Problem 44 - 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. \(\int_{0}^{1} \int_{0}^{x / 2} \frac{x}{x^{2}+y^{2}} d y d x\)

Step-by-Step Solution

Answer

The equivalent polar integral evaluates to \( \frac{1}{2} \ln(5^{1/2}+1) \).

Step 1: Plot Cartesian Region

In the given Cartesian integral, we have the bounds for \( y \) as \( 0 \leq y \leq \frac{x}{2} \) and for \( x \) as \( 0 \leq x \leq 1 \). This means that we are dealing with a triangular region in the first quadrant, where the upper boundary is the line \( y = \frac{x}{2} \) and the left boundary is the line \( y = 0 \) from \( x = 0 \) to \( x = 1 \). Plotting these on the \( xy \)-plane, you get a right triangle with vertices at \((0,0)\), \((0.5,0)\), and \((1,0.5)\).

Step 2: Convert Cartesian Boundaries to Polar

To convert the equation \( y = \frac{x}{2} \) into polar coordinates, recall the relationships \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Substitute these into the equation:\[ r \sin(\theta) = \frac{r \cos(\theta)}{2} \]Simplify to get \( \tan(\theta) = \frac{1}{2} \), which implies that \( \theta = \tan^{-1}(\frac{1}{2}) \). The line \( y = 0 \) corresponds to \( \theta = 0 \), and the line \( x = 1 \) becomes \( r \cos(\theta) = 1 \) in polar coordinates.

Step 3: Plot Polar Region

In the \( r\theta \)-plane, the region is bounded by \( 0 \leq \theta \leq \tan^{-1}\left(\frac{1}{2}\right) \) and the radial line \( \theta = \tan^{-1}(0) = 0 \). The radial limit is where \( r = \frac{1}{\cos(\theta)} \) for \( x = 1 \), thus the polar region is from \( \theta = 0 \) to \( \theta = \tan^{-1}(1/2) \) and \( r \) from \( 0 \) to \( \frac{1}{\cos(\theta)} \).

Step 4: Change Integrand to Polar Form

The integrand \( \frac{x}{x^2 + y^2} \) converts to polar coordinates by replacing \( x \) with \( r \cos(\theta) \) and \( x^2 + y^2 \) with \( r^2 \). Also, remember to multiply by the Jacobian \( r \) when converting to polar coordinates:\[ \frac{r \cos(\theta)}{r^2} \times r = \cos(\theta) \]So the integrand becomes \( \cos(\theta) \).

Step 5: Determine Limits and Evaluate Integral

We will now evaluate the polar integral. The limits of \( \theta \) are from \( 0 \) to \( \tan^{-1}(1/2) \) and for \( r \) from \( 0 \) to \( \frac{1}{\cos(\theta)} \). The integral becomes:\[ \int_{0}^{\tan^{-1}(1/2)} \int_{0}^{1/\cos(\theta)} \cos(\theta) \cdot r \, dr \, d\theta \]Evaluate the inner integral first with respect to \( r \):\[ \int_{0}^{1/\cos(\theta)} r \, dr = \left. \frac{r^2}{2} \right|_0^{1/\cos(\theta)} = \frac{1}{2\cos^2(\theta)} \]Substitute back and evaluate:\[ \int_{0}^{\tan^{-1}(1/2)} \frac{1}{2\cos^2(\theta)} \cdot \cos(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\tan^{-1}(1/2)} \frac{1}{\cos(\theta)} \, d\theta \]This final simplifies to:\[ \frac{1}{2} [ \ln|\sec(\theta) + \tan(\theta)|]_0^{\tan^{-1}(1/2)} \]By evaluating the above expression, we obtain the numerical result (or approximation) for the integral.

Key Concepts

Cartesian to Polar Conversion

To convert Cartesian coordinates to polar coordinates, we use the relationships \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Polar coordinates use the radius \( r \), which is the distance from the origin, and the angle \( \theta \), which is the angle counter clockwise from the positive x-axis.
In the context of integration, converting from Cartesian to polar coordinates allows us to simplify expressions or better understand the geometry of a problem.
When you have a line such as \( y = \frac{x}{2} \), you can substitute the polar relationships into it to find \( r \) and \( \theta \). This specific line becomes \( \tan(\theta) = \frac{1}{2} \), showing how a set boundary in Cartesian form can translate into specific angle limitations in polar coordinates.

Limits of Integration

The limits of integration in polar coordinates differ from Cartesian coordinates because they involve radial and angular boundaries. In polar coordinates, the region of integration is specified by the range of \( r \) and \( \theta \).
For instance, in this problem, \( \theta \) varies from 0 to \( \tan^{-1}(\frac{1}{2}) \), corresponding to the angles formed by the lines from the origin.
Simultaneously, \( r \) varies based on the intersection point with the boundary\( x=1 \), which transforms to \( r \cos(\theta) = 1 \), resulting in \( r \) ranging from 0 to \( \frac{1}{\cos(\theta)} \). These sets of limits form a mapped triangular region within the space defined by the angles and radius.

Polar Coordinates

Polar coordinates provide a crucial tool for understanding and solving integrals, especially when dealing with circular or radial symmetry. Unlike Cartesian coordinates, which give positions on a flat plane through \( (x, y) \), polar coordinates describe a point by how far it is from the origin (\( r \)) and the angle (\( \theta \)) it makes with the positive x-axis.
Polar coordinates are expressed as \((r, \theta)\), where:

• \( r \) (radius) is the distance from the origin.
• \( \theta \) (angle) is measured in radians from the positive x-axis.

These coordinates simplify calculus operations in circular regions, thanks to their straightforward handling of angles and distances from a singular origin point. This also allows for a more natural description of curves and regions, facilitating their use in integration.

Integration Techniques

While solving integrals in a polar coordinate system, it's important to remember changes like multiplying by the Jacobian, \( r \), which accounts for the transformation from two-dimensional Cartesian space to polar areas.
The integrand form transforms as follows: the \( x \) value becomes \( r \cos(\theta) \) and the expression \( x^2 + y^2 \) translates to \( r^2 \). For example, from the original problem, \( \frac{x}{x^2 + y^2} \) converts to \( \cos(\theta) \) when multiplied by the Jacobian \( r \).
The computational steps look as follows:

• Calculate the inner integral over \( r \).
• Substitute back, simplifying to treat the outer integral over \( \theta \).

Mastering polar integration involves recognizing these conversion and substitution techniques, which simplify the evaluation of complex regions.