Question

Find the mass/weight of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\). \(R\) is the annulus in the first and second quadrants bounded by \(x^{2}+y^{2}=9\) and \(x^{2}+y^{2}=36 ; \delta(x, y)=\sqrt{x^{2}+y^{2}} \mid b / f t^{2}\)

Short answer

The mass of the lamina is \( 63\pi \).

Step-by-step solution

Understand the Region R

The region \( R \) is defined as an annulus in the first and second quadrants bounded by the circles \( x^2 + y^2 = 9 \) and \( x^2 + y^2 = 36 \). This implies that \( R \) is the region between these two circles.

Set the Integration Bounds

Since the annulus is in the first and second quadrants, we have the polar coordinate limits for \( \theta \) as \( 0 \) to \( \pi \). The limits for \( r \), which represents the distance from the origin, are from \( 3 \) to \( 6 \) because \( x^2 + y^2 = 9 \) (\( r = 3 \)) and \( x^2 + y^2 = 36 \) (\( r = 6 \)).

Express Density and Set Up the Integral

In polar coordinates, the density function \( \delta(x, y) = \sqrt{x^2 + y^2} \) becomes \( \delta(r) = r \). The mass \( M \) of the lamina can be computed using the integral: \[ M = \int_0^\pi \int_3^6 r (r) \, dr \, d\theta. \]

Simplify and Solve the Inner Integral

Calculate the inner integral: \( \int_3^6 r^2 \, dr \). This evaluates to: \[ \int_3^6 r^2 \, dr = \left[ \frac{r^3}{3} \right]_3^6 = \frac{6^3}{3} - \frac{3^3}{3} = \frac{216}{3} - \frac{27}{3} = 72 - 9 = 63. \]

Solve the Outer Integral

Substitute the result of the inner integral into the outer integral: \[ M = \int_0^\pi 63 \, d\theta = 63 \theta \bigg|_0^\pi = 63(\pi - 0) = 63\pi. \]

Interpret the Result

The mass of the lamina is given by \( M = 63\pi \) (units of mass). Note that the term \( b / ft^2 \) is just a unit for density, implying the mass is in appropriate mass units.

Key concepts

Density Function

The density function is a crucial component in finding the mass of a lamina. It describes how mass is distributed across the region. For any point

• In a lamina, the density might not be uniform.
• It could vary depending on the position within the region.

In this exercise, the density function is given by \( \delta(x, y) = \sqrt{x^2 + y^2} \).

Interestingly, this expression means the density increases with the distance from the origin. In polar coordinates, where the distance from the origin is represented by \( r \), the density simplifies to \( \delta(r) = r \).

This conversion makes it easier to use polar coordinates for integration. Understanding how the density function interacts with the region is key to setting up the correct integral for mass.

Polar Coordinates

Polar coordinates provide a different way of representing points on the plane compared to Cartesian coordinates

• They use a combination of radius \( r \) and angle \( \theta \).
• Radius represents the distance from the origin, while the angle indicates direction.

When dealing with circular or annular regions, polar coordinates often simplify the process significantly.

In our problem, the annulus is described with the following bounds: the radius \( r \) ranges from 3 to 6, and the angle \( \theta \) goes from 0 to \( \pi \) because the region is only in the first and second quadrants.

Instead of using complex Cartesian equations, translating the region into these simple bounds makes setting up and solving integrals far more manageable.

Definite Integration

Definite integration is the process of calculating the net area under a curve within specified limits. It is an essential tool in deriving various physical quantities such as mass, area, and volume.

• For calculating the mass of a lamina, definite integration helps accumulate the density over the desired region.

In this problem, to find the mass, we perform a double integral since we have a two-dimensional region.

The integral setup is: \[M = \int_0^\pi \int_3^6 r^2 \, dr \, d\theta \]
Here, the inner integral is with respect to \( r \) (radius), and the outer integral is with respect to \( \theta \) (angle).

The integration is broken into simpler parts:

• First, we solve \( \int_3^6 r^2 \, dr \).
• Then, substitute the result into \( \int_0^\pi \, d\theta \).

This method allows us to compute the lamina's mass in terms of given bounds and variables.

Annulus Region

An annulus is a ring-shaped area, often defined by two concentric circles. In this problem, the annulus region is bounded by the circles

• \( x^2 + y^2 = 9 \)
• and \( x^2 + y^2 = 36 \).

Understanding the geometry of an annulus is crucial for setting up the integral.

The radii of the circles give us the limits for \( r \) in polar coordinates. Here, the annulus is located in the first and second quadrants, thus \( \theta \) ranges from 0 to \( \pi \).

An annulus is simply a region outside one circle and inside another. This setup leads directly to the concept of a subtraction of areas, where the area of the smaller circle is subtracted from the larger to determine the annular region's area.
By considering the annulus, we focus our calculations only on this specific area of interest, avoiding irrelevant portions of space.