Question

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\iint_{R} \sqrt{x^{2}+y^{2}} d A ; R=\\{(x, y): 0 \leq y \leq x \leq 1\\}$$

Short answer

Question: Determine if the given double integral converges or diverges: $$\iint_{R} \sqrt{x^{2}+y^{2}} d A, \text{ where } R=\\{(x, y): 0 \leq y \leq x \leq 1\\}$$ Answer: The integral diverges.

Step-by-step solution

Sketch the region R

The given region R can be defined by the inequalities: $$0 \leq y \leq x \leq 1$$ If we draw this region, we can see that it is a triangle bounded by the lines y = 0, x = 1, and y=x in the first quadrant of the xy-plane.

Convert the integral into polar coordinates

To evaluate the integral in polar coordinates, we need to rewrite the function and the limits of integration using the transformation formulas: $$x = r\cos(\theta), y = r\sin(\theta), dA = rdrd\theta$$ The function \(\sqrt{x^2+y^2}\) in polar coordinates becomes \(\sqrt{r^2\cos^2(\theta)+r^2\sin^2(\theta)}=r\). Now, we need to find the new limits of integration for the polar coordinates.

Determine the new limits of integration

First, let's determine the limits for r by considering the boundaries in the R region. When y = 0, we have r = r*cos(θ), therefore r ranges from 0 to the line x = 1: $$0 \leq r \leq \frac{1}{\cos(\theta)}$$ Next, we need to find the limits for θ. The region R is in the first quadrant, so we know that θ ranges from 0 to π/4, as we need to consider the range between the y-axis (θ=0) and the line y=x (θ=π/4): $$0 \leq \theta \leq \frac{\pi}{4}$$ Now we have the new limits of integration for r and θ.

Evaluate the integral

Now, we can write the integral in polar coordinates, substituting the function and the limits of integration we found in the previous steps: $$\int_{0}^{\pi/4}\int_{0}^{\frac{1}{\cos(\theta)}} r\cdot rdrd\theta$$ We can now evaluate the inner integral with respect to r: $$\int_{0}^{\pi/4} \left[ \frac{1}{2}r^2 \right]_{0}^{\frac{1}{\cos(\theta)}}d\theta = \int_{0}^{\pi/4} \frac{1}{2\cos^2(\theta)} d\theta$$ Now, let's evaluate the outer integral with respect to θ, using the substitution u = sin(θ), du = cos(θ)dθ: $$\frac{1}{2} \int_{0}^{1/\sqrt{2}} \frac{1-u^2}{u^2} du$$ Integrating this expression, we get: $$\frac{1}{2} \left[ -\ln|u| - \frac{1}{u} \right]_{0}^{1/\sqrt{2}} = \frac{1}{2} \left[ -\ln\left(\frac{1}{\sqrt{2}}\right) - \sqrt{2} + \ln(0) \right]$$ Since \(\ln(0)\) is not defined, our integral diverges. In conclusion, the integral $$\iint_{R} \sqrt{x^{2}+y^{2}} d A$$ diverges.

Key concepts

Double Integrals

Double integrals are a way to integrate over a two-dimensional region. They are used to calculate things like area, volume, and other properties dependent on two variables. In this exercise, we work with an integral over a region \(R\) in the \(xy\)-plane.

• Setup: The integral \(\iint_R \sqrt{x^2 + y^2} \, dA\) represents the accumulation of the function \(\sqrt{x^2 + y^2}\) over the region \(R\).
• Region \(R\): Defined by \(0 \leq y \leq x \leq 1\), this forms a triangular area bounded by the lines \(y = 0\), \(x = 1\), and \(y = x\).

Understanding the specific form of \(R\) helps us visualize where we are integrating and guides us in changing variables when needed.
Double integrals are typically solved by integrating with respect to one variable and then the other. In this case, we change to polar coordinates to simplify the process.

Polar Coordinates

Polar coordinates are often used to simplify integration over certain regions like circles and other radial shapes. They represent points in terms of their distance from the origin ( ) and angle (\(\theta\)). In this problem, converting to polar coordinates helps us tackle the integral.

• Transformation Formulas: Use \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) to express coordinates, and the differential area element as \(dA = r\,dr\,d\theta\).
• Function Simplification: The function \(\sqrt{x^2 + y^2}\) becomes \(r\) in polar coordinates.

This transformation is especially useful because the original integrand \(\sqrt{x^2 + y^2}\) simplifies significantly, allowing for straightforward integration steps. It also alters the region \(R\) into simpler upper and lower bounds when described in \(r\) and \(\theta\).

Limits of Integration

Determining the limits of integration is crucial for setting up a double integral, especially after transforming to a new coordinate system. Here, we define new limits based on the description of region \(R\) in polar coordinates.

• \(r\) Limits: The range of \(r\) originates from 0 (the origin) to \(\frac{1}{\cos(\theta)}\), determined by solving for the edge of the region bounded by \(x = 1\).
• \(\theta\) Limits: Since \(R\) lies in the first quadrant, \(\theta\) ranges from \(0\) to \(\frac{\pi}{4}\), spanning from the positive \(x\)-axis to the line \(y = x\).

By carefully describing these bounds, we ensure the integral perfectly encompasses region \(R\). This setup results in evaluating a simpler integral in polar coordinates, though in our case, it ultimately diverges, highlighting the importance of correctly understood bounds.