Problem 25 - Advanced Engineering Mathematics

Question

In Problems, evaluate the given iterated integral by changing to polar coordinates. $$ \int_{-3}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x $$

Step-by-Step Solution

Answer

The value of the iterated integral is \( 9\pi \).

Step 1: Understand the Region of Integration

The limits of integration for \( y \) are from \( 0 \) to \( \sqrt{9-x^2} \) and for \( x \) from \( -3 \) to \( 3 \). This region describes the upper half of a circle with a radius of 3 centered at the origin. We need to change variables to polar coordinates to simplify the integration.

Step 2: Convert to Polar Coordinates

In polar coordinates, we have \( x = r \cos \theta \) and \( y = r \sin \theta \). The integral \( \sqrt{x^2 + y^2} \) simplifies to \( r \) in polar coordinates. The Jacobian of the transformation is \( r \), so the differential \( dy \, dx \) becomes \( r \, dr \, d\theta \).

Step 3: Determine New Limits

For a circle of radius 3, the region of integration in polar coordinates for \( r \) is from 0 to 3. Since the region is only the upper half of the circle, \( \theta \) will vary from \( 0 \) to \( \pi \).

Step 4: Set Up the Polar Integral

Substitute the variables and limits into the polar integral: \[ \int_{0}^{\pi} \int_{0}^{3} r^2 \, dr \, d\theta \].

Step 5: Evaluate the Inner Integral

Compute the inner integral with respect to \( r \): \( \int_{0}^{3} r^2 \, dr \). The antiderivative of \( r^2 \) is \( \frac{r^3}{3} \), evaluated from 0 to 3 gives: \( \frac{3^3}{3} - \frac{0^3}{3} = 9 \).

Step 6: Evaluate the Outer Integral

Now integrate with respect to \( \theta \): \( \int_{0}^{\pi} 9 \, d\theta \). The antiderivative of 9 is \( 9\theta \), which evaluated from \( 0 \) to \( \pi \) gives \( 9\pi \).

Key Concepts

Iterated Integrals

An iterated integral involves execution of multiple integrals in a nested manner, often encountered in multivariable calculus. Here, we have

1. an inner integral: \( \int_{0}^{\sqrt{9-x^2}} \sqrt{x^2+y^2} \, dy \), which is evaluated first with respect to \( y \),
2. and an outer integral: \( \int_{-3}^{3} \, dx \), evaluated next with respect to \( x \).

The concept of iterated integrals simplifies computation by breaking down complex multivariable functions into manageable calculations. Since these integrals are performed over specific limits, they allow us to compute areas, volumes, and other quantities pertinent to the particular region of interest.
The goal in iterated integration is to evaluate the innermost integral first and work your way outwards. This approach typically requires determining and adjusting limits for each variable based on the region described in the integral.

Jacobians

In transformations such as a switch from Cartesian to polar coordinates, Jacobians play a significant role. The Jacobian is a determinant used when changing variables in multiple integrals, representing how much the area or volume element size changes with the transformation.
In the context of converting Cartesian coordinates to polar coordinates:

• We have the transformations \( x = r \cos \theta \) and \( y = r \sin \theta \).
• The Jacobian of this transformation is \( r \).
• This arises because the differential area \( dx \, dy \) in Cartesian translates into \( r \, dr \, d\theta \) in polar.

Thus, when we perform the integration in polar coordinates, we need to multiply by this Jacobian \( r \) to ensure the area element is correctly transformed. It is essential to always include the Jacobian as it accounts for the area scaling during the transformation.

Coordinate Transformation

Coordinate transformation is the process of converting coordinates from one system to another, such as from rectangular to polar coordinates. This is particularly useful when the geometry of the problem aligns more naturally with another coordinate system.
In polar coordinates:

• The transformation equations are \( x = r \cos \theta \) and \( y = r \sin \theta \).
• The function \( \sqrt{x^2 + y^2} \) simplifies significantly to \( r \).
• This simplification is a reason to transform: it reduces complexity.

Transforming to polar coordinates was beneficial in this exercise because the region of integration was circular, and polar coordinates fit symmetrically within circular regions, often simplifying the integration process.

Limits of Integration

The limits of integration define the boundaries over which the integration occurs. They are essential for specifying the region of interest.
In the original problem:

• The limits for \( y \) ranged from \( 0 \) to \( \sqrt{9-x^2} \), with \( x \) from \( -3 \) to \( 3 \).
• These limits bound the upper half of a circle of radius 3 centered at the origin.
• When transitioning to polar coordinates, the radial limit \( r \) becomes \( 0 \) to \( 3 \), and the angular limit \( \theta \) is \( 0 \) to \( \pi \).

Changing coordinates simplifies the limits of integration and frequently makes solving the integral more straightforward and intuitive, especially when the geometry of the problem implies symmetry, as with circular regions.