Evaluating Double Integrals: A Step-by-Step Guide

When facing a double integral, the primary challenge lies in determining the bounds of integration. This article aims to guide you through the process of evaluating the following double integral, focusing on the algebraic expression of the region of integration and the symmetry properties of the integrand.

Introduction to the Double Integral

Consider the double integral:

[ iint_D x cdot y , dA ]

The region of integration, D, is defined by the curves (y x^2 - 1) and (y -x^2 1). To understand and evaluate this integral, it’s crucial to express D algebraically and determine the bounds of integration.

Expressing the Region D Algebraically

The curves intersect when:

[ x^2 - 1 -x^2 1 ]

Solving for x, we get:

[ 2x^2 2 ] [ x^2 1 ] [ x pm 1 ]

By substituting a point in the interval ([-1, 1]), we confirm that:

[ y leq x^2 - 1 quad text{and} quad y geq -x^2 1 ]

Therefore, the region D can be expressed as:

[ x^2 - 1 leq y leq -x^2 1 quad text{and} quad x in [-1, 1] ]

A plot of the region of integration is given below:

Symmetric Properties and Iterated Integration

Noting that D is symmetric about both the x and y axes, and the integrand is odd in both variables, the double integral can be swiftly evaluated using an iterated integral:

[ iint_D x cdot y , dA int_{-1}^{1} int_{x^2-1}^{-x^2 1} x cdot y , dy , dx ]

Performing the integration with respect to (y) first:

[ int_{x^2-1}^{-x^2 1} x cdot y , dy x left[ frac{y^2}{2} right]_{x^2-1}^{-x^2 1} ]

Evaluating the expression:

[ x left( frac{(-x^2 1)^2}{2} - frac{(x^2-1)^2}{2} right) ]

Simplifying further:

[ x left( frac{x^4 - 2x^2 1}{2} - frac{x^4 - 2x^2 1}{2} right) 0 ]

Therefore, the final result is:

[ boxed{0} ]

Alternative Evaluation Method

Alternatively, one could evaluate the integral by integrating with respect to (x) first:

[ int_{-1}^{1} int_{x^2-1}^{-x^2 1} x cdot y , dy , dx ]

The region is more convenient when integrating with respect to x first due to the cusps in the integrand at (x 0). The lower and upper limits for (x) are (-1) and (1), respectively.

Integrating with respect to (y) first:

[ int_{x^2-1}^{-x^2 1} x cdot y , dy x left[ frac{y^2}{2} right]_{x^2-1}^{-x^2 1} 0 ]

Thus, the overall integral also evaluates to:

[ boxed{0} ]

Exploiting Symmetry

In cases where the domain and integrand share symmetry, you can exploit this to eliminate the need to integrate over the entire region. Here, the integrand (x cdot y) is odd in both (x) and (y), which means the positive and negative contributions will cancel out, confirming the result of zero.

Conclusion

In summary, when evaluating the double integral (iint_D x cdot y , dA), it is essential to first express the region of integration algebraically and consider the symmetry of the integrand. The final result, based on the symmetry and the cancellation of contributions, is (boxed{0}).