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Chapter 9: Problem 43

Find the center of mass of the lamina that has the given shape and density. $$ y=x, x+y=6, y=0 ; \rho(x, y)=2 y $$

Short Answer

Expert verified
The center of mass is \( \left( \frac{3}{2}, \frac{12}{5} \right) \).

Step by step solution

01

Sketch the Region

Begin by sketching the triangular region bounded by the lines. The line \( y = x \) intersects with \( y = 0 \) at \( (0,0) \). The line \( x + y = 6 \) intersects with \( y = 0 \) at \( x = 6 \). Lastly, the line \( y = x \) and \( x + y = 6 \) intersect at \( (3,3) \). Thus, the region is a triangle with vertices at \((0,0)\), \((3,3)\), and \((6,0)\).
02

Set up Integrals for Mass

The mass \(M\) of the lamina is given by the double integral: \[ M = \int_{0}^{6} \int_{0}^{6-x} 2y \, dy \, dx \] Evaluate the inner integral first:
03

Evaluate Inner Integral for Mass

Compute \[ \int_{0}^{6-x} 2y \, dy = \left[y^2 \right]_{0}^{6-x} = (6-x)^2 \] Substitute this result into the mass integral:
04

Evaluate Outer Integral for Mass

Evaluate \[ M = \int_{0}^{6} (6-x)^2 \, dx \] After integration, the result is \[ M = \left[ -\frac{(6-x)^3}{3} \right]_{0}^{6} = 72 \]
05

Compute x-coordinate of Center of Mass

The \(x\)-coordinate of the center of mass is given by \[ \bar{x} = \frac{1}{M} \int_{0}^{6} \int_{0}^{6-x} x \cdot 2y \, dy \, dx \] Solve the inner integral, which is \[ \int_{0}^{6-x} x \cdot 2y \, dy = x(6-x)^2 \] Substitute the result:
06

Evaluate for x-coordinate of Center of Mass

Evaluate \[ \bar{x} = \frac{1}{72} \int_{0}^{6} x(6-x)^2 \, dx \] Solve this integral, which is \[ \bar{x} = \frac{1}{72} \left[ -\frac{x(6-x)^3}{3} + \int \frac{(6-x)^3}{3} \, dx \right]_{0}^{6} = \frac{3}{2} \]
07

Compute y-coordinate of Center of Mass

The \(y\)-coordinate of the center of mass is \[ \bar{y} = \frac{1}{M} \int_{0}^{6} \int_{0}^{6-x} y \cdot 2y \, dy \, dx \] Solve the inner integral, which is \[ \int_{0}^{6-x} 2y^2 \, dy = \frac{2}{3}(6-x)^3 \] Substitute the result:
08

Evaluate for y-coordinate of Center of Mass

Evaluate \[ \bar{y} = \frac{1}{72} \int_{0}^{6} \frac{2}{3}(6-x)^3 \, dx \] After solving, you get \[ \bar{y} = \frac{24}{5} \]
09

Conclusion: Center of Mass Coordinates

The center of mass of the lamina is located at \( \left( \frac{3}{2}, \frac{12}{5} \right) \).

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Key Concepts

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

Lamina Density
When dealing with the concept of the center of mass of a lamina, it is crucial to understand lamina density. A lamina is a flat, two-dimensional figure with a certain thickness, often considered negligible. The density of the lamina is usually a function that describes how mass is distributed over its surface.
For this specific exercise, the density of the lamina is given by \( \rho(x, y) = 2y \). This expression shows that the density varies depending on the \(y\)-coordinate.
  • This density indicates that as we move up in the \(y\)-coordinate, the mass of the lamina increases.
  • A constant density would imply that the entire region is uniformly distributed with mass, but here, it's not uniform.
Understanding the role of density is crucial because it directly influences how the mass integral is set up and evaluated.
Mass Integral
The mass of a lamina can be computed using a mass integral. This involves calculating a double integral over the region that defines the shape of the lamina.
The mass \(M\) for this region is given by the integral: \[ M = \int_{0}^{6} \int_{0}^{6-x} 2y \, dy \, dx \] This equation involves two integrals:
  • The inner integral \(\int_{0}^{6-x} 2y \, dy\) calculates the mass in terms of \(x\) along the vertical line \[y = 0\] to \[y = 6-x\].
  • The result of this integral, \((6-x)^2\), reflects the dependency of mass on the y-dimension.
  • Then, this result is plugged into the outer integral \(\int_{0}^{6} (6-x)^2 \, dx\) to integrate over the entire region along the \(x\)-axis, from \(0\) to \(6\).
The computation of the mass is essential as it is used to find the coordinates of the center of mass.
Coordinate of Center of Mass
The coordinates for the center of mass, \(\bar{x}\) and \(\bar{y}\), represent the point where the entire mass of the lamina might be considered concentrated. This is crucial in understanding how the lamina balances under its weight distribution.### Calculating \(\bar{x}\):The \(x\)-coordinate is determined by:\[ \bar{x} = \frac{1}{M} \int_{0}^{6} \int_{0}^{6-x} x \cdot 2y \, dy \, dx \]Here,
  • The inner integral \(\int_{0}^{6-x} x \cdot 2y \, dy\) becomes \(x(6-x)^2\) after evaluation.
  • The outer integral is \(\int_{0}^{6} x(6-x)^2 \, dx\), which outputs the final value of \(\bar{x} = \frac{3}{2}\).
### Calculating \(\bar{y}\):For the \(y\)-coordinate:\[ \bar{y} = \frac{1}{M} \int_{0}^{6} \int_{0}^{6-x} y \cdot 2y \, dy \, dx \]In this setup,
  • The inner integral is \(\int_{0}^{6-x} 2y^2 \, dy\), which results in \(\frac{2}{3}(6-x)^3\).
  • The outer integral gives \(\bar{y} = \frac{24}{5}\) after evaluation.
The coordinates \(\left( \frac{3}{2}, \frac{12}{5} \right)\) provide the exact placement of the center of mass for this lamina.
Double Integrals in Calculus
Double integrals are a method in calculus to compute a total quantity over a two-dimensional region. They extend the idea of a single integral, which is used over a one-dimensional interval. ### How It Works: The process involves two steps of integration:
  • The inner integral is first computed while holding the other variable constant. This "slices" the region into smaller parts.
  • The result of the inner integral is then used for the outer integral, essentially "stacking" the previous slices to get the total value over the region.
In the context of this problem, the double integral helps in determining both mass and coordinates of the center of mass a lamina. It contextualizes the concept of densities and coordinates over geometric shapes. ### Applications: Double integrals are useful in
  • calculating area, mass, and volume,
  • solving problems in physics, such as electromagnetism and fluid dynamics,
  • evaluating functions over planar regions, more complex than linear integrals.
This makes understanding double integrals a vital skill in effectively solving problems involving two-dimensional regions.

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