Question

Integrals in strips Consider the integral $$I=\iint_{R} \frac{d A}{\left(1+x^{2}+y^{2}\right)^{2}}$$ where \(R=\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq a\\}\) a. Evaluate \(I\) for \(a=1 .\) (Hint: Use polar coordinates.) b. Evaluate \(I\) for arbitrary \(a > 0\) c. Let \(a \rightarrow \infty\) in part (b) to find \(I\) over the infinite strip \(R=\\{(x, y): 0 \leq x \leq 1,0 \leq y < \infty\\}\)

Short answer

Answer: When a approaches infinity, the value of the integral I is $$\frac{\pi}{8}$$.

Step-by-step solution

Change variables to polar coordinates

Since the integral is over the region \(R\), we can change the variables to polar coordinates using the transformations \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). The Jacobian for these transformations is \(r\), which means we need to multiply the integrand by \(r\) when changing variables. The new integral is: $$I=\iint_{R'} \frac{r}{\left(1+r^{2}\right)^{2}} d\theta dr$$ where \(R'\) is the region with \(0 \leq r \leq a\) and \(0 \leq \theta \leq \pi/2\).

Integrate with respect to r

Now, we need to integrate the transformed integrand with respect to r. We can rewrite the integrand as \(\frac{r}{\left(1+r^{2}\right)^{2}} = -\frac{1}{2} \frac{d}{dr} \left(\frac{1}{1+r^{2}}\right)\), and integrate it with respect to r: $$I = -\frac{1}{2}\int_{0}^{\pi/2}\left[\frac{1}{1 + a^{2} \cos(\theta)} - \frac{1}{2}\right] d\theta$$

Integrate with respect to θ

Now, we only need to integrate the resulting expression with respect to θ. Split the integral into two separate integrals to make it easier: $$I = -\frac{1}{2}\int_{0}^{\pi/2}\frac{1}{1 + a^{2} \cos(\theta)} d\theta + \frac{1}{4} \int_{0}^{\pi/2} d\theta$$ The first integral can be solved by substitution, while the second one is straightforward: $$I = -\frac{1}{2a} \left[\arctan \left(\frac{\sqrt{1 - a^{2}}}{a}\right)\right]_{0}^{\pi/2} + \frac{\pi}{8}$$

Evaluate I for a = 1

In this step, we need to evaluate I for a = 1. Substitute a = 1 in the expression from the previous step: $$I = -\frac{1}{2} \left[\arctan \left(\sqrt{1 - 1}\right)\right]_{0}^{\pi/2} + \frac{\pi}{8}$$ $$= \frac{\pi}{8}$$

Evaluate I for arbitrary a > 0

We have the expression for I in terms of a obtained in step 3. Therefore, for arbitrary a > 0, we have $$I = -\frac{1}{2a} \left[\arctan \left(\frac{\sqrt{1 - a^{2}}}{a}\right)\right]_{0}^{\pi/2} + \frac{\pi}{8}$$

Evaluate I for a → ∞

As a approaches infinity, the arctan expression inside the integral will approach zero due to the square roots: $$\lim_{a \to \infty} \frac{\sqrt{1 - a^{2}}}{a} = 0$$ So the integral over the infinite strip \(R\) is: $$I = \lim_{a \to \infty} -\frac{1}{2a} \left[\arctan \left(\frac{\sqrt{1 - a^{2}}}{a}\right)\right]_{0}^{\pi/2} + \frac{\pi}{8} = \frac{\pi}{8}$$

Key concepts

Polar Coordinates

Polar coordinates offer a unique way to describe points in a plane by their distance from the origin and the angle from the positive x-axis, represented as \((r, \theta)\). Unlike Cartesian coordinates \((x, y)\), polar coordinates allow us to handle circular and rotational symmetry more effectively.
Here’s how it works:

• \(x = r \cos(\theta)\) represents the x-component, projecting the distance \(r\) onto the x-axis.
• \(y = r \sin(\theta)\) represents the y-component, projecting the distance \(r\) onto the y-axis.

These transformations are particularly useful when dealing with circular regions or symmetries in problems, such as the double integral in this exercise, involving parts of circular areas.
When converting from Cartesian to polar coordinates, you multiply the integrand by the Jacobian \(r\), as a result of the transformation's scale. This Jacobian accounts for the fact that small changes in \(r\) and \(\theta\) describe radial expansion in the circular plane.

Double Integrals

Double integrals allow us to compute the total accumulation of a function over a two-dimensional region. Simply put, they let us add up values in 2D space. For example, they can calculate mass, area, or volume under a surface.
When dealing with double integrals, you often integrate first with respect to one variable and then the other.
In this exercise:

• We first transformed the region of integration using polar coordinates.
• Then, we integrated in terms of the radial component \(r\) and angular component \(\theta\), which simplifies handling circular region boundaries.

The double integral setup gives us a way to describe a function across a region, accounting for changes in both directions. It’s crucial to choose the right coordinate system, as it can simplify calculations when symmetry or boundaries are involved.

Jacobian Transformation

The Jacobian is a determinant that arises when changing coordinates, such as from Cartesian \((x, y)\) to polar \((r, \theta)\). It helps adjust the integral over a transformed region in terms of area scale factor.
In polar coordinates, the Jacobian is \(r\), which modifies how we interpret area elements in the new coordinate system.
Here's why it's essential:

• When we integrate using polar coordinates, each area element \(dA\) in the plane is represented as \(r\, dr\, d\theta\), rather than \(dx\, dy\) in Cartesian coordinates.
• The factor \(r\) reflects how much a small "patch" expands or shrinks as we move in circles away from or towards the origin.

Understanding the Jacobian ensures that an integral over \((dx, dy)\) in one space correctly corresponds to its counterpart in another space \((dr, d\theta)\), keeping the function’s integral value equivalent despite changing perspectives.