Problem 37 - University Calculus: Early Transcendentals

Question

Converting to a polar integral Integrate \( f(x, y)=\) \(\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}\) over the region \(1 \leq x^{2}+y^{2} \leq e.\)

Step-by-Step Solution

Answer

The integral evaluates to \(4\pi - 2\pi \sqrt{e}\).

Step 1: Understand the Region and Function

The function to integrate is given as \( f(x, y) = \frac{\ln(x^2 + y^2)}{\sqrt{x^2 + y^2}} \). The region is defined by \( 1 \leq x^2 + y^2 \leq e \), which is a circular ring (annular region) in polar coordinates, from radius \( r = 1 \) to \( r = \sqrt{e} \).

Step 2: Convert to Polar Coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). Thus, \( x^2 + y^2 = r^2 \) and the Jacobian \( \frac{\partial(x,y)}{\partial(r,\theta)} = r \), so the differential area element \( dx \ dy = r \, dr \, d\theta \). Thus, the integral becomes: \[\int_{0}^{2\pi} \int_{1}^{\sqrt{e}} \frac{\ln(r^2)}{r} \cdot r \ dr \, d\theta = \int_{0}^{2\pi} \int_{1}^{\sqrt{e}} \ln(r^2) \, dr \, d\theta.\]

Step 3: Simplify the Integrand

The integrand \( \ln(r^2) \) can be simplified using the logarithm property \( \ln(r^2) = 2 \ln(r) \). Thus, the integral becomes:\[2 \int_{0}^{2\pi} \int_{1}^{\sqrt{e}} \ln(r) \, dr \, d\theta.\]

Step 4: Perform the Inner Integration

Perform the integration with respect to \( r \):\[\int_{1}^{\sqrt{e}} \ln(r) \, dr.\]Use integration by parts where \( u = \ln(r) \) and \( dv = dr \). Then, \( du = \frac{1}{r} dr \) and \( v = r \). The integration becomes:\[\left[ r \ln(r) - r \right]_{1}^{\sqrt{e}} = (\sqrt{e} \ln(\sqrt{e}) - \sqrt{e}) - (1 \cdot 0 - 1) = (\sqrt{e} \cdot \frac{1}{2} - \sqrt{e}) + 1.\]

Step 5: Evaluate the Boundary of Inner Integral

Evaluate \( (\sqrt{e} \cdot \frac{1}{2} - \sqrt{e}) + 1 \, = \frac{\sqrt{e}}{2} - \sqrt{e} + 1 \, = -\frac{\sqrt{e}}{2} + 1 \).

Step 6: Perform the Outer Integration

Now, substitute the result from Step 5 into the outer integral:\[2 \int_{0}^{2\pi} (-\frac{\sqrt{e}}{2} + 1) \, d\theta \]Analyse:\[= 2\left(-\frac{\sqrt{e}}{2} + 1\right)\int_{0}^{2\pi} 1 \, d\theta \left|_{0}^{2\pi} = 2(-\frac{\sqrt{e}}{2}+1)(2\pi)\]Simplifying:\[(-\sqrt{e} + 2) \, 2\pi\].

Step 7: Final Simplification

Simplify the expression:\[(-\sqrt{e} + 2) \, 2\pi = (2 - \sqrt{e}) \, 2\pi = 2(2\pi) - 2\pi \sqrt{e}.\]So, the value of the integral is \( 4\pi - 2\pi \sqrt{e} \).

Key Concepts

Polar Coordinates

To make the integration problem simpler, we can use polar coordinates. These are particularly useful when dealing with circular or symmetric regions. Polar coordinates use a radius, \( r \), and an angle, \( \theta \), instead of \( x \) and \( y \).
The conversion is given by:

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

For our integral, the function \( x^2 + y^2 = r^2 \) simplifies in this system. The region of integration, \( 1 \leq x^2 + y^2 \leq e \), in polar becomes \( 1 \leq r^2 \leq e \).
After converting, the integration becomes easy to handle, since the dependency on \( x \) and \( y \) is removed and replaced by \( r \) and \( \theta \). In this transformed setup, we focus directly on \( r \) and \( \theta \), helping in simplifying the computation.

Integration by Parts

Integration by parts is a method used to integrate products of functions. It's particularly helpful when direct integration is challenging.
The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]In our solution, we apply this method to the integral \( \int \ln(r) \, dr \). Here, we choose:

• \( u = \ln(r) \), making \( du = \frac{1}{r} \, dr \)
• \( dv = dr \), resulting in \( v = r \)

The integration transforms to:\[ r \ln(r) - \int r \, \frac{1}{r} \, dr = r \ln(r) - \int dr \]The evaluation of this part yielded the intermediate result used in calculating the final integral, which helped simplify complex algebraic expressions.

Jacobian Determinant

When converting integrals from Cartesian to polar coordinates, the Jacobian determinant is crucial. It scales the area element correctly to account for the transformation.
In this context:

• The Cartesian area element \( dx \ dy \) is converted by multiplying with the Jacobian \( r \), forming \( r \, dr \, d\theta \).

This multiplier, \( r \), arises from the fact that in polar coordinates, the area of a small segment is wider the further it is from the origin. This differential area element ensures that the integration over a region matches between coordinate systems.
Without incorporating the Jacobian, the area and hence the integral would be incorrectly evaluated.

Annular Region

An annular region is essentially a ring-shaped area between two circular boundaries. In our problem, the region is defined by the inequality \( 1 \leq x^2 + y^2 \leq e \).

• The inner boundary is the circle with radius \( r = 1 \).
• The outer boundary is the circle with radius \( r = \sqrt{e} \).

In polar coordinates, rings are simple to handle as an interval on \( r \), which is \( 1 \) to \( \sqrt{e} \) for this case, with full rotation on \( \theta \) from \( 0 \) to \( 2\pi \).
This understanding of the annular region helps in setting the limits for the integration, which simplifies solving the integral using the radial method.