Finding the CM of a Bar of Variable Density

[Lopina]

Homework Statement

We are given a bar, with length of $$d$$, and it's densitiy is given by this formula: $$\lambda=\lambda_{0}+2ax$$, where x is the distance from one side of the bar and a is a constant

Homework Equations

$$\lambda=\lambda_{0}+2ax$$
$$\vec{r}_{CM}=\frac{\sum\vec{r}_{i}\Delta m_{i}}{m}$$

The Attempt at a Solution

Well, I figured, if I have infinitesimal parts of the bar, I should integrate it.
So, this is what I've come up with so far:

$$M=\int^{d}_{0}(\lambda_{0}+2ax)dx=\lambda_{0}x+ax^{2}|^{d}_{0}=\lambda_{0}d+ad^{2}$$

$$X_{cm}=\frac{1}{M}\int^{d}_{0}(\lambda_{0}x+2ax^{2})dx=\frac{1}{M}(\frac{\lambda_{0}x^{2}}{2}+\frac{2ax^{3}}{3})|^{d}_{0}=\frac{3\lambda_{0}d+4ad^{2}}{6(\lambda_{0}+ad)}$$

 

[nickjer]

Where is the question? Your formulas look ok.

 

[Lopina]

Well, I was not sure of this because the solution looks very awful, and I had no one to verify this.
But if it seems OK to you, I guess it's correct then.
Thanks

 

[nickjer]

Well the math works out correctly, although you can pull a 'd' out from the numerator to slightly simplify it a bit more. Unless you misread something, then everything looks correct.