Problem 29 - Advanced Engineering Mathematics

Question

Evaluate the given iterated integral by changing to polar coordinates. $$ \int_{0}^{1} \int_{\sqrt{1-x^{2}}}^{\sqrt{4}-x^{2}} \frac{x^{2}}{x^{2}+y^{2}} d y d x+\int_{1}^{2} \int_{0}^{\sqrt{4}-x^{2}} \frac{x^{2}}{x^{2}+y^{2}} d y d x $$

Step-by-Step Solution

Answer

The value of the iterated integral is \( \frac{3\pi}{8} \).

Step 1: Analyze the Integral Bounds

Observe that the integral is split into two parts. For the first part, the outer limits range from 0 to 1 with the inner limits from \( y = \sqrt{1-x^2} \) to \( y = \sqrt{4-x^2} \). The second part has outer limits from 1 to 2, and inner limits of \( y = 0 \) to \( y = \sqrt{4-x^2} \). These describe a region under two parts of a circle divided at \( x = 1 \).

Step 2: Set Up Polar Coordinates

Recall that in polar coordinates, \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The transformation requires the region to be split as follows: for \( 0 \leq x \leq 1 \), \( r \) ranges from 1 to 2, and for \( 1 \leq x \leq 2 \), \( r \equiv \sqrt{x^2+y^2} \) ranges from 1 to 2. \( \theta \) goes from 0 to \( \pi/2 \) for both parts.

Step 3: Convert the Function to Polar Coordinates

Substitute \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \) into the expression \( \frac{x^2}{x^2+y^2} = \cos^2(\theta) \). The differential element \( dA = dy \, dx \) converts into \( r \, dr \, d\theta \) in polar coordinates.

Step 4: Evaluate the Polar Integral

The integral becomes \( \int_{0}^{\pi/2} \int_{1}^{2} \cos^2(\theta) \, r \, dr \, d\theta \). First, evaluate the integral with respect to \( r \):\[\int_{1}^{2} r \, dr = \left[ \frac{r^2}{2} \right]_{1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}\]Then, the integral with respect to \( \theta \):\[\int_{0}^{\pi/2} \cos^2(\theta) \, d\theta\]

Step 5: Solve the Theta Integral

Use the trigonometric identity \( \cos^2(\theta) = \frac{1}{2}(1+\cos(2\theta)) \) to integrate:\[\int_{0}^{\pi/2} \cos^2(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\pi/2} (1 + \cos(2\theta)) \, d\theta \]Evaluate this:\[\frac{1}{2} \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{0}^{\pi/2} = \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{\pi}{4}\]

Step 6: Calculate the Final Result

Multiply the results of the \( r \) and \( \theta \) integrals: \( \frac{3}{2} \times \frac{\pi}{4} = \frac{3\pi}{8} \). Thus, the integral over the given region is \( \frac{3\pi}{8} \).

Key Concepts

Iterated Integral

An iterated integral refers to the successive integration of a function in multiple variables over designated limits. In the given problem, there are two variables, \(x\) and \(y\). The iterated integral is split into two parts:

• The first part integrates \(y\) from \(\sqrt{1-x^2}\) to \(\sqrt{4-x^2}\) while \(x\) goes from 0 to 1.
• The second part integrates \(y\) from 0 to \(\sqrt{4-x^2}\), with \(x\) ranging from 1 to 2.

These limits represent regions cut out within a circle of radius 2 centered at the origin. By evaluating the intersections along the \(x\)-axis, one segment goes until \(x = 1\), and the other continues from 1 to 2. Iterated integrals simplify the process by allowing one to integrate in stages, starting from the innermost function.

Trigonometric Identity

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides of the identity are defined. They are immensely helpful in calculus, especially during integration.In this exercise, one such identity used is \(\cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta))\). This identity facilitates the transformation of the integral into a form that is easier to compute. It helps break down the squared terms into coefficients and simpler sine and cosine functions that are straightforward to integrate.Applying this identity in the polar integral simplifies the integration over \(\theta\), transforming \(\cos^2(\theta)\) which is straightforward to integrate using the identity. This results in a simple evaluation over a defined interval.

Transformation to Polar Coordinates

Transforming Cartesian coordinates to polar coordinates involves the use of relationships between \(x\), \(y\), and the radius \(r\) along with the angle \(\theta\). These relationships are defined as:

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

The transformation to polar coordinates is advantageous in evaluating integrals over circular regions or in cases where the region can be more naturally expressed in terms of radii and angles. In this problem, expressing the functions in polar form and their respective bounds becomes more manageable. The transformation also simplifies the function \(\frac{x^2}{x^2 + y^2}\) to \(\cos^2(\theta)\), which directly leverages the polar coordinate properties for easier integration.

Integration in Polar Coordinates

Integration in polar coordinates utilizes the radial distance \(r\) and the angle \(\theta\) as variables. In the transformation from Cartesian to polar, the area element \(dA = dy \, dx\) changes to \(r \, dr \, d\theta\).The integral \[\int_{0}^{\pi/2} \int_{1}^{2} \cos^2(\theta) \, r \, dr \, d\theta\] involves evaluating two separate integrals.First, integrating with respect to \(r\). Here, \(r\) ranges from 1 to 2, and as a result the integral becomes:\[\left[ \frac{r^2}{2} \right]_{1}^{2} = \frac{3}{2}\]Next, integrate with respect to \(\theta\). This requires the prior trigonometric identity, leading to:\[\frac{1}{2} \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{0}^{\pi/2} = \frac{\pi}{4}\]Together, these calculations provide the full integral value as \(\frac{3\pi}{8}\), successfully evaluating the region described by the original iterated integral in a much simpler form.