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Magazine Chapter 6: Problem 24
Find the center of mass of a thin plate covering the region bounded below by the parabola \(y=x^{2}\) and above by the line \(y=x\) if the plate's density at the point \((x, y)\) is \(\delta(x)=12 x\)
Short Answer
Expert verified
The center of mass is \((\frac{3}{5}, \frac{1}{2})\).
Step by step solution
01
Determine the Bounded Region
First, let's identify the bounded region by solving the equations of the parabola and the line: \(y = x^2\) and \(y = x\). To find their points of intersection, set the equations equal: \(x^2 = x\). Solving this gives: \(x(x-1) = 0\), hence \(x = 0\) and \(x = 1\). So, the region is bounded between \(x = 0\) and \(x = 1\) on the x-axis.
02
Setup the Mass Calculation
The total mass \(M\) of the plate can be found using the integral of the density function over the region. The mass is given by:\[M = \int_{0}^{1} \int_{x^2}^{x} 12x \; dy \; dx.\] First, integrate with respect to \(y\): \[= \int_{0}^{1} [12xy]_{y=x^2}^{y=x} \; dx = \int_{0}^{1} (12x(x) - 12x(x^2)) \; dx.\] Simplify the expression: \[= \int_{0}^{1} (12x^2 - 12x^3) \; dx.\]
03
Calculate the Total Mass
The expression for the mass becomes: \[M = \int_{0}^{1} (12x^2 - 12x^3) \; dx.\] Integrate term by term: \(\int 12x^2 \; dx = 4x^3\) and \(\int 12x^3 \; dx = 3x^4\).Hence:\[M = \bigg[4x^3 - 3x^4\bigg]_{0}^{1} = (4 \times 1^3 - 3 \times 1^4) - (4 \times 0^3 - 3 \times 0^4) = 1.\]
04
Calculate the x-coordinate of the Center of Mass \\(\bar{x}\\)
The x-coordinate of the center of mass \(\bar{x}\) is given by:\[\bar{x} = \frac{1}{M} \int_{0}^{1} \int_{x^2}^{x} 12x^2 \; dy \; dx.\]First integrate with respect to \(y\):\[= \int_{0}^{1} [12x^2y]_{y=x^2}^{y=x} \; dx = \int_{0}^{1} \left(12x^2(x) - 12x^2(x^2)\right) \; dx.\]Simplify: \[= \int_{0}^{1} (12x^3 - 12x^4) \; dx.\]
05
Calculate \(\bar{x}\)
Now integrate term by term:\(\int 12x^3 \; dx = 3x^4\) and \(\int 12x^4 \; dx = \frac{12}{5}x^5\).Thus,\[\bar{x} = \frac{1}{1} \left([3x^4 - \frac{12}{5}x^5]_{0}^{1}\right) = 3 - \frac{12}{5} = \frac{3}{5}.\]
06
Calculate the y-coordinate of the Center of Mass \\(\bar{y}\\)
The y-coordinate of the center of mass \(\bar{y}\) is calculated using:\[\bar{y} = \frac{1}{M} \int_{0}^{1} \int_{x^2}^{x} 12xy \; dy \; dx.\]First, solve the integral with respect to \(y\):\[= \int_{0}^{1} \left[6xy^2\right]_{x^2}^{x} \; dx = \int_{0}^{1} \left(6x(x^2) - 6x(x^4)\right) \; dx\].This simplifies to \[= \int_{0}^{1} (6x^3 - 6x^5) \; dx.\]
07
Calculate \(\bar{y}\)
Integrate the expression:\(\int 6x^3 \; dx = \frac{6}{4}x^4\) and \(\int 6x^5 \; dx = x^6\).Thus,\[\bar{y} = \frac{1}{1} \left(\left[\frac{6}{4}x^4 - x^6\right]_{0}^{1}\right) = \frac{3}{2} - 1 = \frac{1}{2}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Density Function
In this problem, the density function is given as \( \delta(x) = 12x \). This function tells us how the mass is distributed across the plate. It's essential to understand that a density function can vary across the object, meaning different parts may have different densities. In our exercise, density only depends on the x-coordinate. What this implies is the further out we are along the x-axis, the denser (or heavier) the segment of the plate becomes. The density function is crucial because it affects how we calculate the total mass of the object and subsequently the center of mass. To find the total mass, we must integrate the density over the bounded region, taking into account its variable nature. In simple terms, we multiply the density function by the differential area and sum these up across the whole region.
Integration Techniques
Integration is vital in calculating both the mass and the center of mass. Here, we perform two types of integrations: with respect to \( y \) and \( x \). To begin the calculation of mass, we first integrate the density function \( 12x \) with respect to \( y \). This is called the inner integral. For the limits of this integral, we use the upper and lower bounds from the bounded region: from \( y = x^2 \) to \( y = x \).
After integrating with respect to \( y \), we then move on to integrate with respect to \( x \), using the intersections found earlier, \( x = 0 \) and \( x = 1 \), as the limits. This entire double integration process lets us find the total amount of the property (in this case, mass) across the region and is a common practice for physical applications.
After integrating with respect to \( y \), we then move on to integrate with respect to \( x \), using the intersections found earlier, \( x = 0 \) and \( x = 1 \), as the limits. This entire double integration process lets us find the total amount of the property (in this case, mass) across the region and is a common practice for physical applications.
Bounded Region
Identifying the correct bounded region is the first critical step in finding the center of mass. In this problem, the region is enclosed between a parabola \( y = x^2 \) and a line \( y = x \). To find where they intersect, we set \( x^2 = x \) and solve, which gives us \( x = 0 \) and \( x = 1 \). These x-values form the horizontal limits of our region. The vertical limits for integration are determined by these functions themselves. It is essential to know these limits, as they guide the range over which we integrate. Locating these intersections and determining accurate limits is crucial, not only for integration but for understanding how the shape looks. It ultimately affects how we solve for both the mass and the center of mass.
