How Do You Calculate the Centroid of a Sideways Parabola?

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SUMMARY

The centroid of the region bounded by the curve x=5-y² and the line x=0 is calculated to be (2,0). To find the area A, the integral A = ∫[f(x)-g(x)]dx from 0 to 5 is used, where f(x) is the parabola and g(x) is the line. The x-coordinate of the centroid is determined using x̄ = 1/A ∫ x[f(x)-g(x)]dx from 0 to 5, while the y-coordinate is ȳ = 1/A ∫ 1/2[f(x)²-g(x)²]dx, which simplifies to 0 due to symmetry about the x-axis.

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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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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.
 

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