Problem 13 - Thomas Calculus

Question

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. \(\int_{0}^{2} \int_{0}^{\sqrt{1-(x-1)^{2}}} \frac{x+y}{x^{2}+y^{2}} d y d x\)

Step-by-Step Solution

Answer

The value of the polar integral is \(\pi\).

Step 1: Understand the Region of Integration

The given Cartesian integral is over a region described by \(0 \leq x \leq 2\) and \(0 \leq y \leq \sqrt{1-(x-1)^2}\). This describes a semicircle with center at \( (1,0) \) and radius 1, centered on the x-axis and extending to the right.

Step 2: Convert the Expression to Polar Coordinates

In polar coordinates, \(x = r \cos \theta\), \(y = r \sin \theta\), and \(x^2 + y^2 = r^2\). The integrand \(\frac{x+y}{x^2+y^2}\) becomes \(\frac{r(\cos \theta + \sin \theta)}{r^2}\), simplifying to \(\frac{\cos \theta + \sin \theta}{r}\).

Step 3: Identify Limits for Polar Integration

The semicircle described by the region translates to polar coordinates as \(0 \leq r \leq 2\cos \theta\) and \(\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}\). Here, the angle \(\theta\) varies to cover the semicircle from the negative y-axis to the positive y-axis.

Step 4: Set Up the Polar Integral

The polar integral becomes \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2\cos \theta} \frac{\cos \theta + \sin \theta}{r} \cdot r \, dr \, d\theta\), which simplifies to \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2\cos \theta} (\cos \theta + \sin \theta) \, dr \, d\theta\).

Step 5: Evaluate the Integral in terms of \(r\)

Integrate \((\cos \theta + \sin \theta)\) with respect to \(r\), which results in \((\cos \theta + \sin \theta)r\) evaluated from \(0\) to \(2\cos \theta\). This gives \(2\cos^2 \theta + 2\cos \theta \sin \theta\).

Step 6: Evaluate the Resulting Integral in terms of \(\theta\)

Evaluate the integral \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} (2\cos^2 \theta + 2\cos \theta \sin \theta) \, d\theta\). Using trigonometric identities and standard integrals, find the value for each part and sum them up.

Key Concepts

Cartesian Integral

In mathematics, a Cartesian integral deals with integrating a function over a region in the Cartesian coordinate system. This system uses two perpendicular axes, usually labeled as the x-axis and y-axis, to define points on a plane.
The integral \[ \int_{0}^{2} \int_{0}^{\sqrt{1-(x-1)^{2}}} \frac{x+y}{x^{2}+y^{2}} \, d y \, d x \] represents a Cartesian integral because it involves standard x and y coordinates.When evaluating a Cartesian integral, it's essential to understand the limits of integration and the region they define. In our original problem, the x-limits and y-limits describe a semicircular region within a rectangular boundary, which is where the integration occurs.
By understanding this, you can accurately convert the integral into another coordinate system like polar coordinates, which often simplifies the calculations for circular or radial regions.

Semicircle Region

The concept of a semicircle region is vital in this problem because it defines the area over which to integrate. A semicircle is simply half of a circle; here, described as having a radius of 1, centered at the point \((1,0)\), and extending to the right on the x-axis.
This geometric shape forms the boundary in which the function \(\frac{x+y}{x^2+y^2}\) is integrated.The region in polar coordinates translates into a distinct set of limits for \(r\) and \(\theta\). For this particular semicircle, \(r\) ranges from \(0\) to \(2\cos \theta\) and \(\theta\) spans from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\). Understanding these limits helps you accurately evaluate the integral in polar coordinates.
Visualizing the semicircle also aids in setting up the integral correctly by ensuring that it accurately represents the area in question.

Integration in Polar Coordinates

Polar coordinates offer a powerful tool for solving integrals defined over circular regions. By switching from Cartesian to polar coordinates, complex integrals can become significantly more manageable.
In polar coordinates, each point in the plane is represented by a radius \(r\) and an angle \(\theta\), thus simplifying the descriptions of circular regions.The original problem's function, \(\frac{x+y}{x^2+y^2}\), is converted in polar form to \(\frac{\cos \theta + \sin \theta}{r}\). Upon multiplying by the Jacobian factor \(r\), it simplifies to \(\cos \theta + \sin \theta\). Thus, the integral in polar form becomes:\[\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2\cos \theta} (\cos \theta + \sin \theta) \, dr \, d\theta\]This simplicity is often the main advantage of using polar coordinates for integration in circular regions.

Trigonometric Identities

Trigonometric identities play a crucial role in evaluating integrals, particularly those in polar coordinates.
In this problem, trigonometric identities simplify the integral involving \(\cos \theta\) and \(\sin \theta\).To evaluate:\[\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} (2\cos^2 \theta + 2\cos \theta \sin \theta) \, d\theta\]you will use identities such as:

• \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \)
• \( 2\cos \theta \sin \theta = \sin(2\theta) \)

These identities transform the integral into components that are easier to compute. For example, \(\cos^2\theta\) turns into a sum of constants and simple trigonometric functions, which is straightforward to integrate over the given bounds.
The problem then simplifies to evaluating standard integrals, making the calculation process smoother and less prone to error.