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Chapter 6: Problem 22

In Exercises \(13-24\) , find the center of mass of a thin plate of constant density \(\delta\) covering the given region. a. The region cut from the first quadrant by the circle \(x^{2}+y^{2}=9\) b. The region bounded by the \(x\) -axis and the semicircle \(y=\sqrt{9-x^{2}}\) Compare your answer in part (b) with the answer in part (a).

Short Answer

Expert verified
The center of mass for part (a) is \((\frac{4}{\pi}, \frac{4}{\pi})\) and for part (b) is \((0, \frac{4}{\pi})\).

Step by step solution

01

Understand the Shape and Area of the Region in Part a

For part (a), the region is a quarter of a circle centered at the origin (0, 0) with radius 3, since the equation is given by \(x^{2} + y^{2} = 9\). Since it is in the first quadrant, we only consider \(x\) and \(y\) values from 0 to 3.
02

Calculate the Area for Part a

The total area of the circle would be \(\pi \times 3^{2} = 9\pi\). Since we are only interested in the first quadrant (a quarter circle), the area is \(\frac{1}{4} \times 9\pi = \frac{9\pi}{4}\).
03

Determine the Coordinates of the Center of Mass for Part a

For a quarter-circle centered at the origin, symmetric about the x and y axes, the center of mass \((x_{cm}, y_{cm})\) will lie along the line \(x = y\). The coordinates are given by \((\frac{4R}{3\pi}, \frac{4R}{3\pi})\), where \(R\) is the radius. Here, \(R = 3\), so \(x_{cm} = y_{cm} = \frac{4 \times 3}{3\pi} = \frac{4}{\pi}\).
04

Understand the Shape and Area of the Region in Part b

For part (b), the region is a semicircle centered along the x-axis with a radius of 3 that stretches from \(x = -3\) to \(x = 3\), described by the curve \(y = \sqrt{9 - x^{2}}\).
05

Calculate the Area for Part b

The area of the full circle would be \(9\pi\), but since it is a semicircle, the area is \(\frac{1}{2} \times 9\pi = \frac{9\pi}{2}\).
06

Determine the Coordinates of the Center of Mass for Part b

For a semicircle along the x-axis, the center of mass lies along \(x = 0\). The y-coordinate of the center of mass is given by \(\frac{4R}{3\pi}\) because of symmetry. As \(R = 3\), \(y_{cm} = \frac{4 \times 3}{3\pi} = \frac{4}{\pi}\), and \(x_{cm} = 0\).
07

Compare the Centers of Mass for Parts a and b

In part (a), the center of mass is \((\frac{4}{\pi}, \frac{4}{\pi})\) and in part (b), it is \((0, \frac{4}{\pi})\). The y-coordinates are the same, but the x-coordinates differ due to the shapes' symmetries in different regions.

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Key Concepts

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

Quarter-Circle
A quarter-circle is a section of a full circle. It accounts for one-fourth of the total area and curve. In the context of coordinate geometry, the quarter-circle often sits snug in one quadrant of the Cartesian plane. To find the equations and properties that mark this region,
  • you identify the equation of the full circle, such as \(x^2 + y^2 = r^2\).
  • When talking about the first quadrant, only consider positive \(x\) and \(y\) values, from zero up to the radius.
This ensures that you are focusing on just that one-fourth section. For instance, when the circle's radius is 3, you look at values from 0 to 3. Understanding how the quarter-circle is positioned in the coordinate system helps locate its center of mass more accurately.
Semicircle
A semicircle represents half of a complete circle. When a semicircle is discussed in terms of a Cartesian coordinate system:
  • Its shape is determined by the top half or bottom half of the circle's equation, for example, \(y = \sqrt{r^2 - x^2}\).
  • In coordinate geometry, a semicircle can be aligned along the x-axis, stretching from \(-r\) to \(r\), with a consistent symmetry that affects its center of mass.
For a semicircle with radius 3, the area affected by this shape will be from \(-3\) to \(3\) on the x-axis. The symmetry of a semicircle helps in simplifying the calculations, ensuring that any mass calculations only need to focus deeply on one axis.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, offers a bridge between algebra and geometry through graphs and equations. When dealing with shapes like the quarter and semicircle:
  • We use the x-y coordinate plane, enabling us to define and understand geometrical shapes with equations.
  • Circle equations are paramount in defining where the shapes lie and calculating essential attributes like center of mass.
For instance, the equation \(x^2 + y^2 = 9\) describes a circle with a radius of 3, allowing for breakdowns like quarter and semicircles to be located precisely in the coordinate plane. This approach makes it easier to handle the points, lines, and areas mathematically, which is vital for any geometry-based calculations.
Area Calculation
Area calculation is crucial when finding the center of mass. Understanding how to break down and evaluate different shapes helps in accurate results. For a quarter-circle:
  • You find its area by taking the circle鈥檚 full area and dividing it by four, \(\frac{\pi \times r^2}{4}\).
  • This way, for a circle with a radius 3, the quarter-circle's area equals \(\frac{9\pi}{4}\).
For a semicircle:
  • A similar approach is used, but you divide the full circle's area by two, \(\frac{\pi \times r^2}{2}\).
  • So, with a radius 3, you end up with an area of \(\frac{9\pi}{2}\).
Grasping these calculations is viable when understanding geometric shapes and their spread, making it simpler to determine properties like the center of mass.

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Most popular questions from this chapter

A cone The line segment joining the origin to the point \((h, r)\) is revolved about the \(x\) -axis to generate a cone of height \(h\) and base radius \(r .\) Find the cone's surface area with the parametric equations \(x=h t, y=r t, 0 \leq t \leq 1 .\) Check your result with the geometry formula: Area \(=\pi r(\text { slant height). }\)

Use a theorem of Pappus to find the volume generated by revolving about the line \(x=5\) the triangular region bounded by the coordinate axes and the line \(2 x+y=6 .\) As you saw in Exercise 29 of Section \(6.4,\) the centroid of a triangle lies at the intersection of the medians, one-third of the way from the midpoint of each side toward the opposite vertex.)

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. \(y=\tan x, \quad 0 \leq x \leq \pi / 4 ; \quad x\) -axis

Volume of a bowl \(A\) bowl has a shape that can be generated by revolving the graph of \(y=x^{2} / 2\) between \(y=0\) and \(y=5\) about the \(y\) -axis. a. Find the volume of the bowl. b. Related rates If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep?

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 parabolas \(y=2 x^{2}-4 x\) and \(y=2 x-x^{2}\)
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