Problem 28 - Advanced Engineering Mathematics

Question

In Problems, evaluate the given iterated integral by changing to polar coordinates. $$ \int_{-\sqrt{\pi}}^{\sqrt{\pi}} \int_{0}^{\sqrt{\pi-x^{2}}} \sin \left(x^{2}+y^{2}\right) d y d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Region of Integration

First, we need to identify the region over which we are integrating. The inner integral is with respect to \( y \), and it goes from \( 0 \) to \( \sqrt{\pi - x^2} \). The outer integral with respect to \( x \) ranges from \( -\sqrt{\pi} \) to \( \sqrt{\pi} \). This region corresponds to the upper half of a circle centered at the origin with radius \( \sqrt{\pi} \) since \( y = \sqrt{\pi - x^2} \) is the upper semicircle.

Step 2: Convert to Polar Coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \), and \( x^2 + y^2 = r^2 \). The Jacobian of the transformation from Cartesian to polar coordinates is \( r \). Therefore, \( dA = r \, dr \, d\theta \). Our integral becomes \[ \int_{0}^{\pi} \int_{0}^{\sqrt{\pi}} \sin(r^2) \, r \, dr \, d\theta. \]

Step 3: Integrate with Respect to r

Evaluate the integral with respect to \( r \):\[ \int_{0}^{\sqrt{\pi}} \sin(r^2) \, r \, dr. \] To solve this, use the substitution \( u = r^2 \), thus \( du = 2r \, dr \) or \( r \, dr = \frac{1}{2} du \). The limits change accordingly from \( r = 0 \) to \( r = \sqrt{\pi} \) giving \( u = 0 \) to \( u = \pi \). This becomes:\[ \frac{1}{2} \int_{0}^{\pi} \sin(u) \, du. \]

Step 4: Evaluate Integral with Respect to u

Determine\[ \frac{1}{2} \int_{0}^{\pi} \sin(u) \, du. \]This evaluates to \(-\cos(u)|_0^\pi = -\cos(\pi) + \cos(0) = -( -1 ) + 1 = 2\). Thus, the integral \( \int_{0}^{\sqrt{\pi}} \sin(r^2) \, r \, dr \) becomes \( 1 \).

Step 5: Integrate with Respect to θ

Now the problem is reduced to:\[ \int_{0}^{\pi} 1 \, d\theta, \]which evaluates straightforwardly as \( \theta |_0^\pi = \pi - 0 = \pi \).

Key Concepts

Iterated Integrals

Iterated integrals are used to evaluate a multiple integral by performing integration successively. In this exercise, the iterated integrals are set over specific bounds for variables, starting from the innermost to the outermost.
The given integral is

• First with respect to \( y \) from \( 0 \) to \( \sqrt{\pi - x^2} \)
• Then with respect to \( x \) from \( -\sqrt{\pi} \) to \( \sqrt{\pi} \)

These bounds describe the region of integration as the upper half of a circle centered at the origin with a radius of \( \sqrt{\pi} \).
By breaking the problem down into easier integrals to evaluate sequentially, iterated integrals help in addressing complex problems, especially when combined with coordinate transformations.

Change of Variables

Change of variables is a technique used to simplify the process of integration by transforming the integration variables.
In this problem, we shift from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) using the relationships:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(x^2 + y^2 = r^2\)

This transformation makes evaluating the integral cleaner and more intuitive because of the circular symmetry of the region.
By choosing polar coordinates, the problem reduces to integrating a simpler function over a more natural region, which can ease calculation and study of such problems.

Jacobian in Coordinate Transformation

In the transformation from Cartesian to polar coordinates, the Jacobian determinant accounts for the change in area element.
The Jacobian factor for changing to polar coordinates is \( r \), and it transforms the area element \( dx \, dy \) to \( r \, dr \, d\theta \) in polar coordinates.

• This means that every tiny piece of area in the Cartesian plane expands or contracts by a factor of \( r \).
• The Jacobian adjusts the integral to reflect the skewed perspective of the new coordinate system.

This central idea is crucial, as it preserves the integral’s value during variable transformations.
Understanding the role of the Jacobian helps in correctly applying transformations and evaluating integrals over various coordinate systems.