# Thread: Find centroid of the following double integral

1. ## Find centroid of the following double integral

we have SS_D f(x,y) dxdy=

I have to first change the order of Integration and then find the centroid (x0,y0) of the region D (the center is the center of mass considering constant mass density).

Maybe its an easy question but i'm just starting to study the subject so I didn't get very far in my attempt to solve this.

2. ## Re: Find centroid of the following double integral

Originally Posted by damule
we have SS_D f(x,y) dxdy=

I have to first change the order of Integration and then find the centroid (x0,y0) of the region D (the center is the center of mass considering constant mass density).
The mass of the region

$D \equiv\begin{Bmatrix} 0\leq y \leq 3\\\dfrac{y}{\sqrt{3}}\leq x\leq \sqrt{12-y^2}\end{matrix}$

with density $f(x,y)$ is

$M=\iint_{D}f(x,y)\;dxdy=\int_{0}^3\left(\int_{y/\sqrt{e}}^{\sqrt{12-y^2}}f(x,y)\;dx\right)dy$

and the centroid is

$(x_0,y_0)=\dfrac{1}{M}\left(\iint_{D}xf(x,y)\;dxdy ,\iint_{D}yf(x,y)\;dxdy\right)$

Why do you want to change the order of integration? Have you had any computation difficulties?

3. ## Re: Find centroid of the following double integral

I have the following solution which I didn't understand completely (the bold part):

The region is the sector of the circle of radius √(12) centered at the origin between the lines y = 0 and x = y/√(3). You can change the order of integration, but it's a really bad idea from a practical stand point. The order dydx requires two integrals because the upper curve changes. You'd have the limits

0 ≤ x ≤ √(3) and 0 ≤ y ≤ √(3)x for the first integral and

√(3) ≤ x ≤ √(12) and 0 ≤ y ≤ √(12 - x²) for the second integral.

The mass of the region is the area of a sector of a circle of radius √(12) and inscribed angle π/3. This is

m = ½(√(12))² (π/3) = 2π.

The circle and the line y = 3 meet at the point (√(3), 3) and so the inscribed angle Θ satisfies

tanΘ = 3/√(3) = √(3) ==> Θ = π/3.

The moments about the x and y axes are

3 √(12-y²)
∫ ∫ x dx dy = 12 = M_y
0 y/√3

3 √(12-y²)
∫ ∫ y dx dy = 4√(3) = M_x.
0 y/√3

The centroid is (M_y/m, M_x/m) = (6/π, 2√(3)/π).

How did they find that m? On what do you rely? I hope someone could explain to me in further detail

Is there like 2 ways to calculate the mass? Why didn't he do the double integral to find M?

4. ## Re: Find centroid of the following double integral

Originally Posted by damule
How did they find that m? On what do you rely? I hope someone could explain to me in further detail
$A=\iint_D\;dxdy$ is the area of the region $D$ . If $f(x,y)=k$ (constant) the mass of $D$ is $m=k\iint_D\;dxdy=kA$ ( $A$ area of $D$ , which is a sector). You can choose $k=1$ for finding the centroid ( $k$ appears multiplying in the numerator and denominator and can be canceled).