Problem 18 - Thomas Calculus

Question

In Exercises \(9-22,\) change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$ \int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} \frac{2}{\left(1+x^{2}+y^{2}\right)^{2}} d y d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Region of Integration

The given Cartesian integral is \[ \int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} \frac{2}{\left(1+x^{2}+y^{2}\right)^{2}} d y d x \].Note that the inner integral limits for \(y\) are from \(-\sqrt{1-x^2}\) to \(\sqrt{1-x^2}\), which describes a semicircle in the \(xy\)-plane with radius 1. This suggests a circular region.

Step 2: Convert to Polar Coordinates

In polar coordinates, \(x = r\cos\theta\), \(y = r\sin\theta\), and \(x^2 + y^2 = r^2\). The Jacobian of the transformation is \(r\). Thus, the given integral becomes\[ \int_{0}^{2\pi} \int_{0}^{1} \frac{2}{\left(1+r^2\right)^{2}} r \, dr \, d\theta \].The limits for \(r\) are from 0 to 1 and for \(\theta\) are from 0 to \(2\pi\) to cover the full circle.

Step 3: Compute the Polar Integral

The integral now is \[ \int_{0}^{2\pi} \int_{0}^{1} \frac{2r}{\left(1+r^2\right)^{2}} dr \, d\theta \]. First, solve the inner integral with respect to \(r\):Let \( u = 1 + r^2 \), then \( du = 2r \, dr \). Change limits: when \( r = 0, \ u = 1 \), and when \( r = 1, \ u = 2 \). Thus,\[ \int_{0}^{1} \frac{2r}{\left(1+r^2\right)^{2}} dr = \int_{1}^{2} u^{-2} du = \left[ -u^{-1} \right]_{1}^{2} = -\frac{1}{2} + 1 = \frac{1}{2}. \]

Step 4: Evaluate the Angular Integral

Now solve the angular integral\[ \int_{0}^{2\pi} \frac{1}{2} d\theta = \frac{1}{2} \cdot \theta \Bigg|_0^{2\pi} = \pi. \]

Step 5: Combine Results

The value of the original Cartesian integral after conversion and evaluation is thus \(\pi\). Combine the results from Steps 3 and 4 to conclude the final result of the integral.

Key Concepts

Cartesian integral

A Cartesian integral is an integral expressed in terms of Cartesian coordinates, which utilize the familiar x and y coordinate axes to define points in a plane. This form of the integral is useful when working with rectangular regions in the xy-plane.
For the exercise at hand, the integral is given in Cartesian form:

• The limits of the outer integral for x are from -1 to 1, which describe the width of a semicircle.
• The inner integral is for y, going from -\(\sqrt{1-x^2}\) to \(\sqrt{1-x^2}\), representing the top and bottom halves of the semicircle.

This semicircle forms a simple shape in Cartesian coordinates, but potentially complicates the integral evaluation due to the algebraic square root expressions. To simplify integration and handle circular shapes more naturally, conversion to polar coordinates is often employed.

polar integral

A polar integral makes use of polar coordinates, which are defined by the radius r and the angle \(\theta\). This system is particularly efficient for integrating over circular regions, as it replaces complex square root expressions with simpler trigonometric ones.
By converting our Cartesian integral's region into polar coordinates,

• The relation of x and y becomes: \(x = r\cos\theta\), \(y = r\sin\theta\)
• The expression \(x^2 + y^2\) changes into simply \(r^2\)

This transformation adapts much more fluently to circular shapes, allowing the problem to be tackled with a change of variables that matches the geometry perfectly. The polar integral over the entire circle then changes the domain where

• r varies from 0 to 1, describing radial distance
• \(\theta\) varies from 0 to \(2\pi\) to complete the circular sweep.

transformation Jacobian

In converting a Cartesian integral into a polar integral, the transformation Jacobian is crucial. It's an adjustment factor that takes into account the change from one coordinate system to another.
The Jacobian is computed from the partial derivatives of the transformations:

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

In this case, the transformation Jacobian for polar coordinates is simply r, reflecting the increased area in polar sections as one moves further from the origin.
Integration must therefore include this factor, making the polar integral:

• \(\int_0^{2\pi} \int_0^1 \frac{2r}{(1+r^2)^2} \, dr\, d\theta\)

integral evaluation

The process of evaluating the integral involves computing step-by-step:

• First, tackle the inner integral, concerning the radial part with respect to r.

By substitution, let \(u = 1 + r^2\), which simplifies the expression inside. With \(du = 2r\,dr\), the integration bounds for u are from 1 to 2. The resulting inner integral \(\int_1^2 u^{-2} du\) evaluates to \(-\frac{1}{u}\) evaluated from 1 to 2, resulting in \(\frac{1}{2}\).

• Next, handle the angular part by solving \(\int_0^{2\pi} \frac{1}{2} \, d\theta\).

This integral resolves straightforwardly to \(\frac{1}{2} \cdot \theta \big|_0^{2\pi}\) equaling \(\pi\).
Finally, combining results from the radial and angular parts gives the total area of the original region as \(\pi\). This demonstrates how the transformation to polar coordinates simplifies complex Cartesian integrals, especially over circular regions.