# Math Help - Finding a centroid take 2.

1. ## Finding a centroid take 2.

This is my second problem with centroids.

Find the centroid of the semicircular region bounded by the x axis and the curve $y=\sqrt{1-x^2}$.
$M=\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} \ dy \ dx=\int_{-1}^{1} \sqrt{1-x^2} \ dx$

Let $x=sin \theta \Rightarrow \ \frac{dx}{d \theta}=cos \theta$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-sin^2 \theta} \ cos \theta \ d \theta=\int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^2 \theta \ d \theta.$

$\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1+ cos 2 \theta \ d \theta=\frac{1}{2} \left[ \theta+\frac{1}{2} sin 2 \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2}$

So $M=\frac{\pi}{2}.$

$M_x=\int_{-1}^{1} \int_0^{\sqrt{1-x^2}} y \ dy \ dx=\frac{1}{2} \int_{-1}^{1} 1-x^2 \ dx=\frac{1}{2} \left[x-\frac{x^3}{3} \right]_{-1}^{1}=\frac{1}{2} \left( 0 \right)=0$

$M_y=\int_{-1}^1 \int_0^{\sqrt{1-x^2}} x \ dy \ dx=\int_{0}^1 x \sqrt{1-x^2} \ dx+\int_{-1}^0 x \sqrt{1-x^2} \ dx=2 \int_0^1 x \sqrt{1-x^2} \ dx$

Let $u=1-x^2 \Rightarrow \ \frac{du}{dx}=-2x \Rightarrow \ 2 \int _{1}^0 \sqrt{u} \ \frac{du}{-2}= 2 \int_0^1 u^{\frac{1}{2}} \frac{du}{2}=\left[ \frac{2}{3} u^{\frac{3}{2}} \right]_0^1=\frac{2}{3}$

$\overline{x}=0 \left( \frac{2}{\pi} \right)=0$

$\overline{y}=\frac{2}{3} \left( \frac{2}{\pi} \right)=\frac{4}{3 \pi}$

Apparently $\overline{x}=0$ but I can't see anything wrong with my working!

If anyone can see a problem, I would be very grateful if you could point it out.

2. Originally Posted by Showcase_22
This is my second problem with centroids.

$M=\int_0^1 \int_0^{\sqrt{1-x^2}} \ dy \ dx=\int_0^1 \sqrt{1-x^2} \ dx$

Let $x=sin \theta \Rightarrow \ \frac{dx}{d \theta}=cos \theta$

$\int_0^{\frac{\pi}{2}} \sqrt{1-sin^2 \theta} \ cos \theta \ d \theta=\int_0^{\frac{\pi}{2}} cos^2 \theta \ d \theta.$

$\frac{1}{2} \int_0^{\frac{\pi}{2}} 1+ cos 2 \theta \ d \theta=\frac{1}{2} \left[ \theta+\frac{1}{2} sin 2 \theta \right]_0^{\frac{\pi}{2}}=\frac{\pi}{4}$

So $M=\frac{\pi}{4}.$

$M_x=\int_0^1 \int_0^{\sqrt{1-x^2}} y \ dy \ dx=\frac{1}{2} \int_0^1 1-x^2 \ dx=\frac{1}{2} \left[x-\frac{x^3}{3} \right]_0^1=\frac{1}{2} \left( \frac{2}{3} \right)=\frac{1}{3}$

$M_y=\int_0^1 \int_0^{\sqrt{1-x^2}} x \ dy \ dx=\int_0^1 x \sqrt{1-x^2} \ dx$

Let $u=1-x^2 \Rightarrow \ \frac{du}{dx}=-2x \Rightarrow \int_1^0 \sqrt{u} \ \frac{du}{-2}=\frac{1}{2} \int_0^1 u^{\frac{1}{2}} \ du=\frac{1}{2} \left( \frac{2}{3} \right) \left[u^{\frac{3}{2}} \right]_0^1=\frac{1}{3}$

$\overline{x}=\frac{1}{3} \left( \frac{4}{\pi} \right)=\frac{4}{3 \pi}$

$\overline{y}=\frac{1}{3} \left( \frac{4}{\pi} \right)=\frac{4}{3 \pi}$

Apparently $\overline{x}=0$ but I can't see anything wrong with my working!

If anyone can see a problem, I would be very grateful if you could point it out.

It's a semi-circle. Your theta limits should be 0 to $\pi$, surely. Perhaps it would be easier to change it to a polar double integral straight from the beginning.

And the integrand in the formula for a centroid is y, not 1.

$\displaystyle A\bar{y}= \int \int_A y \, dA = \int \int_A y \,dx \,dy$

$y = r \sin(\theta)$

Hence: $A\bar{y} = \int_0^{\pi} \int_0^1 r\sin(\theta) r \, dr \, d\theta$

You should find that this equates to $\frac{2}{3}$

For the x coordinate it's the same story only we use $x = r\cos(\theta)$, and your integral should turn out to be zero.

As a general rule of thumb, if a body is symmetrical around a particular axis, then the coordinate of the centroid in the perpendicular axis will be in the centre of the body. i.e., your semicircle is centered at the origin, and is symmetric around the y axis. Hence you can jump straight to the conclusion that the x coordinate of the centroid will be in the centre of the body with respect to the x axis... hence at zero.

3. (oops, i'm confusing threads, if you saw anything in this post before I edited it please ignore it!)

That's just to find the mass of the shape so I integrated 1. In one of the next integrals the integrand is y to find the x co-ordinate of the centre of mass.

I think you're right about the theta limits. Let me work it out and post back later.

4. Originally Posted by Showcase_22
(oops, i'm confusing threads, if you saw anything in this post before I edited it please ignore it!)

That's just to find the mass of the shape so I integrated 1. In one of the next integrals the integrand is y to find the x co-ordinate of the centre of mass.

I think you're right about the theta limits. Let me work it out and post back later.
I have edited my post to give a bit more detail.

5. Right, thanks for that. Find the initial thread edited to make it correct!

You rock!

6. Originally Posted by Showcase_22
Right, thanks for that. Find the initial thread edited to make it correct!

You rock!
Isn't your y coordinate also wrong?

7. I checked in the back of the book and apparently that's right.

I've drawn a (correct) picture now and everything is starting to make sense.

(thanks for replying to my other thread btw, you cleared up that problem for me!)

It also took more than I thought to edit the original thread, the limits had a knock-on effect throughout the calculation and I had to take into account the symmetry of the graph!