Problem 49 - Thomas Calculus

Question

In Exercises \(49-52,\) 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. $$ \begin{array}{l}{\text { a. Plot the Cartesian region of integration in the } x y \text { -plane. }} \\ {\text { b. Change each boundary curve of the Cartesian region in part }} \\ {\text { (a) to its polar representation by solving its Cartesian equation for } r \text { and } \theta .}\end{array} $$ $$ \begin{array}{l}{\text { c. Using the results in part (b), plot the polar region of integra- }} \\ {\text { tion in the } r \theta \text { -plane. }} \\\ {\text { d. Change the integrand from Cartesian to polar coordinates. De- }} \\ {\text { termine the limits of integration from your coordinates. De- }} \\ {\text { evaluate the polar integral using the CAS integration utility. }}\end{array} $$ $$ \int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x $$

Step-by-Step Solution

Answer

The evaluated polar integral is \( \frac{\pi}{4} \).

Step 1: Plot the Cartesian Region

We begin by identifying the region of integration in the Cartesian coordinates. The integration limits are \( 0 \leq x \leq 1 \) and \( x \leq y \leq 1 \). Plot the region as a triangle in the first quadrant of the \(xy\)-plane with vertices at \((0,0)\), \((1,1)\), and \((1,0)\).

Step 2: Convert Cartesian Boundary Curves to Polar Coordinates

Identify the boundary lines: \( y = x \) and \( y = 1 \). For \( y = x \), in polar coordinates, this becomes \( r\sin\theta = r\cos\theta \), or \( \theta = \frac{\pi}{4} \).. For \( y = 1 \), in polar coordinates, it becomes \( r \sin \theta = 1 \), or \( r = \frac{1}{\sin\theta} \).

Step 3: Plot the Polar Region

Using the results from Step 2, we plot the polar region defined by \( 0 \leq r \leq \frac{1}{\sin \theta} \) and \( 0 \leq \theta \leq \frac{\pi}{4} \) in the \(r\theta\)-plane.

Step 4: Change Integrand to Polar Coordinates

The integrand is \( \frac{y}{x^2 + y^2} \). To convert this to polar coordinates, substitute: \( x = r \cos \theta \), \( y = r \sin \theta \), and \( x^2 + y^2 = r^2 \). The integrand becomes \( \frac{r \sin \theta}{r^2} = \frac{\sin \theta}{r} \). Additionally, include the Jacobian \( r \) for polar coordinates, yielding \( r \cdot \frac{\sin \theta}{r} = \sin \theta \).

Step 5: Determine Limits and Evaluate Integral

The limits in polar coordinates are \( 0 \leq \theta \leq \frac{\pi}{4} \) and \( 0 \leq r \leq \frac{1}{\sin \theta} \). The integral becomes \( \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{1}{\sin \theta}} \sin \theta \, dr \, d\theta \). To evaluate, first integrate with respect to \( r \) to get \( [r \sin \theta]_{0}^{\frac{1}{\sin \theta}} = 1 \), then integrate with respect to \( \theta \): \( \int_{0}^{\frac{\pi}{4}} 1 \, d\theta = \frac{\pi}{4} \).

Key Concepts

Cartesian Coordinates

Cartesian coordinates form the basics of plotting points in a plane. They use two components, typically labeled as \(x\) and \(y\). Each point can be expressed as \((x, y)\), representing its horizontal (x-axis) and vertical (y-axis) distances from the origin (0,0). In our exercise, the region of integration is defined in Cartesian coordinates with the limits \(0 \leq x \leq 1\) and \(x \leq y \leq 1\).

In this system:

• The x-axis is horizontal and increases to the right.
• The y-axis is vertical and increases upwards.
• Each coordinate point is unique and fixed relative to these axes.

By plotting the given equations \( y = x \) and \( y = 1 \), we identify a triangle in the first quadrant. Understanding how Cartesian coordinates work is essential for interpreting and visualizing algebraic relationships in graphs.

Polar Coordinates

Polar coordinates present an alternative way of describing points in a plane. They use a radius \(r\) and an angle \(\theta\) to indicate a point's position. In polar coordinates, each point is a distance \(r\) from the origin and \(\theta\) radians from the positive x-axis.

Key aspects of the polar coordinate system include:

• \(r\) measures the distance from the origin to the point.
• \(\theta\) is an angle measured in radians, starting from the positive x-axis.
• A point can have multiple representations, as adding multiples of \(2\pi\) to \(\theta\) results in the same point.

In our exercise, the Cartesian equations \( y = x \) and \( y = 1 \) convert to polar forms \( \theta = \frac{\pi}{4} \) and \( r = \frac{1}{\sin \theta} \), respectively. This transforms our integration limits and helps us interpret the region in the \(r\theta\)-plane.

Integral Transformation

The integral transformation involves changing the integrand from one coordinate system to another. This allows easier evaluation of the integral by aligning with the geometry of the region. Transitioning from Cartesian integrals to polar integrals is a common technique.

Steps for transforming include:

• Substitute Cartesian variables \(x = r \cos \theta\) and \(y = r \sin \theta\).
• Recalculate the integrand with these substitutions.
• Adjust the limits of integration based on polar region definitions.

The transformation aligns the integrand and limits with the configuration of areas more naturally suited to circular or radial physics, such as the sector described by \(0 \leq r \leq \frac{1}{\sin \theta}\) and \(0 \leq \theta \leq \frac{\pi}{4}\). This makes the final integration process more straightforward and often simpler.

Jacobian Determinant

The Jacobian determinant is crucial in multivariable calculus when changing variables, especially during coordinate transformations in integrals. It involves a mathematical construct that accounts for the change of "volume" or "area" between coordinate systems. In simpler terms, it adjusts the integrand to reflect the new dimensions of scaling.

In the context of polar coordinates, the Jacobian determinant is:

• For Cartesian to polar conversions, it simplifies to \(r\).
• It accounts for the change in area: \(dA = r \, dr \, d\theta\).
• In our exercise, it modifies the integrand \(\frac{\sin \theta}{r}\) to \(\sin \theta\), making the calculation align accurately with polar scaling.

By incorporating the Jacobian, the transformed integral respects the scale and curvature implicated by the new coordinates, ensuring that results are precise and consistent across different formulations.