Finding the centroid of the region

[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.

 

[rollingstein]

Make an attempt?

 

[thekey]

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

 

[haruspex]

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

 

[thekey]

yeah I did sketched the three curves then ?..

 

[haruspex]

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

 

[thekey]

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

 

[thekey]

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

 

[thekey]

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

 

[thekey]

I got the answer :D >>

thank u : )

 

[HallsofIvy]

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.

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 textbook 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.