \leq 1\}$$ using symmetry. 2. **Understanding the region and integrand:** The region $$R$$ is a rectangle symmetric about the y-axis for $$x$$ from $$-1$$ to $$1$$ and $$y$$ from $$0$$ to $$1$$. 3. **Check the integrand for symmetry:** The integrand is $$f(x,y) = \frac{xy}{x^2 + 1}$$. - The denominator $$x^2 + 1$$ is an even function in $$x$$. - The numerator $$xy$$ is an odd function in $$x$$ because $$x$$ is odd and $$y$$ is independent of $$x$$. Therefore, $$f(-x,y) = \frac{(-x)y}{(-x)^2 + 1} = -\frac{xy}{x^2 + 1} = -f(x,y)$$, so $$f$$ is an odd function in $$x$$. 4. **Using symmetry property:** Since the region $$R$$ is symmetric about the $$y$$-axis and the integrand is odd in $$x$$, the integral over $$x$$ from $$-1$$ to $$1$$ cancels out for each fixed $$y$$. 5. **Evaluate the integral:** $$ \iint_R \frac{xy}{x^2 + 1} \, dA = \int_0^1 \left( \int_{-1}^1 \frac{xy}{x^2 + 1} \, dx \right) dy = \int_0^1 0 \, dy = 0 $$ 6. **Final answer:** $$\boxed{0}$$