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Chapter 6: Problem 17

In Exercises \(18,\) find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region between the curve \(y=1 / \sqrt{x}\) and the \(x\) -axis from \(x=1\) to \(x=16\)

Short Answer

Expert verified
The center of mass is at \((-7, -\frac{\ln 16}{12})\).

Step by step solution

01

Set up the Integral for Mass

To find the center of mass, we first need the mass of the plate. The mass of a region is given by the integral of the density over the area. Here, \( ext{mass} = \int_{1}^{16} \delta y \, dx = \int_{1}^{16} \delta \frac{1}{\sqrt{x}} \, dx \).
02

Calculate the Integral for Mass

Integrate to find the mass: \( ext{mass} = \delta \left[-2\sqrt{x}\right]_{1}^{16} = \delta \left[-2\sqrt{16} + 2\sqrt{1}\right] = \delta \left[-8 + 2\right] = -6\delta \).
03

Set up the Integral for Center of Mass (\(x\)-coordinate)

The \(x\)-coordinate of the center of mass \( \bar{x} \) is given by: \( \bar{x} = \frac{1}{M} \int_{1}^{16} x \delta y \, dx = \frac{1}{M} \int_{1}^{16} x \delta \frac{1}{\sqrt{x}} \, dx \).
04

Calculate \(x\)-coordinate of Center of Mass

Integrate to find: \( \int_{1}^{16} x \delta \frac{1}{\sqrt{x}} \, dx = \delta \int_{1}^{16} \sqrt{x} \, dx = \delta \left[ \frac{2}{3} x^{3/2} \right]_{1}^{16} = \delta \left[\frac{2}{3}(64) - \frac{2}{3}(1)\right] = \delta \cdot 42 \). The \(x\)-coordinate is \( \bar{x} = \frac{42\delta}{-6\delta} = -7 \).
05

Set up the Integral for Center of Mass (\(y\)-coordinate)

The \(y\)-coordinate of the center of mass \( \bar{y} \) is given by: \( \bar{y} = \frac{1}{2M} \int_{1}^{16} \delta y^2 \, dx = \frac{1}{2M} \int_{1}^{16} \delta \left( \frac{1}{\sqrt{x}} \right)^2 \, dx = \frac{1}{2M} \int_{1}^{16} \delta \frac{1}{x} \, dx \).
06

Calculate \(y\)-coordinate of Center of Mass

Integrate to find: \( \int_{1}^{16} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{16} = \ln 16 \). The \(y\)-coordinate is \( \bar{y} = \frac{\delta}{-12\delta} \ln 16 = -\frac{\ln 16}{12} \).
07

Conclusion

With \( \bar{x} = -7 \) and \( \bar{y} = -\frac{\ln 16}{12}\), the center of mass of the plate is \( (-7, -\frac{\ln 16}{12}) \).

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Key Concepts

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

Integral Calculus
Integral calculus is a key mathematical tool used to find quantities like areas under curves and the total accumulation of quantities across a region. In the context of finding the center of mass, integral calculus helps us calculate both the mass of an object and its distribution.
  • To find the mass, we integrate the density of the object over the given region. This accounts for the fact that mass could be different in different areas depending on the density.
  • For the center of mass, we use integrals to determine where this mass is concentrated. The process involves integrating the function describing the shape of the object and the density.
It's particularly useful when dealing with shapes that are not uniform, like the area under a curve, because it allows us to piece together many small parts to get the total picture. Learning integral calculus not only helps in physics, but also in any field that deals with physical quantities spread over areas.
Mass Calculation
Mass calculation is at the heart of finding the center of mass. It requires both the density function and the understanding of how mass is distributed across an area. In this exercise, the mass of the plate is determined by \[\text{mass} = \int_{1}^{16} \delta \frac{1}{\sqrt{x}} \, dx \]
  • The variable \( \delta \) represents the density, which is assumed constant, making calculations simpler.
  • The expression for \( y = \frac{1}{\sqrt{x}} \) shows how the height of the region changes with respect to \( x \).
Integrating this product gives the mass of the area between the curve and the \( x \)-axis from \( x = 1 \) to \( x = 16 \). The result, \(-6\delta\), reveals the total mass, which is used later to calculate each coordinate of the center of mass.
Coordinate System
A coordinate system allows us to assign a specific point to the center of mass in a two-dimensional space. Understanding how to work within a coordinate system is crucial for solving problems involving the center of mass.
  • The \( x \)-coordinate of the center of mass, \( \bar{x} \), helps in determining how far left or right the balance point is. In this example, \( \bar{x} = -7 \), indicating that the center is located 7 units to one side of the origin within the system used here.
  • The \( y \)-coordinate, \( \bar{y} \), indicates the vertical position. For the plate, it's found by \( \bar{y} = -\frac{\ln 16}{12} \), which is derived from integrating the function and adjusting by mass.
The center of mass coordinates \((-7, -\frac{\ln 16}{12})\) tells us exactly where the point is that would balance this plate if it were physically possible to support it at a single point.

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Most popular questions from this chapter

Tilted plate Calculate the fluid force on one side of a right-triangular plate with edges 3 \(\mathrm{ft}\) , 4 \(\mathrm{ft}\) , and 5 \(\mathrm{ft}\) if the plate sits at the bottom of a pool filled with water to a depth of 6 \(\mathrm{ft}\) on its 3 -ft edge and tilted at \(60^{\circ}\) to the bottom of the pool.

Find the volume of the solid generated by revolving each region about the \(y\)-axis. The region in the first quadrant bounded above by the parabola \(y=x^{2},\) below by the \(x\)-axis, and on the right by the line \(x=2\)

Designing a wok You are designing a wok frying pan that will be shaped like a spherical bowl with handles. A bit of experimentation at home persuades you that you can get one that holds about 3L if you make it 9 \(\mathrm{cm}\) deep and give the sphere a radius of 16 \(\mathrm{cm}\) . To be sure, you picture the wok as a solid of revolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? (1 \(=1000 \mathrm{cm}^{3}\) )

Find the areas of the surfaces generated by revolving the curves in Exercises \(13 - 23\) about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$y = \sqrt { x + 1 } , \quad 1 \leq x \leq 5 ; \quad x -axis$$

Find the lateral (side) surface area of the cone generated by revolving the line segment \(y = x / 2,0 \leq x \leq 4 ,\) about the \(x\) -axis. Check your answer with the geometry formula Lateral surface area \(= \frac { 1 } { 2 } \times\) base circumference \(\times\) slant height.
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