Finding Centroid of Region

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In summary: The details of the integration are left to the reader.In summary, the task is to find the centroid of the region bounded by the curve x=5-y2 and x=0, which is located at the point (2,0). The equations and steps for finding the centroid are provided, including the use of thin vertical strips and the lever arm for each "mass" element. The limits of integration for x are 0 and 5.
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Saterial
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Homework Statement


Find the centroid of the region bounded by the curve x=5-y2 and x=0.

Answer is (2,0).

Homework Equations



A = ∫[f(x)-g(x)]dx a->b
xbar = 1/A ∫ x[f(x)-g(x)]dx from a->b
ybar = 1/A ∫ 1/2[f(x)2-g(x)2]dx

The Attempt at a Solution



I know that the graph is a sideway parabola with vertex at (5,0) and bounded at x=0, this means that the graph is symmetrical above and below the x-axis so y value of centroid is 0.

For the x value of the centroid. I don't know if I should be integrating from 0->5? or integrating from the y intercepts -√5 -> √5.

In addition, I was always told it's the upper graph - lower graph. Because this is kind of a sideway graph, would I subtract parabola from line (5-y^2 - 0) or line from parabola (0-5-y^2)?

If anyone could help me understand setting the bounds, that would be great!
 
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  • #2
Saterial said:

Homework Statement


Find the centroid of the region bounded by the curve x=5-y2 and x=0.

Answer is (2,0).

Homework Equations



A = ∫[f(x)-g(x)]dx a->b
xbar = 1/A ∫ x[f(x)-g(x)]dx from a->b
ybar = 1/A ∫ 1/2[f(x)2-g(x)2]dx

The Attempt at a Solution



I know that the graph is a sideway parabola with vertex at (5,0) and bounded at x=0, this means that the graph is symmetrical above and below the x-axis so y value of centroid is 0.

For the x value of the centroid. I don't know if I should be integrating from 0->5? or integrating from the y intercepts -√5 -> √5.
You have to use thin vertical strips whose width is Δx, which means you'll be integrating with respect to x. The limits of integration will necessarily be 0 and 5.
Saterial said:
In addition, I was always told it's the upper graph - lower graph. Because this is kind of a sideway graph, would I subtract parabola from line (5-y^2 - 0) or line from parabola (0-5-y^2)?

If anyone could help me understand setting the bounds, that would be great!
The typical "mass" element is a thin vertical strip whose height is the y-value on the upper part of the parabola - the y-value on the lower part. The width is Δx. As a shortcut, the integrand could be two time the upper half y-value - 0. You also need the lever arm for each "mass" element, which in this case will be x.
 

1. What is a centroid?

A centroid is a point that represents the center of mass or balance of a region. It is the geometric center of a figure, where all the mass is evenly distributed.

2. Why is finding the centroid important?

Finding the centroid allows us to calculate the average position of the mass in a region. This information is useful in various fields such as physics, engineering, and architecture. It also helps us determine the stability and balance of an object or structure.

3. How is the centroid of a region calculated?

The centroid of a region can be calculated by finding the weighted average of the coordinates of all the points in the region. The coordinates are weighted by the mass or density of the points. Alternatively, the centroid can also be calculated using integration techniques in calculus.

4. What is the difference between the centroid and center of gravity?

The centroid is a purely geometric concept, while the center of gravity takes into account the effects of gravity on an object. In simpler terms, the centroid is the center of mass of a figure, while the center of gravity is the point at which the weight of an object can be considered to act.

5. Can the centroid of a region be outside the region?

Yes, the centroid of a region can be outside the region. This can happen when the region has an irregular shape or when the distribution of mass is uneven. However, in most cases, the centroid lies within the region.

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