Question

For what values of \(p\) does the integral \(\iint_{R} \frac{d A}{\left(x^{2}+y^{2}\right)^{p}}\) exist in the following cases? a. \(R=\\{(r, \theta): 1 \leq r < \infty, 0 \leq \theta \leq 2 \pi\\}\) b. \(R=\\{(r, \theta): 0 \leq r \leq 1,0 \leq \theta \leq 2 \pi\\}\)

Short answer

a. In the case where the region R is defined by \(1 \leq r < \infty, 0 \leq \theta \leq 2 \pi\), the double integral exists for \(p>\frac{1}{2}\). b. In the case where the region R is defined by \(0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi\), the double integral exists for \(p<\frac{3}{2}\).

Step-by-step solution

Convert to polar coordinates

In polar coordinates, the given integral is $$\iint_{R} \frac{dA}{\left(x^{2}+y^{2}\right)^{p}} = \iint_{R} \frac{r \, dr \, d\theta}{(r^2)^p}.$$

Set up the integral for case a

For case a, the region \(R\) is defined by \(1 \leq r < \infty, 0 \leq \theta \leq 2 \pi\). Thus, the integral becomes: $$\int_{0}^{2\pi} \int_{1}^{\infty} \frac{r}{r^{2p}} \, dr \, d\theta.$$

Find condition for convergence in case a

To check if the integral converges, we need to find the condition for which \(\frac{r}{r^{2p}} \, dr\) converges. This can be simply written as \(\frac{1}{r^{2p-1}} \, dr\). Now, we look at the r integral's convergence: $$\int_{1}^{\infty} \frac{1}{r^{2p-1}} \, dr.$$ Since this is an improper integral, we can see that it converges if \(2p-1>1\). Solving for \(p\) gives \(p>\frac{1}{2}\).

Case b integration setup

For case b, the region \(R\) is defined by \(0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi\). The integral becomes: $$\int_{0}^{2\pi} \int_{0}^{1} \frac{r}{r^{2p}} \, dr \, d\theta.$$

Find condition for convergence in case b

As before, the main focus is on the convergence of the r integral. The r integral becomes: $$\int_{0}^{1} \frac{1}{r^{2p-1}} \, dr.$$ Since this is also an improper integral, we can determine that it converges if \(2p-1<1\). Solving for \(p\) yields \(p<\frac{3}{2}\).

Combine the conditions

In conclusion, the double integral converges for the following values of \(p\): a. \(p>\frac{1}{2}\) b. \(p<\frac{3}{2}\)

Key concepts

Double Integrals

In multivariable calculus, a double integral allows us to integrate over a two-dimensional area. This is especially useful when dealing with functions of two variables, like in the example given earlier. The expression \(\iint_{R} f(x,y) \, dA\) represents a double integral across the region \(R\), where \(dA\) is a small area element within \(R\). Double integrals can be used to calculate areas, volumes, and average values over regions in the plane.

• Double integrals are similar to single-variable integrals, but they consider areas, not just lengths.
• The choice of integration order (whether integrating \(x\) first or \(y\) first) depends on the ease of computation and the region's shape.
• This function is often given in terms of \(x\) and \(y\), but can be conveniently expressed in polar coordinates for circular or ring-shaped regions.

Understanding double integrals helps build a foundation for evaluating more complex regions and functions.

Convergence of Integrals

Convergence of an integral is a crucial concept that determines whether the integral produces a finite result. When we talk about convergence in terms of improper integrals, such as \(\int_{1}^{\infty} \frac{1}{r^{2p-1}} \, dr\), we refer to the tendency of the integral to approach a limit as it extends over an infinite range.

• To find convergence, especially for improper integrals, you need to assess the behavior of the integrand at the boundaries of the region.
• If the value of \(p\) makes the power of \(r\) negative in the denominator, the integral is more likely to converge because the function decreases faster.
• For case (a), the integral converges if \(2p - 1 > 1\), leading us to \(p > \frac{1}{2}\).
• For case (b), it converges if \(2p - 1 < 1\), implying \(p < \frac{3}{2}\).

Understanding these conditions ensures accurate evaluation of integrals, especially over infinite or unbounded regions.

Polar Coordinates

Polar coordinates provide an alternative way of describing a point in the plane by using the distance from the origin \( r \) and the angle \( \theta \) from the positive x-axis. This method is particularly useful when dealing with circular or radial symmetry in problems, such as the regions in the given exercise.

• Polar coordinates simplify integration in circular or ring-shaped regions, which appear complicated in Cartesian coordinates.
• The transformation from Cartesian to polar coordinates follows the equations \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\).
• The differential area element, \(dA\), becomes \(r \, dr \, d\theta\) in polar coordinates because of the geometry.

Employing polar coordinates in double integrations allows for more straightforward calculations when the problems involve symmetries related to circles or sectors in the plane.