Evaluating a Double Integral over a Complex Region: A Step-by-Step Guide

In the realm of calculus, evaluating double integrals can be challenging, especially when the region of integration is complex. In this article, we will explore the process of evaluating the double integral:

The Integral in Question

We want to evaluate the following double integral:

[ I iint_D frac{1}{x^2 y^2} , dA ]

where (D) is the region bounded by the disk (x^2 y^2 leq 1) and the half plane (x - y geq 1). A plot of this region is given below:

Converting to Polar Coordinates

To tackle this problem, we will convert the integral into polar coordinates, which often simplifies the process of dealing with circular or radial regions.

First, let's understand the equations of the region in polar coordinates. The equation of the line (x - y 1) can be written as:

[ r cos(theta) - r sin(theta) 1 ]

By isolating (r), we get:

[ r frac{1}{cos(theta) - sin(theta)} ]

The unit circle (x^2 y^2 1) transforms to (r 1). Therefore, the line and circle intersect when:

[ frac{1}{cos(theta) - sin(theta)} 1 ]

This simplifies to:

[ cos(theta) - sin(theta) 1 ]

Solving for (theta) yields:

[ theta 0, frac{pi}{2} ]

This result aligns with the plot, as expected.

Applying the Change of Variables

Considering the Jacobian of the polar coordinate transformation, which is (r), the integral becomes:

[ I int_{0}^{frac{pi}{2}} int_{frac{1}{cos(theta) - sin(theta)}}^{1} frac{1}{r^2} cdot r , dr , dtheta int_{0}^{frac{pi}{2}} int_{frac{1}{cos(theta) - sin(theta)}}^{1} r^{-3} , dr , dtheta ]

Evaluating the inner integral, we get:

[ I int_{0}^{frac{pi}{2}} -frac{1}{2} r^{-2} Bigg|_{frac{1}{cos(theta) - sin(theta)}}^{1} , dtheta frac{1}{2} int_{0}^{frac{pi}{2}} left(2 cos^2(theta) - 1right) , dtheta ]

Using the trigonometric identity (2 cos^2(theta) - 1 cos(2theta)), the integral becomes:

[ I frac{1}{2} int_{0}^{frac{pi}{2}} cos(2theta) , dtheta ]

Evaluating this integral, we get:

[ I frac{1}{2} sin(2theta) Bigg|_{0}^{frac{pi}{4}} frac{1}{2} ]

Conclusion

Thus, the value of the double integral is:

[ I boxed{frac{1}{2}} ]

This step-by-step approach demonstrates the effectiveness of converting to polar coordinates in handling complex regions of integration. Understanding and applying the Jacobian and trigonometric identities are crucial in solving such problems.