Centroid of a uniform shape, using area

[J-dizzal]

Homework Statement

Find the coordinates of the centroid of the uniform area.

Homework Equations

equations for centroid coordinates at the top of my paper.

The Attempt at a Solution

 

[Dr. Courtney]

How does k go from the denominator to the numerator of your integrals as you are evaluating them?

 

[J-dizzal]

How does k go from the denominator to the numerator of your integrals as you are evaluating them?

(2/3k)(58.1)
=(.667k)(58.1)
=38.7k
where k is a constant. And i treated the other integral similarly.

 

[Nathanael]

In one of your steps you basically said $\sqrt{\frac{y}{k}} =\frac{1}{k}\sqrt{y}$

Also Dr. Courtney's point still stands... it's not 2/3k it's 2/(3k)

 

[J-dizzal]

In one of your steps you basically said $\sqrt{\frac{y}{k}} =\frac{1}{k}\sqrt{y}$

Also Dr. Courtney's point still stands... it's not 2/3k it's 2/(3k)

Ok thanks Dr. Courtney and Nathanael. I now have 38/k. In my final answer the k's cancel out and I am left with the same 7.5/38.7.

 

[Nathanael]

38.7/k is still not right. It should be 38.7/√k

When you find the x-coordinate of the centroid, you should be integrating with respect to x.
The "dA" in the formula $\bar x=\frac{1}{A}\int xdA$ is the area of the thin strip between x and x+dx.

 

[J-dizzal]

38.7/k is still not right. It should be 38.7/√k

When you find the x-coordinate of the centroid, you should be integrating with respect to x.
The "dA" in the formula $\bar x=\frac{1}{A}\int xdA$ is the area of the thin strip between x and x+dx.

Wouldnt it be easier to integrate with respect to y, because of the shaded region is above the curve.

 

[Nathanael]

It's not about what is easier, it's simply wrong.

When you integrate with respect to y, you are taking horizontal strips of area, right? Well when you find the x-coordinate of the centroid you want to take vertical strips of area. The reason for this is that you want to take strips of area which all have the same x-value, and then multiply them by that x-value. You just can't do this when you integrate w.r.t. y.

 

[J-dizzal]

It's not about what is easier, it's simply wrong.

When you integrate with respect to y, you are taking horizontal strips of area, right? Well when you find the x-coordinate of the centroid you want to take vertical strips of area. The reason for this is that you want to take strips of area which all have the same x-value, and then multiply them by that x-value. You just can't do this when you integrate w.r.t. y.

$\bar x=\frac{1}{A}\int xdA$
= $1/2 \int kx^3dx$ does this look ok?

 

[Nathanael]

$\bar x=\frac{1}{A}\int xdA$
= $1/2 \int kx^3dx$ does this look ok?

kx²dx is the area under the curve. Try to figure out a way to find the area above the curve.

Also where did the 1/A go? And where did this 1/2 come from? (Are you saying the area is 2?)