Question

Show that the integral $$ \iint_{R} \log \sqrt{x^{2}+y^{2}} d x d y $$ converges, where \(R\) is the region \(x^{2}+y^{2} \leq 1\), and find its value. This can be interpreted as minus the logarithmic potential, at the origin, of a uniform mass distribution over the circle.

Short answer

Question: Determine the convergence of the integral \(\iint_{R} \log \sqrt{x^2 + y^2}\, dx \, dy\) over the region \(R: x^2 + y^2\leq 1\), and find its value. Answer: The given integral converges, and its value is \(-\frac{1}{2}\pi\).

Step-by-step solution

Transform to polar coordinates

Let us rewrite the function \(\log(\sqrt{x^2+y^2})\) using polar coordinates, where \(x=r\cos\theta\) and \(y=r\sin\theta\). Since the region \(R\) is a circle with radius \(1\), we have \(0\leq r \leq 1\) and \(0\leq\theta\leq2\pi\). The Jacobian of the transformation is \(r\), so our integral becomes: $$ \iint_{R} \log(\sqrt{x^2+y^2})\, dx\, dy = \int_0^{2\pi}\int_0^1 \log r \cdot r\, dr \, d\theta $$

Solve the inner integral

First, let's solve the inner integral with respect to \(r\): $$ \int_0^1 \log r \cdot r\, dr $$ To solve this, we can use integration by parts, with \(u = \log r\) and \(dv = r\, dr\). Then, \(du = \frac{1}{r}\,dr\) and \(v = \frac{1}{2}r^2\). Now, apply integration by parts formula: $$ \int u\,dv = uv - \int v\,du $$ Thus, $$ \int_0^1 \log r \cdot r\, dr = \left[\frac{1}{2}r^2 \log r\right]_0^1 - \int_0^1 \frac{1}{2}r^2 \cdot \frac{1}{r}\, dr = -\frac{1}{2}\int_0^1 r\, dr $$ Now, compute the integral: $$ -\frac{1}{2}\int_0^1 r\, dr = -\frac{1}{2}\left[\frac{1}{2}r^2\right]_0^1 = -\frac{1}{4} $$

Solve the outer integral

Now, substitute the result of the inner integral into the outer integral: $$ \int_0^{2\pi} (-\frac{1}{4}) \, d\theta = -\frac{1}{4} \int_0^{2\pi} d\theta $$ Then, compute the integral: $$ -\frac{1}{4} \int_0^{2\pi} d\theta = -\frac{1}{4}[\theta]_0^{2\pi} = -\frac{1}{2}\pi $$

Final answer

Thus, the integral converges, and its value is: $$ \iint_{R} \log \sqrt{x^2 + y^2}\, dx \, dy = -\frac{1}{2}\pi. $$

Key concepts

Polar Coordinates Transformation

The transformation from Cartesian coordinates \(x, y\) to polar coordinates \(r, \theta\) is particularly useful when dealing with circular regions. Polar coordinates are defined by:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)

With this substitution, the circle centered at the origin with radius 1 translates to \(0 \leq r \leq 1\) and \(0 \leq \theta \leq 2\pi\). This simplifies our region of integration significantly when dealing with circular symmetry. By expressing \((x^2 + y^2)\) as \(r^2\), the function \(\log(\sqrt{x^2 + y^2})\) becomes \(\log(r)\). This change of variables often makes integration much easier and more intuitive for circular regions.

Jacobian Determinant

The Jacobian determinant is crucial when changing variables in multiple integrals because it provides the scaling factor for the area element. In polar coordinates, the area element \(dx \ dy\) is transformed to \(r \ dr \ d\theta\). The Jacobian of the polar transformation is \(r\), which accounts for the way areas stretch in polar coordinates. This means when transforming an integral to polar coordinates, the integrand must be multiplied by \(r\):\[iint_{R} \log(\sqrt{x^2+y^2})\, dx\, dy = \int_0^{2\pi}\int_0^1 \log r \cdot r\, dr \, d\theta\]This adjustment ensures the integral evaluates correctly by factoring in the non-uniform distance within the circular sectors.

Integration by Parts

Integration by parts is an essential technique in calculus, especially for solving integrals that involve products of functions. The formula for integration by parts is derived from the product rule for differentiation and is expressed as:\[\int u \, dv = uv - \int v \, du\]For the integral of \(\log r \cdot r\), we select \(u = \log r\) and \(dv = r\, dr\). This means \(du = \frac{1}{r}\,dr\) and \(v = \frac{1}{2}r^2\). Plugging these into the integration by parts formula, we solve the inner integral:\[\int \log r \cdot r\, dr = \left[\frac{1}{2}r^2 \log r\right]_0^1 - \int_0^1 \frac{1}{2}r^2 \cdot \frac{1}{r}\, dr = -\frac{1}{4}\]This step simplifies the problem further, enabling us to easily find the value of the integral.

Logarithmic Potential

The concept of a logarithmic potential is intriguing and arises often in physics and potential theory. This integral can be thought of as computing a potential at the origin due to a mass distribution over a disk. The function \(\log \sqrt{x^2+y^2}\) relates to the gravitational or electrostatic potential caused by a circularly symmetric mass distribution with unit density, measured at a point. Its convergence is deduced by considering the symmetry and bounded nature of the region. The negative sign in the final result \(-\frac{1}{2}\pi\) indicates a potential reduction when evaluated at the origin. Understanding how this integral represents a potential can provide deeper insight into related physical phenomena, such as fields and forces, when dealing with similar configurations.