Problem 42 - University Calculus: Early Transcendentals

Question

Evaluate the integral $$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\left(1+x^{2}+y^{2}\right)^{2}} d x d y$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{4} \).

Step 1: Recognize the integral type

The given integral is a double integral over the first quadrant of the xy-plane, extending from 0 to infinity in both directions. We need to solve \( \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\left(1+x^{2}+y^{2}\right)^{2}} \, dx \, dy \).

Step 2: Consider a coordinate transformation

When facing integrals of functions involving \( x^2 + y^2 \), it is beneficial to consider a transformation to polar coordinates where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). The differential area element changes from \( dx \, dy \) to \( r \, dr \, d\theta \).

Step 3: Convert the integral to polar coordinates

The transformation to polar coordinates gives us: \( x^2 + y^2 = r^2 \). Thus, the integral becomes \( \int_{0}^{\pi/2} \int_{0}^{\infty} \frac{r}{(1+r^2)^2} \, dr \, d\theta \).

Step 4: Simplify and integrate with respect to r

Perform the integration with respect to \( r \). Consider the substitution \( u = 1 + r^2 \), yielding \( du = 2r \, dr \). The limits change accordingly and give the integral: \( \frac{1}{2} \int_{1}^{\infty} u^{-2} \, du \). Solving this gives \( -\frac{1}{2} \left[ u^{-1} \right]_{1}^{\infty} = \frac{1}{2} \).

Step 5: Integrate with respect to theta

The integration with respect to \( \theta \) is straightforward as it is independent of \( r \). Thus, the integral becomes \( \int_{0}^{\pi/2} \frac{1}{2} \, d\theta = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \).

Step 6: Conclude the result

The calculation of both integrals gives the final result. The integral of the given function over the specified area is \( \frac{\pi}{4} \).

Key Concepts

Polar Coordinates

Polar coordinates offer an alternative way to represent points on a plane, especially useful when you encounter radial symmetry or terms involving expressions like \( x^2 + y^2 \).

• The \'x\' and \'y\' Cartesian coordinates can be transformed into polar coordinates through \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \).
• \( r \) represents the radius (or distance from the origin), and \( \theta \) represents the angle from the positive x-axis.
• This transformation simplifies the integration of circular or radial functions.

By transforming into polar coordinates, certain double integrals become easier to handle, especially when they involve circular regions. This approach is particularly beneficial in the context of this exercise, where the function \( 1 + x^2 + y^2 \) naturally transforms into \( 1 + r^2 \).
By understanding the role of polar coordinates, we can solve integrals in contexts where Cartesian coordinates might be cumbersome.

Coordinate Transformation

Coordinate transformation is a powerful tool in calculus, allowing us to change the perspective from which we view a problem.

• It is particularly helpful for simplifying complex integrals by taking advantage of symmetries inherent in certain functions.
• In this exercise, transforming from Cartesian to polar coordinates provides this advantage.

The main advantage of a coordinate transformation here is the simplification of the given integral over an infinite plane. When we switch from \( (x, y) \) to \( (r, \theta) \), the differential element \( dx \, dy \) transforms to \( r \, dr \, d\theta \).
This subtly changes the domain of integration into a space where one of the limits, \( 0 \) to \( \pi/2 \), is finite, and the other, \( 0 \) to infinity, follows the radial symmetry that naturally fits the problem structured around \( r^2 \).
Such transformations not only simplify the integration process but also illuminate underlying symmetries that might not be immediately obvious in their original form.

Integral Evaluation

Integral evaluation involves calculating the value of an integral, which can provide information about area, volume, or other accumulated quantities depending on the context.

• In this problem, the evaluation begins by simplifying the expression using polar coordinates and substitution.
• Specifically, \( u = 1 + r^2 \) is used to further transform the integral expression, making the procedure more straightforward.

This substitution leads to a simpler integral \( \frac{1}{2} \int_{1}^{\infty} u^{-2} \, du \), which we can easily evaluate.
Integrating from \( 1 \) to infinity results in \( \frac{1}{2} \), and then integrating with respect to \( \theta \) yields \( \frac{\pi}{4} \).
These computations showcase how strategic choices in integration limits and transformations can yield the final results efficiently, demonstrating the elegance and power of calculus in solving complex real-world problems.