Problem 45 - 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. \(\int_{0}^{1} \int_{-y / 3}^{y / 3} \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y\)

Step-by-Step Solution

Answer

The polar integral evaluates to 0.

Step 1: Plot the Cartesian Region

The given integral is \[\int_{0}^{1} \int_{-y/3}^{y/3} \frac{y}{\sqrt{x^2 + y^2}} \, dx \, dy\]The inner limits \(-y/3\) to \(y/3\) for \(x\) define horizontal strips. These strips run between lines \(x = -y/3\) and \(x = y/3\) as \(y\) varies from 0 to 1. This forms a triangular region in the Cartesian plane with vertices at \((0,0)\), \((1/3,1)\), and \((-1/3,1)\). Plot this triangular region in the \(xy\)-plane with these vertices.

Step 2: Convert Cartesian Boundaries to Polar

The boundary line equations are \(x = -y/3\) and \(x = y/3\). In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). Thus,\[x = -y/3 \implies r\cos\theta = -r\sin\theta/3 \implies \tan\theta = -\frac{1}{3}\]and \[x = y/3 \implies r\cos\theta = r\sin\theta/3 \implies \tan\theta = \frac{1}{3}\]These give angle bounds \(\theta = -\tan^{-1}(1/3)\) and \(\theta = \tan^{-1}(1/3)\). Furthermore, the outer boundary is \(y = 1\), giving \(r\sin\theta = 1\), so \(r = \csc\theta\).

Step 3: Plot the Polar Region

With the bounds \(r = 0\) to \(r = \csc\theta\) and \(\theta\) from \(-\tan^{-1}(1/3)\) to \(\tan^{-1}(1/3)\), we can transform the region. Plot the region in polar coordinates using these bounds, which appears as a sector of a circle.

Step 4: Convert the Integrand

The Cartesian integrand is \(\frac{y}{\sqrt{x^2 + y^2}}\). In polar coordinates, \(x^2 + y^2 = r^2\) and \(y = r\sin\theta\). Therefore, the integrand becomes \(\frac{r\sin\theta}{r} = \sin\theta\). The differential \(dx\,dy\) becomes \(r\,dr\,d\theta\) in polar coordinates.

Step 5: Evaluate the Polar Integral

The integral in polar coordinates is \[\int_{-\tan^{-1}(1/3)}^{\tan^{-1}(1/3)} \int_{0}^{\csc\theta} \sin\theta \, r \, dr \, d\theta\]Evaluate the \(r\) integral first:\[\int_{0}^{\csc\theta} \sin\theta \, r \, dr = \sin\theta \left[\frac{r^2}{2} \right]_0^{\csc\theta} = \sin\theta \left(\frac{(\csc\theta)^2}{2}\right)\]Simplifying gives:\[\frac{1}{2}\cot\theta\]Now evaluate the \(\theta\) integral:\[\int_{-\tan^{-1}(1/3)}^{\tan^{-1}(1/3)} \frac{1}{2}\cot\theta \ d\theta\]This integral evaluates to 0 since it is symmetric about \(\theta = 0\) and involves an odd function within symmetric limits.

Key Concepts

Cartesian to Polar Conversion

To solve integrals involving regions that are best described in polar coordinates, the first important step is conversion from Cartesian to polar coordinates. In this process, we aim to transform the Cartesian limits and integrand into their polar equivalents.

• Start with the Cartesian equations: For instance, lines such as \(x = \frac{-y}{3}\) and \(x = \frac{y}{3}\) need to be converted.
• Utilize the polar coordinate equations: Substitute \(x = r\cos\theta\) and \(y = r\sin\theta\) into the Cartesian equations.
• For lines \(x = y/3\) and \(x = -y/3\), set \(r\cos\theta = \pm r\sin\theta/3\) to find \(\theta\). This gives us the boundary angles in polar coordinates.

This step essentially allows us to reinterpret the original region into a format that is manageable using polar integrals.

Regions of Integration

The concept of regions of integration in polar coordinates involves redefining a set region from the Cartesian plane to polar. This is crucial since it aligns with the transformed limits and boundaries that fit the polar system more naturally.

• In the Cartesian system, describe the region using lines or curves. For example, a triangular region determined by \(y\) and \(x\) bounds.
• In the polar coordinates, redefine boundaries using \(r\) and \(\theta\), such as sectors of a circle or annular shapes.
• Judiciously determine the region by understanding the natural arcs and radial distances that substitute the Cartesian boundaries.

Regions of integration set the stage upon which the limits of integration are applied.

Definite Integrals

Once the regions of integration are defined in polar coordinates, the next step involves setting up the definite integral itself. This requires careful handling of both the integrand and the integration limits.

• Convert the original Cartesian integrand, which might involve \(x\) and \(y\), into polar form using relationships like \(x^2 + y^2 = r^2\) and transformations of functions such as \(r\sin\theta\).
• Substitute and compute the new integrand and limits into the polar integral equation.
• Conduct the integration from the innermost to the outermost variable—typically integrating \(r\) first, followed by \(\theta\).

The accuracy in evaluating these definite integrals solidifies understanding and provides the exact area or volume under the defined region.

Mathematical Plotting

Plotting in mathematical integration serves as both a means of verification and an enhancement to comprehension. It helps visualize the regions and boundaries, facilitating intuition about their forms and calculations.

• Involve initial plotting of the Cartesian region to understand its size and shape.
• Transform such visual plots into polar coordinates, ensuring the sector's bounds are represented correctly on a polar grid.
• Use plots to double-check limits and transformations, such as visually confirming \(r\) boundaries and \(\theta\) angles.

Thus, plotting acts as an invaluable tool that not only aids in solving problems but also reinforces the learning process by connecting abstract concepts to tangible visuals.