# Finding the centroid of the region

1. Feb 1, 2013

### thekey

Hi

VVVVVV

Find the centroid of the region bounded by the curves y = 2x - 4 , y = 2 Sqr x, and x = 1. Make a sketch of the region.

2. Feb 1, 2013

### rollingstein

Make an attempt?

3. Feb 1, 2013

### thekey

I did not try ..because I am not familiar with the way of solving

4. Feb 1, 2013

### haruspex

Did you at least manage to sketch the region? You have three lines. Find out where they meet.

5. Feb 1, 2013

### thekey

yeah I did sketched the three curves then ?..

6. Feb 1, 2013

### haruspex

So where do they intersect? What formula do you know for finding a coordinate of a centroid?

7. Feb 1, 2013

### thekey

they intersect between 1 and ( a unknown point intersection between 2 Sqrt x and x = 1 )

8. Feb 1, 2013

### thekey

also, there is a triangle region under x-axis between 1 and 2 but I think it is not included in the intersection

9. Feb 1, 2013

### thekey

how can I know the second point of the limit integration ?!

10. Feb 1, 2013

### thekey

I got the answer :D >>

thank u : )

11. Feb 1, 2013

### HallsofIvy

Staff Emeritus
I am puzzled by this. Why in the world would you be given a homework problem like this if you had never been given instruction in these and your text book has nothing on it? You are taking Calculus are you not? And every text I have seen has the formulas for "centroid". You also seem to be saying that you cannot solve a simple quadratic equation.

In any case, The line x= 1 intersects $y= 2\sqrt{x}$ at (1, 2) and the line y= 2x- 4 at (1, -2) and forms the left boundary. The line y= 2x- 4 and $y= 2\sqrt{x}$ intersect when $y= 2x- 4= 2\sqrt{x}$. Divide by 2 to get $x- 2= \sqrt{x}$ and square both sides: $(x- 2)^2= x^2- 4x+ 4= x$ or $x^2- 5x+ 4= 0$. That factors easily: (x- 4)(x- 1)= 0. Either x= 1 or x= 4. The point (1, -4) is an "extraneous" root- it is not on $y= 2\sqrt{x}$. So the last vertex, the intersection between $y= 2\sqrt{x}$ and y= 2x- 4, is at (4, 4).

The area of that region is given by the single integral
$$\int_1^4 2\sqrt{x}- (2x- 4)dx$$
which could also be done by the double integral
$$\int_1^4\int_{2x- 4}^{2\sqrt{x}} dydx$$.

I give that double integral (which easily integrates with respect to y to give the first integral) because it is needed for the centroid.

The x coordinate of the centroid is given by the integral of x over that region
$$\int_1^4\int_{2x-4}^{2\sqrt{x}} x dydx= \int_1^4 x(2\sqrt{x}- (2x-4))dx$$
divided by the area and

the y coordinate of the centroid is given by the integral of y over that region
$$\int_1^4\int_{2x-4}^{2\sqrt{x}} y dy dx$$
divided by the area.