Thread: Centroid

Find the exact coordinates of the centroid of the lamina bounded by the curve y=1/(1+x^2), the x-axis, the y-axis, and x=1.

I got the Area to be (Pi/4)

For X-Bar, I got: (2/Pi)*ln(2)

I got stuck on the Y-Bar.
I got;

(4/Pi)*(1/2)*the integaral (1/(1+x^2)^2) dx.

This comes to (2/Pi)*the integral (1/(1+x^2)^2) dx.

I don't know how to integrate that. So can someone tell me if my area and X-Bar are correct and then help me integrate for Y-Bar?

Anybody??? Bueller???

Put:

$\displaystyle I=\int \frac{1}{(1+x^2)^2}~dx$

Split using partial fractions:

$\displaystyle 2I=\int \frac{1}{1+x^2}~dx+\int \frac{1-x^2}{(1+x^2)^2}~dx$

Now split up the second term:

$\displaystyle 2I=\int \frac{1}{1+x^2}~dx+\int \frac{1}{(1+x^2)^2}~dx-\int \frac{x^2}{(1+x^2)^2}~dx$

The middle term is $\displaystyle I$, so:

$\displaystyle I=\int \frac{1}{1+x^2}~dx-\int \frac{x^2}{(1+x^2)^2}~dx$

The first integral on the right you can do, and the second can be done by integration by parts after spliting into the product of $\displaystyle x$ and $\displaystyle \frac{x}{(1+x^2)^2}.$

RonL

Can you walk me through those steps? I am not getting it. I don't understand where the (1-x^2) came from. When I used partial fractions, I didn't get what you did.

You asked a centroid question before and I gave a detailed explanation of finding centroids here. See it again, try to apply it to this question and then tell me where you're stuck.

Originally Posted by thegame189

Can you walk me through those steps? I am not getting it. I don't understand where the (1-x^2) came from. When I used partial fractions, I didn't get what you did.

Do you know how to expand something like $\displaystyle \frac{1}{(1+u)^2}$ using partial fractions?

You are going to write it as:

$\displaystyle \frac{1}{(1+u)^2}=\frac{A}{1+u}+\frac{B+Cu}{(1+u)^ 2}$

then find the $\displaystyle A,\ B$ and $\displaystyle C$ that satisfy this.

Now just replace $\displaystyle u$ with $\displaystyle x^2$ to get the partial fraction expansion of $\displaystyle \frac{1}{(1+x^2)^2}$

RonL

Originally Posted by thegame189

This comes to (2/Pi)*the integral (1/(1+x^2)^2) dx.

I don't know how to integrate that.

Of course you could do what people do in the real world and look it up ibn a table of integrals, or ask a symbolic integrator what it is.

RonL

I'm stuck on that derivative, not the actual centroid.

And why isn't it (Ax+B)/(1+x^2) + (Cx+D)/((1+x^2)^2) ?

Originally Posted by CaptainBlack

Of course you could do what people do in the real world and look it up ibn a table of integrals, or ask a symbolic integrator what it is.

RonL

I have to show work or else I would.

Originally Posted by thegame189

I'm stuck on that derivative, not the actual centroid.

And why isn't it (Ax+B)/(1+x^2) + (Cx+D)/((1+x^2)^2) ?

Try it and see what you get.

RonL

I got B=1 in yours and D=1 in mine. It's the same thing. I'm just not getting it. I hate days like these. Nothing seems to click.

Mine: Ax^3+Bx^2+Ax+Cx+B+D = 1, A=0 from the Ax^3=0, B=0 from the Bx^2=0, C=0 from the Ax+Cx=0, and D=1 from B+D=1 and B=0

Yours: Ax^2+Bx+A+C = 1, A=0 from Ax^2=0, B=0 from Bx=0, and C=1 from A+C=1 and A=0.