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Chapter 15: Problem 35

In polar coordinates, the average value of a function over a region \(R\) (Section 15.3 ) is given by \begin{equation}\frac{1}{\operatorname{Area}(R)} \iint_{R} f(r, \theta) r d r d \theta.\end{equation} \begin{equation} \begin{array}{l}{\text { Average distance from interior of disk to center Find the }} \\ {\text { average distance from a point } P(x, y) \text { in the disk } x^{2}+y^{2} \leq a^{2} \text { to }} \\ {\text { the origin. }}\end{array}\end{equation}

Short Answer

Expert verified
The average distance to the center from any point in the disk is \(\frac{2a}{3}\).

Step by step solution

01

Understanding the Region of Integration

The problem asks us to find the average distance from a point within a disk of radius \(a\) to the origin. The region \(R\) is described by the boundary \(x^2 + y^2 \leq a^2\), which in polar coordinates translates to \(0 \leq r \leq a\) and \(0 \leq \theta \leq 2\pi\).
02

Set Up the Function for Average Distance

The function for the distance from a point \((r, \theta)\) to the origin is simply \(f(r, \theta) = r\). We want to integrate \(r\) over the area of the disk.
03

Calculate Area of the Disk

The area of the disk \(R\) with radius \(a\) is given by \(\pi a^2\). This will be used to normalize the double integral.
04

Set Up the Double Integral for Average Distance

The average value is given by \(\frac{1}{\operatorname{Area}(R)} \iint_{R} r \cdot r \, dr \, d\theta\). Thus, our integral becomes \(\frac{1}{\pi a^2} \int_0^{2\pi} \int_0^a r^2 \, dr \, d\theta\).
05

Evaluate the Inner Integral

First, integrate with respect to \(r\). \[ \int_0^a r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^a = \frac{a^3}{3} \].
06

Evaluate the Outer Integral

Now, integrate with respect to \(\theta\). \[ \int_0^{2\pi} \frac{a^3}{3} \, d\theta = \frac{a^3}{3} \cdot [\theta]_0^{2\pi} = \frac{a^3}{3} \, (2\pi) = \frac{2\pi a^3}{3} \].
07

Calculate the Average Distance

Insert the result of the double integral into the formula for the average value and use the area of the disk: \[ \text{Average distance} = \frac{1}{\pi a^2} \times \frac{2\pi a^3}{3} = \frac{2a}{3} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Value of a Function
To determine the average value of a function over a particular area, you need to use integration, specifically a double integral in polar coordinates. The formula is:
\[ \text{Average Value} = \frac{1}{\operatorname{Area}(R)} \iint_{R} f(r, \theta) \, r \, dr \, d\theta \] The function \(f(r, \theta)\) represents what you are averaging鈥攊n this case, the distance from a point within the disk to the center.
The denominator involves the area of the region \(R\) to normalize the integral. By using this formula, you incorporate the entire region in your calculations, ensuring that the average takes into account all possible positions within the disk.
Double Integral
A double integral extends the idea of integrals to two dimensions. It's used here because we are dealing with a region of a disk, which is two-dimensional.
In polar coordinates, the double integral setup allows you to consider both the radial and angular components. The integration over \(r\) and \(\theta\) means you account for every possible point in the region.
For our problem, the double integral becomes:- Inner integral with respect to \(r\): \( \int_0^a r^2 \, dr \)- Outer integral with respect to \(\theta\): \( \int_0^{2\pi} \)'s result from the inner integral
These two integrals together cover the entire disk, giving you the cumulative effect of all potential distance values.
Region of Integration
The region of integration is the area in which you are interested for the problem鈥攊n this case, the disk described by \(x^2 + y^2 \leq a^2\). In polar coordinates, this becomes a simpler range of \(0 \leq r \leq a\) for the radial part and \(0 \leq \theta \leq 2\pi\) for the angular part.
Visualize this as a full circle with radius \(a\), centered at the origin. All relevant calculations are constrained to this circle.
  • \(r\) represents the distance from the origin, starting at the origin and extending out to the edge of the disk.
  • \(\theta\) represents the angle, encompassing a complete rotation around the circle.
The simplicity of polar coordinates makes dealing with circular regions straightforward and logical.
Disk Area
The area of the disk is crucial for determining the average value of a function because it is used to normalize the double integral. Without this normalization, the integral would not reflect the average over the entire area.
The area \(R\) of a disk with radius \(a\) is calculated by the formula \(\pi a^2\). This classic formula comes from the geometry of circles and confirms the size of the region over which you integrate.
Knowing the length around an entire disk lets you factor this into the average, making sure that when you find the average distance, it truly reflects the full scope of the disk.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(41-44\) by changing the order of integration in an appropriate way. $$ \int_{0}^{2} \int_{0}^{4-x^{2}} \int_{0}^{x} \frac{\sin 2 z}{4-z} d y d z d x $$

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals. $$ \int_{\pi / 6}^{\pi / 3} \int_{\csc \phi}^{2 \csc \phi} \int_{0}^{2 \pi} \rho^{2} \sin \phi d \theta d \rho d \phi $$

Set up the iterated integral for evaluating \(\int \int \int_{D} f(r, \theta, z) d z r d r d \theta\) over the given region \(D .\) $$ \begin{array}{l}{D \text { is the prism whose base is the triangle in the } x y \text { -plane bounded }} \\ {\text { by the } y \text { -axis and the lines } y=x \text { and } y=1 \text { and whose top lies in }} \\ {\text { the plane } z=2-x}\end{array} $$

Converting to polar integrals \begin{equation} \begin{array}{l}{\text { a. The usual way to evaluate the improper integral }} \\\ \quad {I=\int_{0}^{\infty} e^{-x^{2}} d x \text { is first to calculate its square: }}\end{array} \end{equation} \begin{equation} \\ \quad I^{2}=\left(\int_{0}^{\infty} e^{-x^{2}} d x\right)\left(\int_{0}^{\infty} e^{-y^{2}} d y\right)=\int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y \end{equation} \begin{equation} \begin{array}{l} \\ \quad \quad \quad \quad {\text { Evaluate the last integral using polar coordinates and solve }} \\ \quad \quad \quad \quad {\text { the resulting equation for } I .} \\ \quad \quad \quad {\text { b. Evaluate }}\end{array}\end{equation} \begin{equation}\lim _{x \rightarrow \infty} \operatorname{erf}(x)=\lim _{x \rightarrow \infty} \int_{0}^{x} \frac{2 e^{-t^{2}}}{\sqrt{\pi}} d t.\end{equation}

Centroid Find the centroid of the region in the first octant that is bounded above by the cone \(z=\sqrt{x^{2}+y^{2}}\) , below by the plane \(z=0,\) and on the sides by the cylinder \(x^{2}+y^{2}=4\) and the planes \(x=0\) and \(y=0\)
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