thekey said:
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 [itex]y= 2\sqrt{x}[/itex] at (1, 2) and the line y= 2x- 4 at (1, -2) and forms the left boundary. The line y= 2x- 4 and [itex]y= 2\sqrt{x}[/itex] intersect when [itex]y= 2x- 4= 2\sqrt{x}[/itex]. Divide by 2 to get [itex]x- 2= \sqrt{x}[/itex] and square both sides: [itex](x- 2)^2= x^2- 4x+ 4= x[/itex] or [itex]x^2- 5x+ 4= 0[/itex]. 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 [itex]y= 2\sqrt{x}[/itex]. So the last vertex, the intersection between [itex]y= 2\sqrt{x}[/itex] and y= 2x- 4, is at (4, 4).
The area of that region is given by the single integral
[tex]\int_1^4 2\sqrt{x}- (2x- 4)dx[/tex]
which could also be done by the double integral
[tex]\int_1^4\int_{2x- 4}^{2\sqrt{x}} dydx[/tex].
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
[tex]\int_1^4\int_{2x-4}^{2\sqrt{x}} x dydx= \int_1^4 x(2\sqrt{x}- (2x-4))dx[/tex]
divided by the area and
the y coordinate of the centroid is given by the integral of y over that region
[tex]\int_1^4\int_{2x-4}^{2\sqrt{x}} y dy dx[/tex]
divided by the area.