Problem 49 - University Calculus: Early Transcendentals

Question

Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a. Plot the Cartesian region of integration in the \(x y\)-plane. b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for \(r\) and \(\theta\) c. Using the results in part (b), plot the polar region of integration in the \(r \theta\)-plane. d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility. $$\int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x$$

Step-by-Step Solution

Answer

The polar integral evaluates to \(-\frac{1}{2} \ln(2)\).

Step 1: Sketch the Cartesian Region

For the given Cartesian double integral \[ \int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x, \]the region of integration, R, is defined by the inequalities 0 ≤ x ≤ 1 and x ≤ y ≤ 1. This region forms a triangle characterized by vertices at (0,0), (1,1), and (0,1). The boundary includes the line y = x and y = 1.

Step 2: Convert Cartesian Boundary Curves to Polar Coordinates

The boundary y = x in Cartesian coordinates converts to θ = π/4 in polar coordinates. The boundary y = 1 converts to r sin(θ) = 1, or r = 1/sin(θ). In polar coordinates, r ranges from 0 to 1/sin(θ) for a given θ.

Step 3: Plot the Polar Region

With the polar transformation, we plot the region where 0 ≤ r ≤ 1/sin(θ) and 0 ≤ θ ≤ π/4 in the rθ-plane. This plot represents the polar version of the triangular region initially found in Cartesian coordinates.

Step 4: Change the Integrand and Determine Integration Limits

The integrand y/(x²+y²) converts to sin(θ)/(r²) in polar coordinates, as y = r sin(θ) and x² + y² = r². The limits of integration are determined from the polar plot: r from 0 to 1/sin(θ) and θ from 0 to π/4.

Step 5: Evaluate the Polar Integral

The polar integral we need to evaluate is\[ \int_{0}^{\pi/4} \int_{0}^{1/\sin(\theta)} \frac{\sin(\theta)}{r} \, r \, dr \, d\theta. \]This simplifies to:\[ \int_{0}^{\pi/4} \int_{0}^{1/\sin(\theta)} \sin(\theta) \, d\theta. \]Integrating with respect to r first gives us:\[ \int_{0}^{\pi/4} \sin(\theta)\cdot \left[ \ln(r) \right]_0^{1/\sin(\theta)} \, d\theta = \int_{0}^{\pi/4} \sin(\theta) \cdot \ln\left( \frac{1}{\sin(\theta)} \right) \, d\theta. \]Further simplifying and using integration by substitution, we find the definite integral:\[ -\cfrac{1}{2} \ln(2). \]

Key Concepts

Cartesian to Polar conversion

Converting from Cartesian to polar coordinates involves understanding a simple relationship. In the Cartesian system, points are defined by

• the horizontal position, \( x \)
• the vertical position, \( y \)

In polar coordinates, points are described by

• their distance from the origin, \( r \)
• an angle from the positive x-axis, \( \theta \)

To convert, use the following formulas:

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

The boundary line \( y = x \) translates into \( \theta = \frac{\pi}{4} \). This is because \( y = x \) means the angle is typically 45 degrees (or \( \frac{\pi}{4} \) radians), representing when y is equal to x in a right triangle.
The boundary \( y = 1 \) changes into \( r = \frac{1}{\sin(\theta)} \). This conversion uses the fact that \( y = r \sin(\theta) \), and thus \( r \) can be expressed in terms of \( \theta \) as \( 1/\sin(\theta) \). Remember that these conversions are key to evaluating how the regions transform, enabling integration beyond the Cartesian system.

double integral evaluation

Understanding and evaluating double integrals is crucial to dealing with complex regions. A double integral allows you to sum over a two-dimensional area. In this exercise, the integral \[\int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} \, d y \, d x\] is being converted from Cartesian to polar, becoming:\[\int_{0}^{\pi/4} \int_{0}^{1/\sin(\theta)} \sin(\theta) \, dr \, d\theta\]The process involves:

• Changing variables, where \( y = r \sin(\theta) \) for the integrand and \( x^2 + y^2 = r^2 \) for the denominator.
• Adjusting limits of integration based on region mapping in polar coordinates.
• Recognizing that the function converts to \( \frac{\sin(\theta)}{r} \times r = \sin(\theta) \).

Solving the simplified polar integral begins with integrating \( \sin(\theta) \) with respect to \( r \), yielding a logarithmic function:\[\int_{0}^{\pi/4} \sin(\theta) \cdot \left[ \ln(r) \right]_0^{1/\sin(\theta)} \, d\theta\]You'll often use substitution or parts to solve these integrals, simplifying expressions and managing logarithmic terms to find the result.

region of integration in polar coordinates

When working with polar coordinates, it's important to visualize and understand the region over which integration occurs. In polar coordinates, the region is defined by radius \( r \) and angle \( \theta \), unlike conventional x and y coordinates.
For this specific problem, the Cartesian region defined by \( 0 \leq x \leq 1 \) and \( x \leq y \leq 1 \) forms a triangular shape. This shape translates into:

• a radial line from \( r = 0 \) to \( r = 1/\sin(\theta) \)
• bounded angles from \( \theta = 0 \) to \( \theta = \pi/4 \)

This transformation results in a wedge-shaped area in the polar coordinate system. Plotting this out, the region can be visualized in the \( r\theta \)-plane, which provides the clear limits for \( r \) and \( \theta \).
This polar region identifies exactly where the double integral will be evaluated. Successfully plotting and transferring the region ensures accurate integration regardless of the complexity of the original Cartesian region.