# Finding the y-centroid of y=x^3 between the x-axis and x=2

• t.kirschner99
In summary, the problem is to find the y-coordinate of the centroid of the region bounded by y=x3, the x-axis, and the lines x=0 and x=2. This can be solved using the formula yc = integral of (ydA/A), where dA = g(y)dy and yc = h/2 * (n+1)/(2n+1). The incorrect attempt at a solution involved setting up the integral incorrectly, but the correct integral should be yc = integral of (y(2-y^(1/3))/4 dy) evaluated between 0 and 8. The correct answer is 2.28.
t.kirschner99

## Homework Statement

Find the y-centroid of y=x3 between the x-axis, x=2, where area is 4.

## Homework Equations

yc = integral of (ydA/A)
dA = g(y)dy
yc = h/2 * (n+1)/(2n+1)

## The Attempt at a Solution

g(y) = y1/3
A = 4
yc = integral of (y*y1/3/4 dy)
= integral(y4/3 * 1/4 dy)
= 3/7 y7/3 * 1/4
Now since the upper limit is y=8, and lower is y=0, you evaluate this expression between 0 and 8.
= 3/28 * (8)7/3
= 13.71
So that is the answer I get, but another formula I have is the one using h and n above where h = upper limit = 8, and n = order = 3.

My question is, where am I going wrong in the integral?

t.kirschner99 said:

## Homework Statement

Find the y-centroid of y=x3 between the x-axis, x=2, where area is 4.

## Homework Equations

yc = integral of (ydA/A)
dA = g(y)dy
yc = h/2 * (n+1)/(2n+1)

## The Attempt at a Solution

g(y) = y1/3
A = 4
yc = integral of (y*y1/3/4 dy)
Your ##y_c## is not set up correctly. Set up correctly as a double integral it should look like this:$$\frac 1 4 \int_0^8\int_{y^{\frac 1 3}}^2 y~dxdy$$In other words, your ##y## integrand should be ##y(2-y^{\frac 1 3})##.

t.kirschner99 said:
Find the y-centroid of y=x3 between the x-axis, x=2, where area is 4.
This problem statement is somewhat vague. A better problem statement would be the following.
Find the y coordinate of the centroid of the region bounded by the curve y = x3, the x-axis, and the lines x = 0 and x = 2.

## 1. What is the y-centroid of y=x^3 between the x-axis and x=2?

The y-centroid, or the center of mass in the y-direction, can be found by calculating the integral of x^3 from 0 to 2, dividing it by the area under the curve, and then multiplying it by the width of the area (2 units in this case). This results in a y-centroid of 1.6 for the given function and interval.

## 2. How is the y-centroid related to the x-axis and x=2?

The y-centroid is the point on the y-axis where the area under the curve is evenly distributed on both sides. In this case, the x-axis and the line x=2 create the boundaries for the area under the curve, with the y-centroid being the midpoint between them.

## 3. Why is finding the y-centroid important?

Finding the y-centroid is important in many applications, such as engineering and physics, as it helps determine the balance and stability of an object or system. It also plays a crucial role in calculating moments of inertia and other physical properties.

## 4. Can the y-centroid be outside of the given interval?

No, the y-centroid can only be within the given interval of the function. This is because the boundaries of the interval determine the limits for the area under the curve, and the y-centroid represents the point where this area is evenly distributed.

## 5. How does the shape of the curve affect the location of the y-centroid?

The shape of the curve plays a significant role in determining the location of the y-centroid. In general, a wider and flatter curve will have a lower y-centroid, while a narrow and steep curve will have a higher y-centroid. This is because the area under the curve is distributed differently based on its shape.

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