Problem 46 - Thomas Calculus

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 limits of integration from your plot in part (c) utility. \(\int_{0}^{1} \int_{y}^{2-y} \sqrt{x+y} d x d y\)

Step-by-Step Solution

Answer

Use polar coordinates to solve the integral with limits: \( 0 \leq r \leq \frac{2}{\cos \theta + \sin \theta}, 0 \leq \theta \leq \frac{\pi}{4} \).

Step 1: Plot the Cartesian Region

First, identify the region of integration in the Cartesian coordinate system. The given double integral \[ \int_{0}^{1} \int_{y}^{2-y} \sqrt{x+y} \, dx \, dy \] suggests the region is bounded by the curves \( y = 0 \), \( y = 1 \), \( x = y \), and \( x = 2-y \). Plot these lines on the \(xy\)-plane, where the region of integration forms a triangle with vertices at \((0, 0)\), \((1, 1)\), and \((0, 1)\).

Step 2: Convert Cartesian Boundaries to Polar

Next, convert each boundary to polar coordinates. Using the transformations \( x = r \cos \theta \) and \( y = r \sin \theta \):- The line \( x = y \) becomes \( r \cos \theta = r \sin \theta \) or \( \tan \theta = 1 \), so \( \theta = \frac{\pi}{4} \).- The line \( x = 2 - y \) becomes \( r \cos \theta = 2 - r \sin \theta \). Solving for \( r \), \( r = \frac{2}{\cos \theta + \sin \theta} \).Since \( r \) and \( \theta \) need to vary consistent with these constraints over the region bounded in Step 1, identify limits: \( 0 \leq r \leq \frac{2}{\cos \theta + \sin \theta} \) and \( 0 \leq \theta \leq \frac{\pi}{4} \).

Step 3: Plot the Polar Region

Plot the polar region identified from Step 2. The radial dimension ranges from \( r = 0 \) to \( r = \frac{2}{\cos \theta + \sin \theta} \), while the angular dimension spans \( 0 \leq \theta \leq \frac{\pi}{4} \). This describes a sector of a circle on the \(r\theta\)-plane.

Step 4: Change Integrand to Polar Coordinates

Convert the integrand \( \sqrt{x+y} \) to polar coordinates:\[ x + y = r(\cos \theta + \sin \theta) \]Therefore, the integrand becomes \( \sqrt{r(\cos \theta + \sin \theta)} \). The transformation introduces a Jacobian of \( r \) in the integration, so the new integrand is:\[ r \sqrt{r(\cos \theta + \sin \theta)} \].

Step 5: Set up and Evaluate the Polar Integral

The polar integral is set up with:\[ \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{2}{\cos \theta + \sin \theta}} r \sqrt{r(\cos \theta + \sin \theta)} \, dr \, d\theta \]To solve, integrate with respect to \( r \) over the prescribed limits and then with respect to \( \theta \). By evaluating or using a CAS, the result of the polar integral gives you the final solution to the original spatial integral.

Key Concepts

Cartesian to Polar Conversion

When converting from Cartesian to polar coordinates, we transform the points in the Cartesian plane into polar coordinates using the relationships:

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

In the original exercise, the boundaries of the integral in Cartesian form were determined by the lines \( x = y \), and \( x = 2 - y \).
To convert these into polar coordinates, you equate and solve them accordingly:

\( x = y \) transforms into \( \tan \theta = 1 \) or \( \theta = \frac{\pi}{4} \).

\( x = 2 - y \) transforms by substituting the polar equations,leading to \( r = \frac{2}{\cos \theta + \sin \theta} \).

These transformations describe the boundaries of the region in the polar coordinate system, which are vital for setting the limits in an integral.

Double Integrals in Polar Coordinates

Double integrals allow us to calculate areas, volumes, and more within given regions. When these integrals are expressed in polar coordinates, a Cartesian double integral \(\int\int_R{f(x,y)dxdy}\)becomes \(\int\int_R{f(r\cos \theta, r\sin \theta) \, r \, dr \, d\theta}\).
The additional \(r\) in the integrand is the Jacobian determinant for the polar coordinate transformation, which reflects the change in scale between the coordinate systems.
For the provided exercise, this transformation changes the Cartesian integral into \[ \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{2}{\cos \theta + \sin \theta}} r \sqrt{r(\cos \theta + \sin \theta)} \, dr \, d\theta \]This integral now reflects the chosen boundaries and the polar coordinate system's distinct characteristics.

Integration Limits in Polar Coordinates

Setting integration limits in polar coordinates is slightly different than in Cartesian coordinates.
The limits for \( \theta \) are generally angles measured counterclockwise from the positive x-axis, whereas the limits for \( r \) define the radial distances from the origin.
For the triangle region described in Cartesian coordinates, the polar limits were established as:

• \( 0 \leq \theta \leq \frac{\pi}{4} \), derived from the angles of the boundary lines.
• \( 0 \leq r \leq \frac{2}{\cos \theta + \sin \theta} \), determined by solving the transformed boundary equations.

This highlights the fundamental nature of polar integration, where areas are explored radially around a center point (origin), allowing us to frame complex regions more intuitively.

Polar Area Integration

Polar area integration uses the characteristics of polar coordinates to compute areas within polar-curved regions.
By expressing areas using polars, integrations around circular arcs and sectors become more straightforward compared to Cartesian coordinates.
In practice, the area of a region in polar coordinates can be integrated using the formula:
\[ A = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} r \, dr \, d\theta \]
In the exercise, the region was a triangular section in the \(xy\)-plane, mapped into a polar sector by:

• Converting each point to its polar equivalent using trigonometric identities.
• Establishing limits based on radial distance and angular direction.

Evaluating the resulting polar integral offers a precise computation of area within defined curvaceous boundaries, demonstrating the strength of this approach for managing complex shapes.