# How do you find the centroid of this?

• Manwe Sulimo
In summary, the student is trying to solve an equation for the centroid of a shape formed by x- and y-coordinates.

## Homework Statement

Find the centroid of the shape formed by the equation y2=x3-x4, the x-axis, and the y-axis.

## Homework Equations

A=∫f(x)dx
Mx=∫(1/2)[f(x)]2dx
My=∫x[f(x)]dx

## The Attempt at a Solution

I'm stuck on the integral.
I attempted u-substitution and got du=(1/2)(x3-x4)-1/2dx; "parts," trigonometric substitution/identities, and partial fractions don't seem to apply.

Manwe Sulimo said:

## Homework Statement

Find the centroid of the shape formed by the equation y2=x3-x4, the x-axis, and the y-axis.

## Homework Equations

A=∫f(x)dx
Mx=∫(1/2)[f(x)]2dx
My=∫x[f(x)]dx

## The Attempt at a Solution

I'm stuck on the integral.
I attempted u-substitution and got du=(1/2)(x3-x4)-1/2dx; "parts," trigonometric substitution/identities, and partial fractions don't seem to apply.
Try polar coordinates.

SammyS said:
Try polar coordinates.
I haven't worked in those for a while (I'm not sure I remember how to use them); but I'm fairly certain my teacher wants me to stick with x-y coordinates.

Manwe Sulimo said:
I haven't worked in those for a while (I'm not sure I remember how to use them); but I'm fairly certain my teacher wants me to stick with x-y coordinates.
You've already tried trig substitutions. Think of polar coordinates as a form of trig substitution.

Besides, what will your teacher prefer? That you kept working with cartesian coordinates and didn't solve the problem, or you converted to polar coordinates and got an answer?

Manwe Sulimo said:

## Homework Statement

Find the centroid of the shape formed by the equation y2=x3-x4, the x-axis, and the y-axis.

## Homework Equations

A=∫f(x)dx
Mx=∫(1/2)[f(x)]2dx
My=∫x[f(x)]dx
You swapped ##M_x## and ##M_y##.

## The Attempt at a Solution

I'm stuck on the integral.
I attempted u-substitution and got du=(1/2)(x3-x4)-1/2dx; "parts," trigonometric substitution/identities, and partial fractions don't seem to apply.
Try starting like this:
$$\int x\sqrt{x^3-x^4}\,dx = \int x^2\sqrt{x-x^2}\,dx = \int x^2\sqrt{\frac 14-\left(x-\frac 12\right)^2}\,dx.$$ The last step comes from completing the square. Then try a few more substitutions and see if you get anywhere.

SammyS

## 1. What is a centroid?

A centroid is the geometric center of a two-dimensional shape or the center of mass of a three-dimensional object. It is often used to represent the average location of all points in a shape.

## 2. How do you calculate the centroid of a shape?

The centroid of a shape can be calculated by finding the mean of all the x-coordinates and the mean of all the y-coordinates of the points that make up the shape. This can be done using a formula or by graphically finding the intersection point of the medians of the shape.

## 3. What is the importance of finding the centroid?

Finding the centroid of a shape is important in many fields, including mathematics, physics, and engineering. It can help determine the center of mass and balance point of an object, as well as aid in the calculation of important values such as moment of inertia and center of pressure.

## 4. How does the number of points in a shape affect the centroid?

The number of points in a shape does not affect the calculation of the centroid. As long as the x and y coordinates of all the points are known, the centroid can be calculated regardless of the number of points in the shape.

## 5. Can the centroid be outside of the shape?

Yes, the centroid can be outside of the shape. This can happen if the shape is irregular or has a concave section. In such cases, the centroid will still be the point of intersection of the medians, but it may fall outside of the shape.