Question

Find the mass \(m\) and center of mass \((\bar{x}, \bar{y})\) of the lamina bounded by the given curves and with the indicated density. \(r=2 \sin \theta ; \delta(r, \theta)=r\)

Short answer

The mass is 4, and the center of mass is (0, 3/8).

Step-by-step solution

Find the Mass of the Lamina

The mass of a lamina in polar coordinates is found using the formula \( m = \int \int_R \delta(r, \theta) \, r \; dr \; d\theta \). Here, \( \delta(r, \theta) = r \) is the density function, and the region \( R \) is defined by \( r = 2 \sin \theta \), which describes a circle of radius 1 centered at (0,1). The limits for \( \theta \) are from 0 to \( \pi \) and for \( r \) from 0 to \( 2 \sin \theta \). The mass is computed as: \[m = \int_0^{\pi} \int_0^{2 \sin \theta} r^2 \; dr \; d\theta.\]

Evaluate Inner Integral for Mass

First, evaluate the inner integral with respect to \( r \):\[ \int_0^{2 \sin \theta} r^2 \; dr = \left[ \frac{r^3}{3} \right]_0^{2 \sin \theta} = \frac{(2 \sin \theta)^3}{3} = \frac{8 \sin^3 \theta}{3}. \]

Evaluate Outer Integral for Mass

Now, evaluate the outer integral with respect to \( \theta \):\[ m = \int_0^{\pi} \frac{8 \sin^3 \theta}{3} \; d\theta. \]Use the identity \( \sin^3 \theta = \frac{3\sin \theta - \sin(3\theta)}{4} \) to simplify and integrate:\[ m = \frac{8}{3} \cdot \frac{1}{4} \int_0^{\pi} (3 \sin \theta - \sin(3\theta)) \; d\theta. \]

Solve for Mass Using Integration Techniques

Continue by splitting the integral:\[ m = \frac{8}{12} \left[ \int_0^{\pi} 3 \sin \theta \; d\theta - \int_0^{\pi} \sin(3\theta) \; d\theta \right]. \]Each integral can be solved directly:\( \int \sin \theta \, d\theta = -\cos \theta \) and \( \int \sin(3\theta) \; d\theta = -\frac{1}{3} \cos(3\theta) \), thus:\[ m = \frac{2}{3} \left[ [-3\cos \theta]_0^{\pi} + \left. \frac{1}{3}\cos(3\theta) \right|_0^{\pi} \right] = \frac{2}{3} \cdot (6 + 0) = 4. \]

Express Center of Mass Formula

The center of mass \((\bar{x}, \bar{y})\) in polar coordinates is given by \(\bar{x} = \frac{1}{m} \int \int_R r \cos \theta \, \delta(r, \theta) \, r \; dr \; d\theta \) and \(\bar{y} = \frac{1}{m} \int \int_R r \sin \theta \, \delta(r, \theta) \, r \; dr \; d\theta \). For our problem, substitute \(\delta(r, \theta) = r \).

Calculate \(\bar{x}\)

Substitute \(\delta(r, \theta) = r\) to obtain:\[ \bar{x} = \frac{1}{4} \int_0^{\pi} \int_0^{2 \sin \theta} r^3 \cos \theta \; dr \; d\theta. \]First solve the inner integral: \[ \int_0^{2 \sin \theta} r^3 \; dr = \left[ \frac{r^4}{4} \right]_0^{2 \sin \theta} = \frac{16 \sin^4 \theta}{4} = 4 \sin^4 \theta. \]So, \[ \bar{x} = \frac{1}{4} \int_0^{\pi} 4 \sin^4 \theta \cos \theta \; d\theta = 0 \] using odd function symmetry for \(\cos \theta\) on limits \(0\) to \(\pi\).

Calculate \(\bar{y}\)

Substitute \(\delta(r, \theta) = r\) for \(\bar{y}\):\[ \bar{y} = \frac{1}{4} \int_0^{\pi} \int_0^{2 \sin \theta} r^3 \sin \theta \; dr \; d\theta. \]After solving the inner integral, we get:\[ 4 \sin^5 \theta. \]Thus:\[ \bar{y} = \frac{1}{4} \int_0^{\pi} 4 \sin^5 \theta \; d\theta = \int_0^{\pi} \sin^5 \theta \; d\theta. \]Solve using integration by parts or identities to find the average value:\[ \bar{y} = \frac{3}{8}. \]

Conclusion

The mass of the lamina is \(4\). The center of mass coordinates are \((\bar{x}, \bar{y}) = (0, \frac{3}{8})\).

Key concepts

Polar Coordinates

Polar coordinates are a way of representing points on a plane using a distance and an angle. Instead of the usual Cartesian coordinates \(x, y\), polar coordinates use \(r\) and \( heta\):
\(r\) is the distance from the origin to the point.
\(\theta\) is the angle measured from the positive x-axis to the line connecting the origin with the point.

This system is particularly useful when dealing with circular or spiral patterns as seen in this problem's curve, \(r = 2 \sin \theta\), which describes a circle centered at (0,1).

When working with polar coordinates in calculus, remember the differential area element is \(r \, dr \, d\theta\), which differs from the Cartesian \(dx\, dy\). This is crucial for integrating and finding areas or volumes.

Mass Calculation

To calculate the mass of a lamina using polar coordinates, we integrate the product of the density function \(\delta(r, \theta)\) and the area element \(r \, dr \, d\theta\).
In this exercise, the mass \(m\) is determined by evaluating the double integral:
\[m = \int_0^{\pi} \int_0^{2 \sin \theta} r^2 \, dr \, d\theta.\]
This integral accounts for the density function \(\delta(r, \theta) = r\), indicating that mass increases with distance from the origin.

The limits of integration for \(r\) and \(\theta\) are derived from the shape of the lamina, here confined by \(r = 2 \sin \theta\) and \(\theta\) ranging from 0 to \(\pi\).

Lamina

In mathematics and physics, a lamina refers to a thin, flat, two-dimensional object. We analyze its geometrical and physical properties, such as area, perimeter, and mass. In this problem, the lamina is part of a circle described by the polar equation \(r = 2 \sin \theta\).

The lamina has a variable density, impacting calculations for mass and the center of mass. Its properties can vary based on position, determined by the problem's density function \(\delta(r, \theta)\). Understanding the shape and bounds of a lamina is essential for applying calculus to analyze its characteristics.

Density Function

A density function \(\delta(r, \theta)\) in this context describes how mass is distributed throughout the lamina. It's often a function of the position, either by explicit variables like \(r\) and \(\theta\) in polar coordinates.

Here, our density function is \(\delta(r, \theta) = r\), indicating the mass density increases linearly with \(r\). This means points further from the origin have greater mass. When calculating integrals to find mass or center of mass, the density function is multiplied by \(r\) to contribute to the combined effect of mass distribution.

This influences numerical outcomes like total mass and the location of the center of mass, enabling comprehensive analysis of the lamina's physical properties.