Centroid

**thegame189** · Mar 17th 2008, 01:10 PM

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?

**thegame189** · Mar 17th 2008, 02:45 PM

Anybody??? Bueller???

**CaptainBlack** · Mar 18th 2008, 02:42 AM

Put:

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

Split using partial fractions:

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

Now split up the second term:

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

The middle term is $I$ , so:

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

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 $x$ and $\frac{x}{(1+x^2)^2}.$

RonL

**thegame189** · Mar 18th 2008, 07:08 AM

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.

**wingless** · Mar 18th 2008, 07:14 AM

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.

**CaptainBlack** · Mar 18th 2008, 07:15 AM

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 $\frac{1}{(1+u)^2}$ using partial fractions?

You are going to write it as:

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

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

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

RonL

**CaptainBlack** · Mar 18th 2008, 07:21 AM

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

**thegame189** · Mar 18th 2008, 07:27 AM

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.

**CaptainBlack** · Mar 18th 2008, 10:26 AM

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

**thegame189** · Mar 18th 2008, 10:44 AM

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.

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