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Magazine Chapter 5: Problem 9
As needed, use a computer to plot graphs and to check values of integrals. Find the centroid of the area above \(y=x^{2}\) and below \(y=c(c > 0)\).
Short Answer
Expert verified
The centroid is \( (0, \frac{3c}{5}) \)
Step by step solution
01
- Understand the Region
Identify the region bounded by the curves. The region of interest is the area above the parabola \(y=x^{2}\) and below the horizontal line \(y=c\) (where \(c>0\)).
02
- Find Intersection Points
Solve for the intersection points of the curves \(y=x^{2}\) and \(y=c\). Set \[x^{2}=c\] and solve for \(x\). The intersection points are \[x=-\sqrt{c}\] and \[x=\sqrt{c}\].
03
- Area of the Region
Find the area of the region bounded by the two curves. The area \(A\) can be found by integrating the difference of the functions: \[A = \int_{-\sqrt{c}}^{\sqrt{c}} (c - x^{2}) \, dx\].
04
- Calculate the Area
Compute the integral to find the area: \[A = \int_{-\sqrt{c}}^{\sqrt{c}} c \, dx - \int_{-\sqrt{c}}^{\sqrt{c}} x^{2} \, dx\]. This simplifies to \[A = 2c\sqrt{c} - \frac{2c^{3/2}}{3} = \frac{4c^{3/2}}{3}\].
05
- Find the Centroid Coordinates
The x-coordinate of the centroid \(\bar{x}\) is given by \[\bar{x} = \frac{1}{A} \int_{-\sqrt{c}}^{\sqrt{c}} x(c - x^{2}) \, dx\]. Due to symmetry, \[\bar{x} = 0\]. The y-coordinate of the centroid \(\bar{y}\) is given by \[\bar{y} = \frac{1}{A} \int_{-\sqrt{c}}^{\sqrt{c}} (c - x^{2})^{2} \, dx\].
06
- Calculate the Centroid鈥檚 Y-Coordinate
Compute the integral for \(\bar{y}\): \[\bar{y} = \frac{1}{A} \int_{-\sqrt{c}}^{\sqrt{c}} (c - x^{2})x^{2} \, dx\]. This simplifies to \[\bar{y} = \frac{1}{\frac{4c^{3/2}}{3}} \left(\frac{4c^{3/2}}{5}\right) = \frac{3c}{5}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is a crucial mathematical tool used to find areas under curves and other relevant properties. In this exercise, we use integration to determine the area between the parabola and the horizontal line. This involves setting up an integral that represents the total area above the parabola and below the line.
We specifically use definite integration, meaning we calculate the integral from one boundary to another, which in this case are the intersection points.
Remember, integration is essentially the process of summing infinitely small pieces to find a whole.
Here, it helps us in accurately determining the area needed to locate the centroid.
We specifically use definite integration, meaning we calculate the integral from one boundary to another, which in this case are the intersection points.
Remember, integration is essentially the process of summing infinitely small pieces to find a whole.
Here, it helps us in accurately determining the area needed to locate the centroid.
Intersection Points
Intersection points are where two functions meet on a graph. Finding these points is crucial because they define the boundaries for integration.
In this problem, we identify the intersection points of the parabola, \(y = x^2\), and the horizontal line, \(y = c\).
Setting these equations equal, \(x^2 = c\), we solve for \(x\), finding \(x = -\forall c\) and \(x = \forall c\).
These points, \((-\forall c, c)\) and \((\forall c, c)\), form the limits of integration for calculating the area and centroids.
In this problem, we identify the intersection points of the parabola, \(y = x^2\), and the horizontal line, \(y = c\).
Setting these equations equal, \(x^2 = c\), we solve for \(x\), finding \(x = -\forall c\) and \(x = \forall c\).
These points, \((-\forall c, c)\) and \((\forall c, c)\), form the limits of integration for calculating the area and centroids.
Area Calculation
Calculating the area of a region involves integrating the difference between the upper and lower functions within the determined boundaries.
In this exercise, the upper function is \(y = c\) and the lower function is \(y = x^2\).
The area \(A\) is found by integrating \((c - x^2)\) from \(x = -\forall c\) to \(x = \forall c\).
Splitting this into simpler integrals: \(A = \forall (c\forall) - 2\forall \frac{c^{3/2}}{3} = \frac{4c^{3/2}}{3}\).
This gives us the total area, necessary for calculating centroid coordinates.
In this exercise, the upper function is \(y = c\) and the lower function is \(y = x^2\).
The area \(A\) is found by integrating \((c - x^2)\) from \(x = -\forall c\) to \(x = \forall c\).
Splitting this into simpler integrals: \(A = \forall (c\forall) - 2\forall \frac{c^{3/2}}{3} = \frac{4c^{3/2}}{3}\).
This gives us the total area, necessary for calculating centroid coordinates.
Symmetry in Functions
Symmetry simplifies many mathematical problems, particularly those involving centroids.
A function is symmetric if it looks the same on both sides of a central axis. In this problem, the parabola \(y = x^2\) is symmetric about the y-axis, meaning that the area above it is mirrored evenly on both sides.
Because of this symmetry, we know that the x-coordinate of the centroid \(\forall x\) is zero, as the centroid lies directly on the y-axis.
Using symmetry often reduces complexity and aids in direct computational solutions, like determining the centroid in this exercise.
A function is symmetric if it looks the same on both sides of a central axis. In this problem, the parabola \(y = x^2\) is symmetric about the y-axis, meaning that the area above it is mirrored evenly on both sides.
Because of this symmetry, we know that the x-coordinate of the centroid \(\forall x\) is zero, as the centroid lies directly on the y-axis.
Using symmetry often reduces complexity and aids in direct computational solutions, like determining the centroid in this exercise.
