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Mobile App Chapter 5: Problem 33
A lamina covering the quarter circle \(x^{2}+y^{2} \leq 4, x>0, y>0\), has (area) density \(x+y\). Find the mass of the lamina.
Short Answer
Expert verified
The mass of the lamina is \( \frac{16}{3} \).
Step by step solution
01
- Define the Region
The lamina covers a quarter circle given by the inequality: \[ x^2 + y^2 \leq 4, \ x \geq 0, \ y \geq 0 \]This represents the part of the circle of radius 2 in the first quadrant.
02
- Convert to Polar Coordinates
In polar coordinates, the equations become \[ r^2 \leq 4 \]where the radius range is \[0 \leq r \leq 2\] and the angle range is \[0 \leq \theta \leq \pi/2\].
03
- Set Up the Integral for Mass
The mass can be found using the double integral of the density \(x + y\):\[ M = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (rcos\theta + rsin\theta) \cdot r \ dr \ d\theta \]where \(x = r cos \theta \) and \(y = r sin \theta \).
04
- Simplify the Integral
Simplify and separate the integral:\[ M = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (r^2 cos\theta + r^2 sin\theta) \ dr \ d\theta \]\[ M = \int_{0}^{\frac{\pi}{2}} \( cos\theta \int_{0}^{2} r^2 \ dr \ + sin\theta \int_{0}^{2} r^2 \ dr \) \ d\theta \].
05
- Integrate with Respect to r
Integrate each part with respect to \(r\):\(\int_{0}^{2} r^2 \ dr = \left[ \frac{r^3}{3} \right]_{0}^{2} \ = \ \frac{8}{3}\)Thus, the integral becomes:\[ M = \frac{8}{3} \int_{0}^{\frac{\pi}{2}} (cos\theta + sin\theta) \ d\theta \].
06
- Integrate with Respect to \(\theta\)
Now integrate with respect to \(\theta\):\( \int_{0}^{\frac{\pi}{2}} cos\theta \ d\theta = [sin\theta]_{0}^{\frac{\pi}{2}} = 1 \)and \( \int_{0}^{\frac{\pi}{2}} sin\theta \ d\theta = [-cos\theta]_{0}^{\frac{\pi}{2}} = 1 \).So,\[ M = \frac{8}{3} (1 + 1) = \frac{8}{3} \cdot 2 = \frac{16}{3} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
To solve the problem of finding the mass of a lamina, we first need to understand polar coordinates. Polar coordinates are an alternative to Cartesian coordinates (x, y) for specifying points on a plane. In polar coordinates, each point is determined by a distance from a reference point (usually the origin) and an angle from a reference direction (usually the positive x-axis).
The distance is denoted by \(r\), which represents how far the point is from the origin. The angle is denoted by \(\theta\) and measures the counterclockwise angle from the positive x-axis to the point. For example, the point with Cartesian coordinates \((x, y)\) can be expressed in polar coordinates as \((r, \theta)\), where:
The distance is denoted by \(r\), which represents how far the point is from the origin. The angle is denoted by \(\theta\) and measures the counterclockwise angle from the positive x-axis to the point. For example, the point with Cartesian coordinates \((x, y)\) can be expressed in polar coordinates as \((r, \theta)\), where:
- \( r = \sqrt{x^2 + y^2} \)
- \( \theta = \tan^{-1}(y/x) \)
Double Integral
To find the mass of the lamina, we need to use a double integral. Double integrals allow us to sum values over a two-dimensional area. This is essential when we want to integrate a function that depends on both \(x\) and \(y\) over a specified region.
In our case, we use a double integral to account for the varying density of the lamina over the quarter circle. The general form of a double integral over a region \(D\) is: \[ \text{Mass} \,=\, \ \ \int_{\text{Region D}} \text{Density Function} \, dA \= \int_{\text{Region D}} \rho(x, y) \, dA \ \ \] Here, \( dA \) represents an infinitesimally small area element. In polar coordinates, this becomes \( r \, dr \, d\theta \), where \(r\) and \(\theta\) vary over their respective ranges.
For our lamina problem, the region of integration is a quarter circle of radius 2, and the density function is given by \( \rho(x, y) \, = \, x \, + \, y \). We will use these elements to set up our double integral for finding the mass.
In our case, we use a double integral to account for the varying density of the lamina over the quarter circle. The general form of a double integral over a region \(D\) is: \[ \text{Mass} \,=\, \ \ \int_{\text{Region D}} \text{Density Function} \, dA \= \int_{\text{Region D}} \rho(x, y) \, dA \ \ \] Here, \( dA \) represents an infinitesimally small area element. In polar coordinates, this becomes \( r \, dr \, d\theta \), where \(r\) and \(\theta\) vary over their respective ranges.
For our lamina problem, the region of integration is a quarter circle of radius 2, and the density function is given by \( \rho(x, y) \, = \, x \, + \, y \). We will use these elements to set up our double integral for finding the mass.
Density Function
In the context of the lamina problem, the density function \( \rho(x, y) \, = \, x + y \) represents how the density varies at different points on the lamina.
A density function determines how packed or spread out mass is over an area. In mathematical problems involving mass distribution, it is crucial in calculating properties like mass, center of mass, and moments of inertia.
To handle a varying density, we express the function in terms of polar coordinates. For our problem, this density function is converted as follows: \[ \rho(r, \, \theta) \,=\, r \cos\theta + r \sin\theta \]
To find the mass of the lamina, we integrate the density function over the specified region. This integral accounts for all the infinitesimal pieces of mass over the quarter circle. As we have converted our region and density function into polar coordinates, we can integrate with respect to \(r\) and \(\theta\) to determine the lamina's mass accurately.
The final integral setup and simplification help us compute the total mass by summing the contributions from each small element over the entire region.
A density function determines how packed or spread out mass is over an area. In mathematical problems involving mass distribution, it is crucial in calculating properties like mass, center of mass, and moments of inertia.
To handle a varying density, we express the function in terms of polar coordinates. For our problem, this density function is converted as follows: \[ \rho(r, \, \theta) \,=\, r \cos\theta + r \sin\theta \]
To find the mass of the lamina, we integrate the density function over the specified region. This integral accounts for all the infinitesimal pieces of mass over the quarter circle. As we have converted our region and density function into polar coordinates, we can integrate with respect to \(r\) and \(\theta\) to determine the lamina's mass accurately.
The final integral setup and simplification help us compute the total mass by summing the contributions from each small element over the entire region.
