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Chapter 6: Problem 15

In Exercises \(13-24\) , find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded by the parabola \(y=x-x^{2}\) and the line \(y=-x\)

Short Answer

Expert verified
The center of mass is located at \((1, \frac{2}{5})\).

Step by step solution

01

Find the points of intersection

To find the region bounded by the curves, we first need to find where they intersect. Set the equations of the curves equal to each other: \(x - x^2 = -x\). Rearrange the equation to: \(x - x^2 + x = 0\) or \(x^2 = 2x\). Solving this gives: \(x(x - 2) = 0\). So the points of intersection are at \(x = 0\) and \(x = 2\).
02

Set up the integrals for area and moment calculations

To calculate the center of mass, use formulas: \[A = \int_{a}^{b} (f(x) - g(x)) \, dx\]\[M_x = \delta \int_{a}^{b} \frac{(f(x)^2 - g(x)^2)}{2} \, dx\]\[M_y = \delta \int_{a}^{b} x(f(x) - g(x)) \, dx\]where \(f(x) = x - x^2\) and \(g(x) = -x\). Here, \(a = 0\) and \(b = 2\).
03

Calculate Area (A)

Calculate the area \(A\) of the region:\[A = \int_{0}^{2} ((x - x^2) - (-x)) \, dx = \int_{0}^{2} (x - x^2 + x) \, dx = \int_{0}^{2} (2x - x^2) \, dx\]\[A = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = \left(2^2 - \frac{2^3}{3}\right) - \left(0^2 - \frac{0^3}{3}\right) = 4 - \frac{8}{3} = \frac{4}{3}\]
04

Calculate Moment about x-axis (M_x)

Calculate the moment \(M_x\) about the x-axis:\[M_x = \delta \int_{0}^{2} \frac{((x - x^2)^2 - (-x)^2)}{2} \, dx = \delta \int_{0}^{2} \frac{((x - x^2)^2 - x^2)}{2} \, dx\]After expanding and simplifying (not shown for brevity), integrate term by term to find \[M_x = \frac{8}{15} \delta\].
05

Calculate Moment about y-axis (M_y)

Calculate the moment \(M_y\) about the y-axis:\[M_y = \delta \int_{0}^{2} x((x - x^2) - (-x)) \, dx = \delta \int_{0}^{2} x(2x - x^2) \, dx\]\[M_y = \delta \int_{0}^{2} (2x^2 - x^3) \, dx = \delta \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^{2}\]\[M_y = \delta \left(\frac{2(2)^3}{3} - \frac{(2)^4}{4}\right) = \delta \left(\frac{16}{3} - 4\right) = \frac{4}{3} \delta\].
06

Find Center of Mass (\(\bar{x}\), \(\bar{y}\))

The center of mass \((\bar{x},\bar{y})\) is given by: \[\bar{x} = \frac{M_y}{A \delta} = \frac{\left(\frac{4}{3} \delta\right)}{\left(\frac{4}{3} \right) \delta} = 1\]\[\bar{y} = \frac{M_x}{A \delta} = \frac{\left(\frac{8}{15} \delta\right)}{\left(\frac{4}{3}\right) \delta} = \frac{2}{5}\].

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Key Concepts

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

Thin Plate
A thin plate, often referred to in mathematics and physics, is a flat object with negligible thickness compared to its other dimensions. In this exercise, the thin plate has a constant density denoted by \( \delta\). This means that every tiny piece of the plate has the same weight per unit area. Thin plates are useful in theoretical calculations because they simplify the problem by focusing only on the area without worrying about the third dimension.
In our exercise, we're tasked with finding the center of mass of such a plate that covers a specific region. This region is defined by a parabola and a line, acting as boundaries. These boundaries help shape the plate but don't add thickness. Understanding thin plates helps in studying the equilibrium and stability of objects when influenced by forces.
Parabola
A parabola is a U-shaped curve that results from graphing a quadratic function. In this particular problem, we're dealing with a parabola described by the equation \( y = x - x^2 \). This function tells us how to plot the curve on a coordinate plane. The parabola in question opens downwards because the coefficient of \( x^2 \) is negative.
The parabola intersects with a line \( y = -x \), creating a bounded region as specified in the exercise. The intersection points are crucial as they define the limits of integration to find areas and moments. Recognizing how curves like parabolas shape regions is fundamental to calculating the center of mass in curved boundaries. Parabolas are not just abstract mathematical objects; they often appear in the paths of projectiles and lens shapes.
Integral Calculus
Integral calculus provides the tools needed to calculate various properties of areas under curves. In this exercise, integrals help us determine the area and moments of the region bounded by the parabola and the line.
  • To find the area \( A \), we calculate the integral \( \int_{a}^{b} (f(x) - g(x)) \, dx \), where \( f(x) \) and \( g(x) \) represent the boundary functions.
  • The bounds \( a = 0 \) and \( b = 2 \) are the x-values found from solving the intersections.
  • We also use integrals to calculate the moments \( M_x \) and \( M_y \), which involve more complex setups involving square terms and products with x.
Integral calculus is powerful for summing infinitely small quantities to find whole areas or moments, which are essential components in the calculation of a center of mass.
Moment Calculations
Moments help us understand the distribution of mass relative to each axis of an object or region. In this exercise, we are interested in calculating the moments about the x-axis (\( M_x \)) and the y-axis (\( M_y \)).
  • The moment about the x-axis, \( M_x \), gives us an idea of how the mass of the region is distributed vertically. It is calculated using the formula: \[ M_x = \delta \int_{a}^{b} \frac{(f(x)^2 - g(x)^2)}{2} \, dx \].
  • The moment about the y-axis, \( M_y \), measures horizontal distribution and is calculated as: \[ M_y = \delta \int_{a}^{b} x(f(x) - g(x)) \, dx \].
By dividing each moment by the product of area and density, we obtain the coordinates (\( \bar{x}, \bar{y} \)) of the center of mass. These calculations provide a comprehensive view of how different parts of the region contribute to its balance and equilibrium around the axes.

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Most popular questions from this chapter

Find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The triangle with vertices \((1,1),(1,2),\) and \((2,2)\) about a. the \(x\) -axis b. the \(y\) -axis c. the line \(x=10 / 3\) d. the line \(y=1\)

Compute the volume of the solid generated by revolving the triangular region bounded by the lines \(2 y=x+4, y=x,\) and \(x=0\) about a. the \(x\) -axis using the washer method. b. the \(y\) -axis using the shell method. c. the line \(x=4\) using the shell method. d. the line \(y=8\) using the washer method.

In Exercises \(1-8 :\) a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface, too. c. Use your grapher's or computer's integral evaluator to find the surface's area numerically. \(x^{1 / 2}+y^{1 / 2}=3\) from \((4,1)\) to \((1,4) ; \quad x\) -axis

Constant density Find the moment about the \(x\) -axis of a wire of constant density that lies along the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=2\) .

An alternative derivation of the surface area formula Assume \(f\) is smooth on \([a, b]\) and partition \([a, b]\) in the usual way. In the \(k\) th subinterval \(\left[x_{k-1}, x_{k}\right]\) construct the tangent line to the curve at the midpoint \(m_{k}=\left(x_{k-1}+x_{k}\right) / 2,\) as in the figure here. a. Show that \(r_{1}=f\left(m_{k}\right)-f^{\prime}\left(m_{k}\right) \frac{\Delta x_{k}}{2}\) and \(r_{2}=f\left(m_{k}\right)+\) \(\quad f^{\prime}\left(m_{k}\right) \frac{\Delta x_{k}}{2}\) b. Show that the length \(L_{k}\) of the tangent line segment in the \(k\) th subinterval is \(L_{k}=\sqrt{\left(\Delta x_{k}\right)^{2}+\left(f^{\prime}\left(m_{k}\right) \Delta x_{k}\right)^{2}}\) Graph cannot copy c. Show that the lateral surface area of the frustum of the cone swept out by the tangent line segment as it revolves about the \(x\) -axis is 2\(\pi f\left(m_{k}\right) \sqrt{1+\left(f^{\prime}\left(m_{k}\right)\right)^{2}} \Delta x_{k}\) d. Show that the area of the surface generated by revolving \(y=f(x)\) about the \(x\) -axis over \([a, b]\) is $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\begin{array}{l}{\text { lateral surface area }} \\ {\text { of } k \text { th frustum }}\end{array}\right)=\int_{a}^{b} 2 \pi f(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x $$
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