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Chapter 15: Problem 34

In polar coordinates, the average value of a function over a region \(R\) (Section 15.3\()\) is given by $$\frac{1}{\text { Area(R) }} \iint_{R} f(r, \theta) r d r d \theta$$ Average height of a cone Find the average height of the (single) cone \(z=\sqrt{x^{2}+y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\) -plane.e.

Short Answer

Expert verified
The average height of the cone above the disk is \( \frac{2a}{3} \).

Step by step solution

01

Convert to Polar Coordinates

The cone is defined as \( z = \sqrt{x^2 + y^2} \) and in polar coordinates, \( x = r \cos \theta \), \( y = r \sin \theta \). Thus, \( z = r \).
02

Set Up the Integral for Average Value

Calculate the average height as \[ \frac{1}{\text{Area(R)}} \iint_{R} r^2 \, dr \, d\theta \].
03

Find the Area of the Disk

The area of the disk \( R \) is \( \pi a^2 \) since it is a circle with radius \( a \).
04

Integrate Over the Disk

Evaluate the integral \[ \int_{0}^{2\pi} \int_{0}^{a} r^2 \, dr \, d\theta \]. First solve the inner integral: \[ \int_{0}^{a} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^a = \frac{a^3}{3} \].
05

Complete the Double Integral

Now solve the outer integral: \[ \int_{0}^{2\pi} \frac{a^3}{3} \, d\theta = \frac{a^3}{3} \left[ \theta \right]_0^{2\pi} = \frac{a^3}{3} \times 2\pi = \frac{2\pi a^3}{3} \].
06

Calculate the Average Height

The average height is obtained by dividing the result of the integral by the area of the disk: \[ \frac{1}{\pi a^2} \times \frac{2\pi a^3}{3} = \frac{2a}{3} \].

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Key Concepts

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

Average Value
The concept of average value of a function over a region is essential in calculating characteristics over a specified area. In polar coordinates, this average value is calculated using a double integral.
Consider a region \( R \) with area \( \text{Area(R)} \). If we have a function \( f(r, \theta) \), its average value over \( R \) is given by
  • \( \frac{1}{\text{Area(R)}} \iint_{R} f(r, \theta) \, r \, dr \, d\theta \)
This formula allows us to divide the total effect of \( f \) by the area, providing the mean effect per unit area in the region.
The average height of a cone can be understood as its mean elevation above the base, which is what the exercise is trying to find using these concepts.
Double Integral
Double integrals are a powerful tool in calculus for computing over two-dimensional regions. Specifically, they allow us to accumulate quantities over an area. This is done by integrating a function twice, first with respect to one variable, and then the other.
In the problem context, double integration over polar coordinates is used:
  • Set up as \( \int_{0}^{2\pi} \int_{0}^{a} r^2 \, dr \, d\theta \)
  • First, integrate with respect to \( r \) from 0 to \( a \), giving \( \frac{r^3}{3} \) evaluated at the limits.
  • Then integrate the outer \( d\theta \) from 0 to \( 2\pi \), giving the total \( \frac{2\pi a^3}{3} \).
This structured step of integration helps compute volume-like entities or area-related computations within certain bounds. In this case, it determines how high the cone stands on average above its circular base.
Area of a Disk
In terms of geometry, a disk is essentially a circle with all the interior points included. To find the area of a disk, the well-known formula \( \pi r^2 \) is employed where \( r \) is the radius of the circle.
For this exercise:
  • The disk in the \( xy \)-plane is defined as \( x^2 + y^2 \leq a^2 \).
  • The area of the disk \( R \) turns out to be \( \pi a^2 \), as the radius here is \( a \).
Understanding the area of the disk is fundamental because it's used as a divisor when you calculate the average value. It reflects the area over which a property of the cone is averaged, giving the average height.
Cone in Polar Coordinates
Converting equations into polar coordinates is particularly useful in cases involving symmetry around a point, like a circular disk.
Here, the cone is defined as the surface \( z = \sqrt{x^2 + y^2} \). In polar coordinates, this transforms because:
  • The position \( (x, y) \) translates to \( (r \cos \theta, r \sin \theta) \).
  • This makes \( \sqrt{x^2 + y^2} = r \), simplifying many calculations.
By treating the problem in terms of polar coordinates, we utilize the natural radial symmetry of the circle, aligning with the cone's geometric properties. This clever transformation makes the integral setup and computation much smoother and more intuitive for problems involving rotational symmetry.

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Most popular questions from this chapter

The Parallel Axis Theorem Let \(L_{\mathrm{cm}}\) be a line through the center of mass of a body of mass \(m\) and let \(L\) be a parallel line \(h\) units away from \(L_{c . m .}\) The Parallel Axis Theorem says the moments of inertia \(I_{\mathrm{cm}}\) and \(I_{L}\) of the body about \(L_{\mathrm{cm}}\) and \(L\) satisfy the equation $$ I_{L}=I_{\mathrm{c.m.}}+m h^{2} $$ As in the two-dimensional case, the theorem gives a quick way to calculate one moment when the other moment and the mass are known. Proof of the Parallel Axis Theorem a. Show that the first moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the \(y z\) -plane. What does the formula \(\overline{x}=M_{y z} / M\) then tell you?) b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line \(L_{c . m . \text { along the }}\) \(z\) z-axis and the line \(L\) perpendicular to the \(x y\) -plane at the point \((h, 0,0) .\) Let \(D\) be the region of space occupied by the body. Then, in the notation of the figure, $$ I_{L}=\iiint_{D}|\mathbf{v}-h \mathbf{i}|^{2} d m $$ Expand the integrand in this integral and complete the proof.

In Exercises \(49-52,\) use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. $$ \begin{array}{l}{\text { a. Plot the Cartesian region of integration in the } x y \text { -plane. }} \\ {\text { b. Change each boundary curve of the Cartesian region in part }} \\ {\text { (a) to its polar representation by solving its Cartesian equation for } r \text { and } \theta .}\end{array} $$ $$ \begin{array}{l}{\text { c. Using the results in part (b), plot the polar region of integra- }} \\ {\text { tion in the } r \theta \text { -plane. }} \\\ {\text { d. Change the integrand from Cartesian to polar coordinates. De- }} \\ {\text { termine the limits of integration from your coordinates. De- }} \\ {\text { evaluate the polar integral using the CAS integration utility. }}\end{array} $$ $$ \int_{0}^{1} \int_{-y / 3}^{y / 3} \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y $$

Sketch the region of integration and evaluate the integral. \begin{equation} \int_{0}^{\pi} \int_{0}^{\sin x} y d y d x \end{equation}

Use Fubini's Theorem to evaluate $$\int_{0}^{1} \int_{0}^{3} x e^{x y} d x d y$$

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation} \int_{0}^{2 \sqrt{\ln 3}} \int_{y / 2}^{\sqrt{\ln 3}} e^{x^{2}} d x d y \end{equation}
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