Problem 23 - Calculus Early Transcendentals

Question

Sketch the given region of integration \(R\) and evaluate the integral over \(R\) using polar coordinates. $$\iint_{R}\left(x^{2}+y^{2}\right) d A ; R=\\{(r, \theta): 0 \leq r \leq 4,0 \leq \theta \leq 2 \pi\\}$$

Step-by-Step Solution

Answer

Question: Sketch the region of integration R and evaluate the given double integral over R using polar coordinates, given the limits of integration for r and θ are \(0 \leq r \leq 4\) and \(0 \leq \theta \leq 2 \pi\), respectively. Answer: The region of integration R is a full circle centered at the origin with radius 4. By converting the integral to polar coordinates and evaluating it, we find that the value of the double integral over R is \(128\pi\).

Step 1: Sketch the region of integration R

To sketch the region of integration R, we need to understand the limits of integration for r and θ. It is given that \(0 \leq r \leq 4\) and \(0 \leq \theta \leq 2 \pi\). These limits tell us that the region of integration is the full circle centered at the origin with radius 4.

Step 2: Convert the integral from Cartesian to polar coordinates

The integral given in the problem is: $$\iint_{R}\left(x^{2}+y^{2}\right) d A$$ To convert this integral to polar coordinates, we can make the following substitutions: 1. Replace \(x^2 + y^2\) with \(r^2\), since \(x^2 + y^2 = r^2\) in polar coordinates. 2. Replace \(dA\) with \(r \ dr \ d\theta\), since the area element in polar coordinates is \(r \ dr \ d\theta\). So the integral in polar coordinates becomes: $$\iint_{R}r^2 \cdot r \ dr\ d\theta$$

Step 3: Evaluate the integral using the limits of integration for r and θ

Now, let's evaluate the integral using the given limits of integration: $$\int_{0}^{2\pi}\int_{0}^{4} r^3 dr\ d\theta$$ We first integrate with respect to r: $$\int_{0}^{2\pi}\left[\frac{1}{4}r^4\right]_{0}^{4}\ d\theta = \int_{0}^{2\pi} 64 \ d\theta$$ Now, we integrate with respect to θ: $$\left[64\theta\right]_{0}^{2\pi} = 64(2\pi) - 64(0) = 128\pi$$ So, the final answer is: $$\iint_{R}\left(x^{2}+y^{2}\right) d A = 128\pi$$.

Key Concepts

Region of Integration

In the context of problems involving integrals, understanding the region of integration is crucial. The exercise provides the region of integration in the form of polar coordinates: \( R = \{(r, \theta) : 0 \leq r \leq 4, 0 \leq \theta \leq 2\pi \} \). This means that we're dealing with a circular region. The circle is centered at the origin \((0, 0)\) with a radius of 4 units.
In polar coordinates:

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

Here:

• The limits \( 0 \leq r \leq 4 \) indicate that the radius varies from 0 to 4, filling up a complete disk.
• \( 0 \leq \theta \leq 2\pi \) denotes a full rotation around the circle, covering 360 degrees.

When translating this into a sketch, the picture becomes a complete circle, and this picture is vital for visualizing how the integration will proceed. Understanding this helps in correctly setting up the limits for the integral.

Converting Integrals

Converting integrals from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) is a useful method, especially when dealing with circular regions.
The expression given in the integral \( \iint_{R} (x^2 + y^2) \ dA \) needs conversion:

• Given that in Cartesian, \( x^2 + y^2 = r^2 \) in polar, this forms the basis for substitution.
• The differential area element in Cartesian coordinates \( dA \) converts to \( r \ dr \ d\theta \) in polar, reflecting the area swept out by changing \( r \) and \( \theta \).

The integral thus becomes \( \iint_{R} r^2 \cdot r \ dr \ d\theta \) or \( \iint_{R} r^3 \ dr \ d\theta \). This change simplifies the integration greatly when the region and the function itself have circular symmetry. Understanding and performing these conversions helps tackle more complex integration problems efficiently.

Double Integrals

Double integrals allow us to calculate the volume under surfaces, across two-dimensional regions. In the exercise, we begin with an integral \( \iint_{R} (x^2 + y^2) \ dA \) over a circular region defined in polar coordinates.
With the conversion complete, the double integral is set up as follows:

• First, integrate in terms of \( r \), \([0, 4]\), resulting in \( \int_{0}^{4} r^3 \ dr \).
• This simplifies to \( \frac{1}{4}r^4 \) evaluated from 0 to 4, which computes to a surface area of 64.
• Next, integrate in terms of \( \theta \), \( [0, 2\pi] \), with the integrand 64 as a constant.

Performing this final integration gives \( 128\pi \), representing the volume under the radial surface described by \( (x^2 + y^2) \).
Showing these calculations step by step ensures a clear understanding of how double integrals work, and why switching to polar is often advantageous in symmetric regions.